February 2011 - Career Point

Volume - 6 Issue - 8 

February, 2011 (Monthly Magazine) 

Editorial / Mailing Office : 

112-B, Shakti Nagar, Kota (Raj.) 

Tel. : 0744-2500492, 2500692, 3040000 

e-mail : xtraedge@gmail.com 

Editor : 

Pramod Maheshwari 

[B.Tech. IIT-Delhi] 

Cover Design 

Om Gocher, Govind Saini 

Layout 

Rajaram Gocher 

Circulation & Advertisement 

Praveen Chandna 

Ph 0744-3040000, 9672977502 

Subscription 

Sudha Jaisingh Ph. 0744-2500492, 2500692 

© Strictly reserved with the publishers 

• No Portion of the magazine can be 

published/ reproduced without the 

written permission of the publisher 

• All disputes are subject to the 

exclusive jurisdiction of the Kota 

Courts only. 

Every effort has been made to avoid errors or 

omission in this publication. In spite of this, 

errors are possible. Any mistake, error or 

discrepancy noted may be brought to our 

notice which shall be taken care of in the 

forthcoming edition, hence any suggestion is 

welcome. It is notified that neither the 

publisher nor the author or seller will be 

responsible for any damage or loss of action to 

any one, of any kind, in any manner, there from. 

Unit Price ` 20/- 

Special Subscription Rates 

6 issues : ` 100 /- [One issue free ] 

12 issues : ` 200 /- [Two issues free] 

24 issues : ` 400 /- [Four issues free] 

Owned & Published by Pramod 

Maheshwari, 112, Shakti Nagar, 

Dadabari, Kota & Printed by Naval 

Maheshwari, Published & Printed at 112, 

Shakti Nagar, Dadabari, Kota. 

Editor : Pramod Maheshwari 

Dear Students, 

It's the question you dreamed about when you were ten years old. It's the 

question our parents nagged you about during high school. It's the question that 

stresses most of us out more and more the older we get. "What do you want to 

be when you grow up?" 

There are people who are studying political science but hate politics, nursing 

majors who hate biology, and accounting majors who hate math. Obviously, a 

lot of people are confused about what exactly it is that they want to spend their 

life doing. Think about it. if you work for 10 hours each day, you're going to 

end up spending over 50% of your awake life at work. Personally, I think it's 

important that we spend that 50% of your awake life at work. Personally, I think 

it's important that we spend that 50% wisely. But how can you make sure that 

you do? Here are some cool tips for how to decide that you really want to be 

when you grow up. 

• Relax and Keep an Open Mind: Contrary to popular belief, you don't have 

to "choose a career" and stick with it for the rest of your life. You never 

have to sign a contract that says, "I agree to force myself to do this for the 

rest of my life" You're free to do whatever you want and the possibilities are 

endless. So relax, dream big, and keep an open mind. 

• Notice Your Passions: Every one of us is born with an innate desire to do 

something purposeful with our lives. We long to do something that we're 

passionate about; something that will make a meaningful impact on the 

world. 

• Figure Out How to Use Your Passions for a Larger Purpose: You notice that 

this is one of your passions, so you decide to become a personal trainer. 

Making a positive impact on the world will not only ensure that you are 

successful financially, it will also make you feel wonderful. It's proven 

principle: The more you give to the world, the more the world will give you 

in return. 

• Figure Our How You Can Benefit: Once you've figured out what your 

passions are and how you can use those passions to add value to the world 

& to yourself, it's time to take the last step: figure out how you can make 

great success doing it. my most important piece of advice about this last 

step is to remember just that: It's the last part of the decision process. I feel 

sorry for people who choose an occupation based on the average income for 

that field. No amount of money can compensate for a life wasted at a job 

that makes you miserable. However, that's not to say that the money isn't 

important. Money is important, and I'm a firm believer in the concept that 

no matter what it is that you love doing, there's at least one way to make 

extraordinary money doing it. So be creative! 

No matter how successful you become, how great your life is, or how beautiful 

you happen to be... there will still be times when you simply feel like you're an 

ugly mess. But when those times come, remember that all you need to get 

yourself back on track is a positive outlook, a dash of self confidence, and the 

willingness to make yourself feel better as soon as you know how. 

Simply discover your passions, figure out how to use your passions to make an 

impact on the world & to yourself. 

Presenting forever positive ideas to your success. 

Yours truly 

Pramod Maheshwari, 

B.Tech., IIT Delhi 

If you can't make a mistake, 

you can't make anything. 

Editorial 

XtraEdge for IIT-JEE 1 FEBRUARY 2011


Volume-6 Issue-8 

February, 2011 (Monthly Magazine) 

NEXT MONTHS ATTRACTIONS 

Much more IIT-JEE News. 

INDEX 

CONTENTS 

Regulars .......... 

PAGE 

S 

Know IIT-JEE With 15 Best Questions of IIT-JEE 

Challenging Problems in Physics,, Chemistry & Maths 

Key Concepts & Problem Solving strategy for IIT-JEE. 

IIT-JEE Mock Test Paper with Solution 

AIEEE & BIT-SAT Mock Test Paper with Solution 

Success Tips for the Months 

• "The way to succeed is to double your error 

rate." 

• "Success is the ability to go from failure to 

failure without losing your enthusiasm." 

• "Success is the maximum utilization of the 

ability that you have." 

• We are all motivated by a keen desire for 

praise, and the better a man is, the more he 

is inspired to glory. 

• Along with success comes a reputation for 

wisdom. 

• They can, because they think they can. 

• Nothing can stop the man with the right 

mental attitude from achieving his goal; 

nothing on earth can help the man with the 

wrong mental attitude. 

• Keep steadily before you the fact that all 

true success depends at last upon yourself. 

NEWS ARTICLE 4 

Dr. Abdul Kalam's Message to Every Indian 

Two Mumbai CAT toppers are from IIT-Bombay 

IITian ON THE PATH OF SUCCESS 6 

Mr. Vineet Buch 

KNOW IIT-JEE 7 

Previous IIT-JEE Question 

Study Time........ 

DYNAMIC PHYSICS 15 

8-Challenging Problems [Set# 9] 

Students’ Forum 

Physics Fundamentals 

Matter Waves, Photo-electric Effect 

Thermal Expansion, Thermodynamics 

CATALYSE CHEMISTRY 31 

Key Concept 

Carbonyl Compounds 

Co-ordination Compound & 

Metallurgy 

Understanding : Physical Chemistry 

DICEY MATHS 38 

Mathematical Challenges 

Students’ Forum 

Key Concept 

Integration 

Trigonometrical Equation 

Test Time .......... 

XTRAEDGE TEST SERIES 50 

Class XII – IIT-JEE 2011 Paper 

Class XI – IIT-JEE 2012 Paper 

Mock Test-3 (CBSE Board Pattern) [Class # XII] 72 

Solution of Mock Test-2 (CBSE Pattern) 

Solution of Mock Test-3 (CBSE Pattern) 


Dr. Abdul Kalam’s Message to 

Every Indian 

What does a system consist of? Very 

conveniently for us it consists of our 

neighbours, other households, other 

cities, other communities and the 

government. But definitely not me and 

YOU. When it comes to us actually 

making a positive contribution to the 

system we lock ourselves along with 

our families into a safe cocoon and 

look into the distance at countries far 

away and wait for a Mr.Clean to come 

along & work miracles for us with a 

majestic sweep of his hand or we 

leave the country and run away. 

‘ASK WHAT WE CAN DO FOR 

INDIA AND DO WHAT HAS TO BE 

DONE TO MAKE INDIA WHAT 

AMERICA AND OTHER WESTERN 

COUNTRIES ARE TODAY’ 

Two Mumbai CAT toppers are 

from IIT-Bombay 

Three people, who appeared for CAT 

from the city, scored 100 percentile. 

Two of them are from the computer 

science department of the IIT-B. The 

other is a faculty member of a citybased 

coaching institute. 

Shashank Samant, 22, who scored 

99.98 percentile on his last attempt in 

2008, gave up the IIM seat to take up 

a job. “After a year and a half job in 

an investment firm, I was finally 

prepared to get in to an IIM. Though 

the 100 percentile was unexpected.” 

“After clearing the CAT, I spoke to 

my peers and seniors at IIT-B and 

decided that work experience would 

be important before getting in to the 

IIMs,” said Samant. He also added 

that IITs help in developing the 

aptitude to clear any competitive 

exams. Samant had graduated in 

computer science from IIT-B in 2009. 

About his choice of IIMs, Samant 

said, “I am ready to get in to any of 

the IIMs though I prefer IIM- 

Ahmedabad and Bangalore over 

others.” 

Gaurav Malpani, a fourth-year student 

of computer science at IIT-B has 

appeared for the entrance exam from 

Mumbai, he is originally from Kolkata. 

He managed to score100 percentile, 

without any coaching. 

“I have never focused on developing 

my technical knowledge. I was only 

polishing my problem solving skills. I 

also focused on my vocabulary,” said 

21-year-old Malpani. He insisted that 

he had never studied exclusively for 

CAT during the year. “I always knew 

that I had the aptitude to score well, but 

scoring 100 percentile was not 

expected,” he added. 

“I would love to join IIM-Ahmedabad 

or Bangalore. Since I am from Kolkata, 

I will also consider seeking admission 

there. I am interested in pursuing an 

MBA in either finance or human 

resources,” he added. The faculty 

member of a city-based coaching 

institute Jose D’Abreu also got a 

perfect score. 

IIT Techfest 2011: The Robots 

Raged 

It was the perfect way to end a very 

well-planned event. On the first two 

days of Techfest, IIT Bombay was 

buzzing with exhibits of some cool 

robots and a few other inventions as 

well. On the final day, the robots 

became restless and just wanted to have 

a go at each other. What followed was a 

lengthy battle fought hard and long. 

Mars Manoeuvre This tournament had 

two robots moving around a grid 

collecting blocks. The team that 

collected the most blocks won. The last 

battle was between C2R and Black 

Beast, from Thailand and Australia 

respectively. 

Black Beast: In all its glory 

Black Beast was the creation of second 

year students from the Department of 

Electrical & Computer Science at 

Swinburne University of Technology 

in Melbourne. Plagued with exams 

and other academic diseases, they 

still managed to build this in about 

two weeks. 

Kids at Swinburne Uni, came 

second with a smile 

Their robots communicated through 

radio frequency waves and 

everything, right from the wheels to 

the circuit boards was custom made. 

One robot would go and measure the 

dimensions of the grid. The second, 

after getting the information would 

start its mission of picking up blocks. 

Can it say, "Thai, Robot"? 

C2R was made by third year students 

of the computer engineering 

department of the Kasetsart 

University in Kapmphaeng Saen, 

Thailand. These kids spent two 

months and about Rs. 1,47,000 

(100,000 Thai Bath) to make these 

robots. 

Kasetsart University students with 

the C2R 


Unlike the kids of Oz, they chose to 

use only one robot which would find 

its way and collect blocks. These 

robots also used radio frequency 

technology to communicate and had 

sensors, so that the robot never drifted 

away from the grid lines. The battle 

ensued and it was clear that spending 

more time with your robot makes 

them strong and obedient. In fact, they 

can also win you competitions! C2R 

won the Mars Manoeuvre competition 

and prize money of Rs. 1,50,000. The 

kids demonstrated how they won the 

battle. Check out the video below. 

IIT-Bombay gets $3 million gift 

More than 40 years ago a quiet 

student named Victor Menezes 

graduated from the Indian Institute of 

Technology Bombay (IIT-B). He 

went on to become, among other 

things, the senior vice-chairperson of 

Citigroup Inc. His “small way to say 

thank you” to the institute has 

translated into a $3 million 

(about Rs 13.5 crore) towards a stateof-the-art 

convention centre on the 

institute’s Powai campus 

“I received priceless education from 

IIT Bombay and this is a small way 

to say thank you”, said Menezes. “I 

hope the centre will help support the 

exchange of ideas at IIT Bombay.” 

IIT-JEE candidates to get 

performance cards now 

Students appearing for the next Joint 

Entrance Examination (JEE) for 

admission to IITs will get 

performance cards specifying marks 

and the ranks secured by them in the 

test. However, as per the new 

provision, they cannot seek regrading 

or re-totalling. 

For the first time, the JEE Board 

would issue performance cards which 

can be considered as certificates by 

many other institutions wanting to 

give admission to JEE candidates. The 

board will also put out the answers of 

the questions on its website to help 

students make assessment of their 

performance. 

IIT Guwahati Director Prof Gautam 

Baruah said the board had urged for 

issuing such performance cards which 

would serve as certificates for the 

students. “Many other institutes, which 

want to take JEE candidates, can give 

admission to students on the basis of 

these performance cards,” Baruah said. 

Indian institute of Science will 

start management course 

Indian Institute of Science (IISc.) is 

planning to start a two-year Master 

programme in management from this 

academic session. The new courses will 

be very advance as it will concentrate 

more on technology management and 

business analytics. 

IISc registrar R Mohan Das said the 

course would concentrate on synergies 

between managing science and 

technology. Das said, “India, in recent 

times, has emerged as one of the global 

hubs of technology and research and 

development (R&D) units. Such 

technology-based and R&D-intensive 

industries need executives with 

exposure and training in technology 

management and business analytics. 

The program has been specially 

designed to train students in technology 

management and business analytics.” 

The course will be conducted by the 

department of management studies, 

which was established in the year 1848, 

and is one of the oldest schools in the 

country. An official at the dept. said 

that application forms for the course 

will be available from the month of 

February. Candidates who have passed 

the Joint Entrance Test (JMET) with 

first class BE/B.Tech degree/equivalent 

is eligible for the course. The 

department will conduct group 

discussion and personal interview 

before selecting students for the course. 

IIT Mandi to formulate plan for 

solving technical problems 

pertaining to agriculture in 

Himachal Pradesh 

Shimla: Shri Ram Subhag Singh, 

Secretary, Agriculture and Information 

and Public Relations said that H.P. 

Agriculture Department and IIT Mandi 

would formulate a long term scheme 

for solving technical problems 

pertaining to agriculture and a Joint 

Working Group at State level would 

be formed for solving the problems 

relating to farm technology. 

Star Donor of the Month - Mr. 

Rajesh Achanta [1987/BT/ME] 

I have been donating off and on as a 

way of keeping the connection with 

IITM going & also to express 

gratitude for the many ways in which 

the institute shaped me in my 

formative years. I'll be transiting 

through Chennai in early January - I 

would like to stop by at IITM & 

relive old memories for a little while! 

Orissa CM confers award to 

IIT-Kanpur Prof. Dr Devi 

Prasad Mishra 

In recognition of his research work, 

Dr. Mishra received Sir Rajendranath 

Mookerjee Memorial and Aerospace 

Engineering Division Prize from The 

Institution of Engineers (India), 

Kolkata, India. Dr. Mishra has more 

than 15 years of teaching and 

research experience. He has served 

as Visiting Professor in 2002 at the 

Tokyo-Denki University, Japan. 

Presently, he is working as an 

Associate Professor in the 

Department of Aerospace 

Engineering at Indian Institute of 

Technology (IIT) Kanpur, Kanpur, 

India where he was instrumental in 

establishing a combustion laboratory. 

His areas of research interest include 

combustion, computational fluid 

dynamics, atomization, nanomaterial 

synthesis etc. He is an Associate 

Editor, Journal of Natural Gas 

Science and Engineering, Elsevier, 

USA and Assistant Editor, 

International Journal of Hydrogen 

Energy, Elsevier, USA. Currently he 

is serving as Editor, Asia Pacific 

Conference on Combustion, 2010. 

Dr. Mishra has four Indian patents 

and more than 154 publications in 

referred Journals and in conference 

proceedings to his credit. 


Success Story 

This article contains storie/interviews of persons who succeed after graduation from different IITs 

Mr. Vineet Buch 

B-Tech from IIT-Kanpur 

(A venture capitalist based in san francisco) 

Vineet Buch still remembers 10 June 1987. Bhopal. The 

Indian Institute of Technology All India Joint Entrance 

Exam (IIT-JEE) results were announced. Buch, then a 15- 

year-old dabbling with career choices, scanned through the 

rank-holders list. Then he scanned it again. Soon he made 

up his mind. He would try and finish No. 1 in the entrance 

exam. “It seemed like a cool thing to do.” 

Every year thousands of Indian students aspire to get into 

an IIT. Close to 400,000 candidates lined up this year. One 

in 65 made the cut. Twenty years ago, the number of 

applicants wasn’t as staggering but there were fewer seats. 

Golfers will tell you that the odds of an amateur pulling off 

a hole-in-one are 1 in 12,750. Still, that’s a doddle 

compared to what Buch was up against. 

“Hardly anyone in Bhopal even wrote the JEE, let alone 

got in,” says Buch, 37, a venture capitalist based in San 

Francisco. “I found it tough to get the right books, like a 

Russian physics book by IE Irodov. My parents [who were 

IAS officers] requested the Indian embassy in Moscow to 

photocopy the book and send it across.” 

In June 1989, Buch was declared No. 1 in the IIT JEE 

exam, arguably the most challenging and competitive 

exam in the world. Only around 50 Indians have 

experienced the feeling—the numbness, the ecstasy, the 

dizziness. 

Once every year, JEE toppers appear on television and 

newspapers carry congratulatory messages. You see mug 

shots of students, interviews with parents, and 

advertisements for coaching centres. We spend a lot of 

time celebrating their success, but rarely do we look 

further. 

What becomes of these brilliant 17-year-olds? What are 

the challenges they encounter? Do any of them pursue 

unconventional careers? These were some of the questions 

Open set out with while tracking down the very elite group 

of JEE toppers. 

IT HELPS TO BE NO. 1 

During his days in IIT Kanpur, Buch was a long-distance 

athlete, weightlifter and footballer. He competed in both 

the 5,000 and 10,000 metres. But in August 1993, a doctor 

at Delhi’s All India Institute of Medical Sciences 

diagnosed the 20-year-old with ankylosing spondylitis, a 

progressively crippling disease without a cure. 

Buch suffered inflammation of the eyes and internal 

organs. “Sometimes it was so hard for me to even sit, stand 

or sleep,” he recalls. Things got progressively worse over 

his two-year graduate program at Cornell University. 

“When I finished in 1995, I was immobilised throughout 

much of my body. A doctor advised me to stop working 

and apply for disability payments.” 

Buch refused. He moved to San Francisco and started a 

self-directed rehabilitation programme. He began with 

long sessions of swimming and gradually started to walk, 

bike and hike. In 2001, he successfully undertook the 

Death Ride over five alpine passes on the Sierra Nevada 

mountain range in California, US. But biking hurt his 

knees. Searching for a sport that didn’t tax his legs, he 

discovered surf skiing, one that uses a long, narrow, 

lightweight kayak with an open cockpit and a foot-pedal 

controlled rudder. On 17 May, Buch took part in the 2009 

Molokai Challenge in Hawaii, a 32-mile surf ski race 

between Molokai and Oahu, in rough waters swarming 

with tiger sharks. He finished the race. 

“I thought being No 1 in JEE was tough,” says Buch. “But 

overcoming this disease has been something else. The JEE 

effort definitely helped with this—I knew the levels of 

determination I was capable of and refused to give up.” 


KNOW IIT-JEE 

By Previous Exam Questions 

PHYSICS 

1. Two narrow cylindrical pipes A and B have the same 

length. Pipe A is open at both ends and is filled with a 

monoatomic gas of molar mass M A . Pipe B is open at 

one end and closed at the other end, and is filled with a 

diatomic gas of molar mass M B . Both gases are at the 

same temperature. [IIT- 2002] 

(a) If the frequency of the second harmonic of the 

fundamental mode in pipe A is equal to the frequecy of 

the third harmonic of the fundamental mode in pipe B, 

determine the value of M A /M B . 

(b) Now the open end of pipe B is also closed (so that 

the pipe is closed at bout ends). FInd the ratio of the 

fundamental frequency in pipe A to that in pipe B. 

Sol. (a) Second harmonic in pipe A = 2 [(v 0 )A] Third 

harmonic of pipe B = 3 [(v 0 )B] 

⎡ v ⎤ 

⎡ v ⎤ 

= 2 ⎢ ⎥ = 3 

⎣2l 

⎢ ⎥ ⎦ ⎣ 4l ⎦ 

= l 

1 

γ ART 

M 

A 

A 

Gas (Monoatomic) 

M A 

3 

= 4l 

B 

Gas (Diatomic) 

M B 

γ BRT 

M 

l 

l 

Given that second harmonic in pipe A = Third 

harmonic of pipe B 

⇒ 

⇒ 

1 

l 

γ ART 

M 

A 

M A 400 = 

M B 189 

(b) (v 0 ) A = 

∴ 

( v0 

) 

( v ) 

0 

A 

B 

= 

= 

γ ART 

M 

A 

γ 

M 

A 

A 

3 

4l 

M 

× 

γ 

γ BRT 

M 

B 

[γ A = 1.67 and γ β = 1.4] 

(v 0 ) B = 

B 

B 

= 4 

3 

B 

γ BRT 

M 

2. A non-conducting disc of radius a and uniform positive 

surcface charge density σ is placed on the ground, with 

its axis vertical. A particle of mass m and positive 

charge q is dropped, along the axis of the disc, from a 

height H with zero initial velocity. The particle has 

q/m = ε 0 g/σ. 

B 

(a) Find the value of H if the particle just reaches the 

disc. 

(b) Sketch the potential energy of the particle as a 

function of its height and find its equilibirum 

position. [IIT- 1999] 

Sol. (a) Given that : a = radius of disc, σ = surface 

charge density, q/m = 4ε 0 g/σ 

The K.E. of the particle, when it reaches the disc can 

be taken as zero. 

Potential due to a charged disc at any axial point 

situated at a distance x from 0. 

σ 2 2 

V(x) = [ a + x – x] 

] 

2ε 0 

Hence, 

V(H) 

σ 

2ε 0 

[ 

2 

a + H 

2 

– H ] 

σa 

and V(O) = 

2ε 0 

According to law of conservation of energy, Loss of 

gravitation potential energy = gain in electric 

potential energy 

H (m,q) 

H 

O 

mgH = qDV = q[V(0) – V(H)] 

2 2 σ 

mgH = g[a – { ( a + H ) – H}] 

…(1) 

2ε 0 

σq 

From the given relatuion : = 2 mg (given) 

2ε 0 

Putting this is equation (1), we get, 

MgH = 2mg[a – { 

a 

( a 2 + 2 

H ) – H }] 

or H = 2[a + H – ( a + H ) ] 

or H = 2a + 2H – 2 ( a + H ) 

2 

2 

or 2 ( a + H ) = H + 2a 

or 4a 2 + 4H 2 = H 2 + 4a 2 + 4aH 

or 3H 2 4a 

+ 4aH or H = 

3 

[Q H = O is not valid] 

2 

2 

2 

2 


(b) Total potential energy of the particle at height h 

qσ 

2 2 

U(x) = mgx + qV(x) = mgx + ( a + x – x) 

2ε 

= mgx + 2mg [ a + x – x] 

2 

2 

2 

2 

= mg [2 a + x ) – x] 

…(2) 

dU 

For equilibrium : = 0 dx 

a 

This gives : x = 

3 

From equation (2), graph between U(x) and x is as 

shown above. 

U 

2 mga 

3 mga 

0 

When the ring is rotating, we can treat it as a current 

carrying loop. The magnetic mement of this loop 

M = iA = T 

Q × πr 

2 

= 

Q ω × πr 

2 

2π 

This current carrying loop will create its own 

magnetic field which will interact with the given 

vertical magnetic field in such a way that the 

tensions in the strings will become unequal. Let the 

tension in the string be T 1 and T 2 . 

For translational equilibrium 

T 1 + T 2 = mg 

…(2) 

For rotational equilibrium 

Torque acting on the ring about the centre of ring 

→ → → 

τ = M × B 

t = M × B × sin 90º 

= 

Q ω × πr 2 × B = 

2π 

2 

QωBr 

2 

For rotational equilibrium, the torque about the 

centre of ring should be zero. 

O a / 3 H = 4a/3 X 

D D QωBr 

∴ T 1 × – T2 × = 2 2 2 

2 

3. A wheel of radius R having charge Q, uniformly 

distributed on the rim of the wheel is free to rotate 

about a light horizontal rod. The rod is suspended by 

ligh inextensible strings and a magnetic field B is 

applied as shown in the figure. The initial tensions in 

the strings are T 0 . If the breaking tension of the stringas 

3T 

are 0 , find the maximum angular velocity ω0 

2 

with which the wheel can be rotated. [IIT-2003] 

d 

⇒ T 1 – T 2 = 

QωBr 

D 

On solving (2) and (3) we get 

T 1 = 

mg + 

2 

2 

QωBr 

2D 

But the maximam tension is 

2 

3T 

0 

2 

…(3) 

∴ 

3T 0 = T0 + 

2 

Q 

ω max 

Br 

2D 

2 

⎡ 

⎢Q 

T 

⎣ 

0 

mg ⎤ 

= 2 

⎥ 

⎦ 

T 0 

ω 0 

Sol. From above figure, when the ring is not rotating wt. 

of ring = Tension in string mg = 2T 0 

∴ T 0 = 

mg 

2 

B 

T 0 

…(1) 

∴ ω max = 

DT 

BQr 

0 

2 

4. An object is moving with velocity 0.01 m/s towards a 

convex lens of focal length 0.3 m. Find the magnitude 

of rate of separation of image from the lens when the 

object is at a distance of 0.4m From the lens. Also 

calculated the magnitude of the rate of change of the 

lateral magnification. 

[IIT-2004] 

Sol. f = 0.3 m, u = – 0.4 m 

Using lens formula 

1 1 1 

– = 

v – 0. 4 0.3 

⇒ v = 1.2 m 


Now we have 

1 1 1 – = , differentiating w.r.t. t 

v u f 

we have – 

1 

2 

v 

dv 1 + 

dt 

2 

u 

du = 0 

dt 

du 

given 

= 0.01 m/s 

dt 

2 

⎛ dv ⎞ (120) 

⇒ ⎜ ⎟ = × 0.01 = 0.09 m/s 

⎝ dt ⎠ 

2 

(0.4) 

So, rate of seperation of the image (w.r.t. the lens) = 

0.09 m/s 

udv vdu 

– 

v dm 

Now, m = ⇒ = 

dt dt 

u dt 

2 

u 

(0.4)(0.09) – (1.2)(0.01) 

= – 0.35 

2 

(0.4) 

So magnitude of the rate of change of lateral 

magnification = 0.35. 

5. A particle of charge equal to that of an electron, –e, and 

mass 208 times the mass of the electron (called a numeson) 

moves in a circular orbit around a nucleus of 

charge + 3e. (Take the mass of the nucleus to be 

infinite). Assuming that the bohr model of the atom is 

applicable to this system. 

(i) Derive an expression for the radius of the nth Bohr 

orbit. 

(ii) Find the value of n for which the radius of the orbit 

is approximately the same as that of the first Bohr 

orbit for the hydrogen atom. 

(iii) Find the wavelength of the radiation emitted when 

the mu-meson jumps from the third orbit of the first 

orbit. 

[IIT-1988] 

Sol. (i) Let m be the mass of electron. Then the mass of 

mu-meson is 208 m. According to Bohr's 

postualte, the angular momentum of mu-meson 

should be an integral multiple of h/2π. 

e 

r 

+3e 

nh 

∴ (208 m) vr = 

2π 

nh nh 

∴ v = 

= 

…(1) 

2 π× 208mr 

416πmr 

Since mu-meson is moving in a circular path 

therefore it needs centripetal force which is 

provided by the electrostatic force between the 

nucleus and mu-meson. 

∴ 

(208m)v 

r 

2 

= 

1 

4πε 

0 

× 

3e× e 

r 

2 

2 

3e 

∴ r = 

2 

4πε0 

× 208mv 

Substituting the value of v from (1) we get 

r = 

3e 

2 

× 416πmr 

× 416πmr 

4πε 

0 

2 2 

h ε0 

2 

2 

× 208n 

h 

2 

n 

⇒ r = 

…(2) 

624πme 

(ii) The radius of the first orbit of the hydrogen atom 

(iii) 

⇒ 

⇒ 

2 

ε0h 

= 

…(3) 

2 

πme 

To find the value of n for which the radius of the 

orbit is approximately the same as that of the 

first Bohr orbit for hydrogen atom, we equate 

equation (2) and (3) 

n 

2 2 

h ε0 

2 

624πme 

= 

ε 

0 

h 

2 

πme 

1 = 208 R × z 

2 

λ 

2 

⎡ 1 

⎢ 

2 

⎢⎣ 

n1 

⇒ n = 624 ≈ 25 

1 ⎤ 

– 

2 

⎥ 

n2 

⎥⎦ 

1 = 208 × 1.097 × 10 

7 

× 3 2 ⎡ 1 1 ⎤ 

λ 

⎢ – 

2 2 ⎥ 

⎣1 

3 ⎦ 

λ = 5.478 × 10 –11 m 

CHEMISTRY 

6. A metallic element crystallizes into a lattice 

containing a sequence of layers of ABABAB ..... Any 

packing of spheres leaves out voids in the lattice. 

What percentage by volume of this lattice is empty 

space ? 

[IIT-1996] 

Sol. A unit cell of hcp structure is a hexagonal cell, which 

is shown in fig. Three such cells form one hcp unit. 

For hexagonal cell, a = b ≠ c; α = β = 90º and 

γ = 120º. It has 8 atoms at the corners and one inside, 

hence 

Number of atoms per unit cell = 8 

8 + 1 = 2 

O 

a 

60º 

N b 

3 

Area of the base = b × ON = b × a sin 60º = a 

2 

2 

( Q b = a) 

Volume of the hexagonal cell 

= Area of the base × height = 

3 a 2 . c 

2 


But c = 

2 

3 

2 

a 

Q [Ag + ] = [I – ] 

∴ K sp of AgI = [Ag + ] 2 

∴ [Ag + ] of AgI = 

K sp of AgI 

c 

β α 

b 

a γ 

∴ Volume of the hexagonal cell 

3 

= a 2 2 2 

. a = a 3 2 

2 3 

and radius of the atom, 

r = a/2 

Hence, fraction of total volume of atomic packing 

Volume of 2 atoms 

factor = 

Volume of the hexagonal cell 

4 4 

3 

⎛ a ⎞ 

2× 

πr 2× 

π⎜ 

⎟⎠ 

= 

3 3 2 

= 

⎝ π 

= 

3 

3 

a 2 a 2 3 2 

= 0.74 = 74% 

∴ The percentage of void space = 100 – 74 

= 26% 

7. (The standard reduction potential of Ag + /Ag 

electrode at 298 K is 0.799V. Given that for AgI, 

K sp = 8.7 × 10 –17 , evaluate the potential of Ag + /Ag 

electrode in a saturated solution of AgI. Also 

calculate the standard reduction potential of 

I – electrode. 

[IIT-1994] 

Sol. In the saturated solution of AgI, the half cell 

reactions are 

At anode : Ag ⎯→ Ag + + e – 

At cathode : AgI + e – ⎯→ Ag + I – 

Cell reaction AgI ⎯→ Ag + + I – 

On applying Nernst equation 

0.0591 

E cell = Eº cell – log [Ag + ] [I – ] 

n 

For electrode 

Ag + + e – → Ag 

∴ 

E = E – 

+ + 

Ag / Ag 

º 

Ag 

/ Ag 

K sp of AgI = [Ag + ] [I – ] 

3 

0.0591 

n 

1 

log 

+ 

[Ag ] 

−17 

[Ag + ] = 8.7× 

10 

= 9.3 × 10 –9 M 

So E = 0.799 – 0.0591 1 

log 

Ag + / Ag 

−9 

1 9.3× 

10 

= + 0.799 + 0.0591 log 9.3 – 0.0591 × 9 log 10 

= + 0.799 + 0.0591 × 0.9785 – 0.0591 × 9 

= 0.325 V 

For above cell reaction 

0.0591 

E cell = Eº cell – log [Ag + ] [I – ] 

n 

0.0591 

= Eº cell – log (K sp of AgI) 

n 

At equilibrium E cell = 0 

0.0591 

∴ Eº cell = log(8.7 × 10 –17 ) = –0.95 volt 

1 

Eº cell = Eº cathode + Eº anode 

– 

–0.95 = –0.799 + Eº Ag/AgI/I 

(In form of cell reaction) 

Eº – Ag/AgI/I = – 0.95 + 0.799 = –0.151 V 

– 

or Eº I /AgI/Ag = + 0.151 V 

8. An organic compound A, C 6 H 10 O, on reaction with 

CH 3 MgBr followed by acid treatment gives 

compound B. The compound B on ozonolysis gives 

compound C, which in presence of a base gives 

1-acetyl cyclopentene D. The compound B on 

reaction with HBr gives compound E. Write the 

structures of A, B, C and E. Show how D is formed 

from C. 

[IIT-2000] 

Sol. The given reactions are as follows. 

O 

OMgBr 

CH 3 

CH 3 

CH 3Br 

CH 3MgBr 

H + 

–H 2O 

HBr 

(A) (B) (E) 

COCH 3 

(D) 

Base 

COCH 3 

O 

CH 3 

(C) 

O O 


The conversion of C into D may involve the 

following mechanism. 

COCH 3 COCH 3 COCH 3 

CH 2 O HC 

B + O HC O – BH + 

–BH + 

–B 

(C) 

COCH 3 

OH 

+B 

–BH + 

COCH 3 

– 

OH 

–OH – COCH 3 

9. A colourless solid (A) on heating gives a white solid 

(B) and a colourless gas (C). (B) gives off reddishbrown 

fumes on treating with H 2 SO 4 . On treating 

with NH 4 Cl, (B) gives a colourless gas (D) and a 

residue (E). The compound (A) on heating with 

(NH 4 ) 2 SO 4 gives a colourless gas (F) and white 

residue (G). Both (E) and (G) impart bright yellow 

colour to Bunsen flame. The gas (C) forms white 

powder with strongly heated Mg metal which on 

hydrolysis produces Mg(OH) 2 . The gas (D) on 

heating with Ca gives a compound which on 

hydrolysis produces NH 3 . Identify compounds (A) to 

(G) giving chemical equations involved. 

Sol. The given information is as follows : 

(i) A ⎯ Heat ⎯⎯ → B + C 

Colourless Solid Colourless 

Solid 

gas 

(ii) B + H 2 SO 4 ⎯⎯→ 

∆ Reddish brown gas 

(iii) B + NH 4 Cl ⎯⎯→ 

∆ D + E 

Colourless gas 

(iv) A + (NH 4 ) 2 SO 4 ⎯⎯→ 

∆ F + G 

olourless gas White 

Residue 

(v) E and G imparts yellow colour to the flame. 

(vi) C + Mg ⎯ Heat ⎯⎯ →White powder ⎯ 

H 2 

⎯⎯ O →Mg(OH) 2 

(vii) D + Ca ⎯ Heat ⎯⎯ →Compound ⎯ 

H 2 

⎯⎯ O →NH 3 

Information of (v) indicates that (E) and (G) and also 

(A) are the salts of sodium because Na + ions give 

yellow coloured flame. Observations of (ii) indicate 

that the anion associated with Na + in (A) may be 

NO – 3 . Thus, the compound (A) is NaNO 3 . 

The reactions involved are as follows : 

(i) 2NaNO 3 ⎯⎯→ 

∆ 2NaNO 2 + O 2 ↑ 

(A) (B) (C) 

(ii) 2NaNO 2 + H 2 SO 4 ⎯→ Na 2 SO 4 + 2HNO 2 

(B) Dil. 

3HNO 2 ⎯→ HNO 3 + H 2 O + 2NO↑ 

2NO + O 2 ⎯→ 2NO 2 ↑ 

Reddish brown 

Fumes 

(D) 

(iii) NaNO 2 + NH 4 Cl ⎯→ NaCl + N 2 ↑ + 2H 2 O 

(B) (E) (D) 

(iv) 2NaNO 3 + (NH 4 ) 2 SO 4 ⎯⎯→ 

∆ Na 2 SO 4 + 2NH 3 

(A) (G) (F) 

2HNO 3 

(v) O 2 + 2Mg ⎯⎯→ 

∆ 2MgO ⎯ 

H 2 

⎯⎯ O →Mg(OH) 2 

(C) 

(vi) N 2 + 3Ca ⎯⎯→ 

∆ Ca 3 N 2 

(D) 

Ca 3 N 2 + 6H 2 O ⎯→ 3Ca(OH) 2 + 2NH 3 ↑ 

Hence, 

(A) is NaNO 3 , 

(B) is NaNO 2 , 

(C) is O 2 , 

(D) is N 2 , 

(E) is NaCl, 

(F) is NH 3 and (G) is Na 2 SO 4 . 

10. An alkyl halide X, of formula C 6 H 13 Cl on treatment 

with potassium t-butoxide gives two isomeric alkenes 

Y and Z(C 6 H 12 ). Both alkenes on hydrogenation give 

2, 3-dimethyl butane. Predict the structures of X, Y 

and Z. 

[IIT-1996] 

Sol. The alkyl halide X, on dehydrohalogenation gives 

two isomeric alkenes. 

C 6H 

Cl 

13 

X 

K−t−butoxide 

⎯ ⎯⎯⎯⎯→ 

∆; 

–HCl 

Y + Z 

C 6 H 12 

Both, Y and Z have the same molecular formula 

C 6 H 12 (C n H 2n ). Since, both Y and Z absorb one mol of 

H 2 to give same alkane 2, 3-dimethyl butane, hence 

they should have the skeleton of this alkane. 

H 

Y and Z (C 6 H 12 ) ⎯ ⎯→ 2 CH 3 – CH – CH – CH 3 

Ni 

CH 3 CH 3 

2,3-dimethyl butane 

The above alkane can be prepared from two alkenes 

CH 3 – C = C – CH 3 and CH 3 – CH – C = CH 2 

CH 3 CH 3 

CH 3 CH 3 

2,3-dimethyl 

2,3-dimethyl butene-1 

butene-2 

(Z) 

(Y) 

The hydrogenation of Y and Z is shown below : 

CH 3 – C = C – CH 3 

CH 3 CH 3 

(Y) 

H 2 

Ni 

CH 3 – CH – CH – CH 3 

CH 3 CH 3 

CH 3 – CH – C = CH 2 

H 2 

Ni 

CH 3 – CH – CH – CH 3 

CH 3 CH 3 

CH 3 CH 3 

(Z) 

Both, Y and Z can be obtained from following alkyl 

halide : 


Cl 

K-t-butoxide 

CH 3 – C – CH – CH 3 

∆; –HCl 

CH 3 CH 3 

2-chloro-2,3-dimethyl butane 

(X) 

CH 2 = C — CH – CH 3 

CH 3 CH 3 

Hence, X, CH 3 – C – CH – CH 3 

+ CH 3 – C = C – CH 3 

CH 3CH 3 

(Z) 20% (Y) 80% 

Cl 

CH 3 CH 3 

Y, CH 3 – C = C – CH 3 

CH 3 CH 3 

Z, CH 3 – CH – C = CH 2 

CH 3 CH 3 

MATHEMATICS 

11. The curve y = ax 3 + bx 2 + cx + 5, touches the x-axis at 

P(–2, 0) and cuts the y axis at a point Q, where its 

gradient is 3. Find a, b, c. 

[IIT-1994] 

Sol. It is given that y = ax 3 + bx 2 + cx + 5 touches x-axis at 

P(–2, 0) which implies that x-axis is tangent at 

(–2, 0) and the curve is also passes through (–2, 0). 

The curve cuts y-axis at (0, 5) and gradient at this 

point is given 3 therefore at (0, 5) slope of the tangent 

is 3. 

dy 

Now, = 3ax 2 + 2bx + c 

dx 

since x-axis is tangent at (–2, 0) therefore 

dy 

= 0 

dx 

x=−2 

⇒ 0 = 3a(–2) 2 + 2b(–2) + c 

⇒ 0 = 12a – 4b + c ...(1) 

again slope of tangent at (0, 5) is 3 

dy 

⇒ = 3 

dx 

(0,5) 

⇒ 3 = 3a(0) 2 + 2b(0) + c 

⇒ 3 = c ...(2) 

Since, the curve passes through (–2, 0), we get 

0 = a(–2) 3 + b(–2) 2 + c(–2) + 5 

0 = – 8a + 4b – 2c + 5 ...(3) 

from (1) and (2), we get 

12a – 4b = –3 ...(4) 

from (3) and (2), we get 

– 8a + 4b = 1 ...(5) 

adding (4) and (5), we get 

4a = –2 

⇒ a = –1/2 

Putting a = –1/2 in (4), we get 

12(–1/2) – 4b = –3 

⇒ – 6 – 4b = –3 

⇒ 

– 3 = 4b 

⇒ b = –3/4 

Hence, a = –1/2, b = –3/4 and c = 3 

12. In a triangle ABC, the median to the side BC is of 

1 

length 

and it divides the angle A into 

11− 

6 3 

angles 30º and 45º. Find the length of the side BC. 

[IIT-1985] 

Sol. Let AD be the median to the base BC = a of ∆ABC 

let ∠ADC = θ then 

⎛ a a ⎞ a a 

⎜ + ⎟ cot θ = cot 30º – cot 45º 

⎝ 2 2 ⎠ 2 2 

3 −1 

⇒ cot θ = 

2 

Applying sine rule in ∆ADC, we get 

A 

30º45º 

θ 

B 

C 

a/2 D a/2 

AD DC 

= 

sin( π − θ − 45º ) sin 45º 

AD a / 2 

⇒ 

= 

sin( θ + 45º ) 1/ 2 

a 

⇒ AD = {sin 45º cosθ + cos45ºsinθ} 

2 

a ⎛ cosθ + sin θ 

⇒ AD = 

⎟ ⎞ a ⎜ 

= (cos θ + sin θ) 

2 ⎝ 2 ⎠ 2 

1 a 

⎛ 

⎞ 

⇒ 

= 

⎜ 3 −1 

2 

+ 

⎟ 

⎜ 

⎟ 

11− 

6 3 2 

⎝ 8 − 2 3 8 − 2 3 ⎠ 

⇒ a = 

⇒ a = 

( 

2 8 − 2 

3 

3 + 1) 11− 

6 

( 

2 8 − 2 

3 + 1) 

2 

3 

3 

11− 

6 

3 


⇒ a = 

⇒ a = 2 

(4 + 2 

2 8 − 2 

44 − 24 

3 

3)(11− 

6 

8 − 2 

3 

3 + 22 

3) 

3 − 36 

13. Without using tables, prove that 

(sin 12º) (sin 48º) (sin 54º) = 8 

1 

Sol. (sin 12º) (sin 48º) (sin 54º) 

= 2 

1 (2 sin 12º sin 48º) sin 54º 

8 − 2 

= 2 

8 − 2 

3 

= 2 

3 

[IIT-1982] 

15. Evaluate 

∫ π −π 

Sol. Let, 

I = 

∫ π −π 

/3 

/ 3 

/ 3 

/ 3 

π + 4x 

dx 

⎛ π ⎞ 

2 − cos⎜| 

x | + ⎟ 

⎝ 3 ⎠ 

πdx 

⎛ π ⎞ 

2 − cos⎜| 

x | + ⎟ 

⎝ 3 ⎠ 

3 

+ 4 

∫ π −π 

/3 

/ 3 

[IIT-2004] 

3 

x dx 

⎛ π ⎞ 

2 − cos⎜| 

x | + ⎟ 

⎝ 3 ⎠ 

a ⎡ 0, f ( −x) 

= − f ( x) 

⎤ 

Using 

∫ 

f ( x) 

dx = ⎢ a 

⎥ 

− a ⎢2 

− = ⎥ 

⎣ 

∫ 

f ( x) 

dx, 

f ( x) 

f ( x) 

0 

⎦ 

= 2 

1 {cos (36º) – cos (60º)}sin 54º 

= 2 

1 

⎧ 1 ⎫ 

⎨cos36º− ⎬ sin 54º 

⎩ 2 ⎭ 

= 4 

1 {2 cos 36º sin 54º – sin 54º} 

= 4 

1 (sin 90º + sin 18º – sin 54º) 

1 ⎪⎧ 5 −1 

5 + 1⎪⎫ 

= ⎨1 

+ − ⎬ 

4 ⎪⎩ 4 4 ⎪⎭ 

1 ⎪⎧ 

5 −1− 

5 −1⎪⎫ 

= ⎨1 

+ 

⎬ 

4 ⎪⎩ 4 ⎪⎭ 

= 4 

1 

⎧ 1 ⎫ 1 

⎨1 − ⎬ = 

⎩ 2 ⎭ 8 

14. An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6, 

is thrown n times and the list on n numbers showing 

up is noted. What is the probability that among the 

numbers 1, 2, 3, 4, 5, 6 only three numbers appear in 

this list ? 

[IIT-2001] 

Sol. Let us define at onto function F from A : [r 1 , r 2 ... r n ] 

to B : [1, 2, 3] where r 1 r 2 .... r n are the readings of n 

throws and 1, 2, 3 are the numbers that appear in the 

n throws. 

Number of such functions, 

M = N – [n(1) – n(2) + n(3)] 

where N = total number of functions and 

n(t) = number of function having exactly t elements 

in the range. 

Now, N = 3 n , n(1) = 3.2 n , n(2) = 3, n(3) = 0 

⇒ M = 3 n – 3.2 n + 3 

Hence the total number of favourable cases 

= (3 n – 3.2 n + 3). 6 C 3 

n n 6 

( 3 − 3.2 + 3) × C3 

⇒ Required probability = 

n 

6 

∴ I = 2 

∫ π 

0 

/ 3 

πdx 

+ 0 

⎛ π ⎞ 

2 − cos⎜| 

x | + ⎟ 

⎝ 3 ⎠ 

⎧ 

⎪ 

⎨as 

⎪ 

⎪⎩ 

∫ π −π 

/3 

/ 3 

I = 2π 

∫ π /3 dx 

0 2 − cos( x + π / 3) 

⎫ 

3 

x dx 

⎪ 

is odd ⎬ 

⎛ π ⎞ 

2 − cos⎜| 

x | + ⎟ ⎪ 

⎝ 3 ⎠ ⎪⎭ 

2 

= 2π 

∫ π / 3 dt 

π 

, where x + = t 

π / 3 2 − cost 

3 

2 

= 2π 

∫ π 

/ 3 

2 t 

sec dt 

2 

t 

1+ 

3tan 

2 

π/ 3 

2 

3 2du 

= 2π 

∫ 1/ 3 1+ 

3u 

= 

∴ 

∫ π −π 

3 

2 

= 

4π . 

−1 

{ 3 tan 3u} 3 1/ 3 

3 

4π (tan –1 3 – tan –1 1) = 

/ 3 

/ 3 

3 

4π tan 

–1 ⎛ 1 ⎞ 

⎜ ⎟ 

3 ⎝ 2 ⎠ 

π + 4x 

4π 

dx = tan 

–1 ⎛ 1 ⎞ 

⎜ ⎟ . 

⎛ π ⎞ 

2 − cos⎜| 

x | + 

3 ⎝ 2 ⎠ 

⎟ 

⎝ 3 ⎠ 



Physics Challenging Problems 

Set # 10 

This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety 

of possible twists and turns of problems in physics that would be very helpful in facing IIT 

JEE. Each and every problem is well thought of in order to strengthen the concepts and we 

hope that this section would prove a rich resource for practicing challenging problems and 

enhancing the preparation level of IIT JEE aspirants. 

By : Dev Sharma 

Solutions will be published in next issue 

Director Academics, Jodhpur Branch 

1. Two capacitors C 1 and C 2 , can be charged to a 

potential V/2 each by having 

C 1 

C 2 

V R R 

S 1 S 2 

(A) S 1 closed and S 2 open 

(B) S 1 open and S 2 closed 

(C) S 1 and S 2 both closed 

(D) cannot be charged at V/2 

2. Energy liberated in the de-excitation of hydrogen 

atom from 3 rd level to 1 st level falls on a photocathode. 

Later when the same photo-cathode is 

exposed to a spectrum of some unknown hydrogen 

like gas, excited to 2 nd energy level, it is found that 

the de-Broglie wavelength of the fastest 

photoelectrons, now ejected has decreased by a 

factor of 3. For this new gas, difference of energies 

of 2 nd Lyman line and 1 st Balmer line if found to be 3 

times the ionization potential of the hydrogen atom. 

Select the correct statement(s) 

(A) The gas is lithium 

(B) The gas is helium 

(C) The work function of photo-cathode is 8.5eV 

(D) The work function of photo-cathode is 5.5eV 

3. In the figure shown there exists a uniform time 

varying magnetic field B = [(4T/s) t + 0.3T] in a 

cylindrical region of radius 4m. An equilateral 

triangular conducting loop is placed in the magnetic 

field with its centroide on the axis of the field and its 

plane perpendicular to the field. 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ + + + 

B + + + + C 

(A) e.m.f. induced in any one rod is 16V 

A 

(B) e.m.f. induced in the complete ∆ ABC is 48 3V 

(C) e.m.f. induced in the complete ∆ ABC is 48V 

(D) e.m.f. induced in any one rod is 16 3V 

O 

4. 6 parallel plates are arranged as shown. Each plate 

has an area A and distance between them is as 

shown. Plate 1-4 and plates 3-6 are connected 

equivalent capacitance across 2 and 5 can be writted 

nA ∈ 

as 0 . Find mininum value of n. (n, d are 

d 

natural numbers) 

1 

2 d 

3 d 

4 2d 

d 5 

d 6 

5. Match the following 

Column – I 

Column – II 

(A) A light conducting (P) Magnetic field B 

circular flexible 

is doubled. 

loop of wire of 

radius r carrying 

current I is placed 

in uniform magnetic 

field B, the tension 

in the loop is doubled if 

(B) Magnetic field at a (Q) Inductance is 

point due to a long increased by four 

straight current 

times. 

carrying wire at a 

point near the wire 

is doubled if 

(C) The energy stored (R) Current I is 

in the inductor will doubled 

become four times 

(D)The force acting on a (S) Radius r is 

moving charge, 

doubled 

moving in a constant 

magnetic field will be 

doubled if 

(T) Velocity v is 

Doubled 


Passage # (Q. No. 6 to Q. No. 8 ) 

A solid, insulating ball of radius ‘a’ is surrounded by 

a conducting spherical shell of inner radius ‘b’ and 

outer radius ‘c’ as shown in the figure. The inner ball 

has a charge Q which is uniformly distribute 

throughout is volume. The conducting spherical shell 

has a charge –Q. 

Answer the following questions. 

b 

Q 

a 

–Q 

6. Assuming the potential at infinity to be zero, the 

potential at a point located at a distance a/2 from the 

centre of the sphere will be : 

(A) 

(B) 

Q 

4πε 

Q 

4πε 

0 

0 

⎡2 

1 ⎤ 

⎢ − ⎥ 

⎣a 

b ⎦ 

⎡11 

1 ⎤ 

⎢ − ⎥ 

⎣8a 

b ⎦ 

Q ⎡1 

1 ⎤ 

(C) ⎢ − 

4πε 

⎥ 

0 ⎣a 

b ⎦ 

(D) None of these 

7. Work done by external agent in taking a charge q 

slowly from inner surface of the shell to surface of 

the sphericalball will be : 

⎡1 

1⎤ 

(A) kQq ⎢ − ⎥ 

⎣a 

c⎦ ⎡1 

1⎤ 

(B) kQq ⎢ − ⎥ 

⎣b 

a ⎦ 

⎡1 

1 ⎤ 

(C) kQq ⎢ − ⎥ 

⎣a 

b ⎦ 

⎡1 

1⎤ 

(D) kQq ⎢ − ⎥ 

⎣c 

a ⎦ 

8. Now the outer shell is grounded, i.e., the outer 

surface is fixed to be zero. Now the charge on the 

inner ball will be : 

(A) zero 

(B) Q 

(C) 

Q ⎛ 1 

⎜ + 

C ⎝ a 

1 

c 

1 ⎞ 

− ⎟ 

b ⎠ 

(D) 

c 

Q ⎛ 1 

⎜ + 

b ⎝ a 

1 

c 

1 ⎞ 

− ⎟ 

b ⎠ 

Regents Physics 

You Should Know 

Nuclear Physics : 

• Alpha particles are the same as helium nuclei and 

have the symbol . 

• The atomic number is equal to the number of 

protons (2 for alpha) 

• Deuterium ( ) is an isotope of hydrogen ( ) 

• The number of nucleons is equal to protons + 

neutrons (4 for alpha) 

• Only charged particles can be accelerated in a 

particle accelerator such as a cyclotron or Van 

Der Graaf generator. 

• Natural radiation is alpha ( ), beta ( ) and 

gamma (high energy x-rays) 

• A loss of a beta particle results in an increase in 

atomic number. 

• All nuclei weigh less than their parts. This mass 

defect is converted into binding energy. (E=mc 2 ) 

• Isotopes have different neutron numbers and 

atomic masses but the same number of protons 

(atomic numbers). 

• Geiger counters, photographic plates, cloud and 

bubble chambers are all used to detect or observe 

radiation. 

• Rutherford discovered the positive nucleus using 

his famous gold-foil experiment. 

• Fusion requires that hydrogen be combined to 

make helium. 

• Fission requires that a neutron causes uranium to 

be split into middle size atoms and produce extra 

neutrons. 

• Radioactive half-lives can not be changed by heat 

or pressure. 

• One AMU of mass is equal to 931 meV of energy 

(E = mc 2 ). 

• Nuclear forces are strong and short ranged. 


8 Questions were Published in January Issue 

1. Option [C] is correct 

3. A → Q B → R 

Magnetic field due to infinite current carrying sheet 

C → P 

D → Q 

µ J 

is given by B = 0 

, where J is linear current 

2 

density. 

µ 0 J 

2 

I 

(a) 

µ 0 J 

2 

µ 0 J 

2 

IV 

(b) 

µ 0 J 

2 

Fig. (a) and (b) represent the direction of magnetic 

field due to current carrying sheets. For x < a, 

µ 0J 

µ 0J(2J) 

µ 0 (3J) µ 0 (4J) 

Bresul tant = − − + 

2 2 2 2 

For a < x < 2a, 

µ 0J 

µ 0 (2J) µ 0 (3J) µ 0 (4J) 

Bresul tant = − − + = −µ 

0J 

2 2 2 2 

For 2a < x < 3a, 

µ 0J 

µ 0 (2J) µ 0 (3J) µ 0 (4J) 

Bresul tant = + − − = 0 

2 2 2 2 

So, the required curve is 

Solution 

Set # 9 

Physics Challenging Problems 

i. At t = 1s, flux is increasing in the inward 

direction, hence induced e.m.f. will be in 

anticlockwise direction. 

ii. At t = 5s, there is no change in flux, so induced 

e.m.f. is zero 

iii. At t = 9s, flux is increasing in upward direction 

hence induced e.m.f. will be in clockwise 

direction. 

iv. At t = 15s, flux is decreasing in upward direction, 

so induced e.m.f. will be in anticlockwise 

direction. 

4. Option [A,B,D] is correct 

Rate of work done by external agent is 

de/dt = BIL.dx/dt = BILv and thermal power 

dissipated in resistor = eI = (BvL) I clearly both are 

equal, hence (A). 

If applied external force is doubled, the rod will 

experience a net force and hence acceleration. As a 

result velocity increase, hence (B). 

Since, I = e/R 

On doubling R, current and hence required power 

become half. 

Since, P = BILv 

Hence (D) 

5. Option [A] is correct 

2. A → P,Q,S ; B → P,Q,R,S 

C → P,Q,R,S ; D → Q 

i. Velocity of the particle may be constant, if forces 

of electric and magnetic fields balance each other. 

Then, path of particle will be straight line. Also, 

path of particle may be helical if magnetic and 

electric fields are in same direction. But path of 

particle cannot be circular. Path can be circular if 

only magnetic field is present, or if some other 

forces is present which can cancel the effect of 

electric field. 

ii. Here, all the possibilities are possible depending 

upon the combinations of the three fields. 

iii. This situation is similar to part (i) 

iv. In a uniform electric field, path can be only 

straight line or parabolic. 

→ 

∧ 

→ 

1.5( 

µ 1× 

j) = 2( µ 2× 

j) 

∧ 

∧ 

∧ 

1.5(a i + b j) × j = 2[(c i + d j) × j] 

∧ 

= 

1 .5a k 

a 

c 

= 

20 

1.5 

∧ 

2c k 

4 

= 

3 

∧ 

∧ 

∧ 

∧ 



I 2 

f sin60º 

f 

f – 

f cos60º 

+ 

WHAT ARE EARTHQUAKES? 

I 2 

∴ 

x 

u = -f cos60º 

f = +f 

1 1 1 

= − 

f v − f cos 60º 

1 1 2 

= + 

f v f 

1 2 1 

− = 

f f v 

v = -f 

f = cos 60º 

x 

f 

= x 

cos 60º 

x = 2f 

final image will formed at optical centre of first 

lens. 

7. Option [C] is correct 

C v = (3 + 2T)R 

dQ = dU + PdV 

adiabatic process dQ = 0 

0 = Rn (3 + 2T)dT + PdV 

nRT 

0 = Rn(3 + 2T)dT + dV 

V 

dV ⎛ 3 + 2T ⎞ 

∫− 

= ∫⎜ 

⎟dT 

V ⎝ T ⎠ 

-log V = 3 logT + 2T + C 

-logV – logT 3 = 2T + C 

log VT 3 = 2T + C 

VT 3 = e 2T 

VT 3 e -2T = C 


2 

P = P 0 − αV 

PV = RT 

RT 

2 

= P 0 − αV 

V 

3 

P0 V αV 

T = − 

R R 

dT = 0 

dV 

2 

P0 

3αV 

− = 

R R 

0 

P 

V = 0 

3 α 

Now put V in T. 

Earthquakes like hurricanes are not only super 

destructive forces but continue to remain a mystery 

in terms of how to predict and anticipate them. To 

understand the level of destruction associated with 

earthquakes you really need to look at some 

examples of the past. 

If we go back to the 27th July 1976 in Tangshan, 

China, a huge earthquake racked up an official 

death toll of 255,000 people. In addition to this an 

estimated 690,000 were also injured, whole 

families, industries and areas were wiped out in the 

blink of a second. The scale of destruction is hard to 

imagine but earthquakes of all scales continue to 

happen all the time. 

So what exactly are they ? Well the earths outer 

layer is made up of a thin crust divided into a 

number of plates. The edges of these plates are 

referred to as boundaries and it’s at these 

boundaries that the plates collide, slide and rub 

against each other. Over time when the pressure at 

the plate edges gets too much, something has to 

give which results in the sudden and often violent 

tremblings we know as earthquakes. 

The strength of an earthquake is measured using a 

machine called a seismograph. It records the 

trembling of the ground and scientists are able to 

measure the exact power of the quake via a scale 

known as the richter scale. The numbers range from 

1-10 with 1 being a minor earthquake (happen 

multiple times per day and in most case we don’t 

even feel them) and 7-10 being the stronger quakes 

(happen around once every 10-20 years). There’s a 

lot to learn about earthquakes so hopefully we’ll 

release some more cool facts in the coming months. 


PHYSICS 

Students'Forum 

Expert’s Solution for Question asked by IIT-JEE Aspirants 

1. A trolley initially at rest with a solid cylinder placed 

on its bed such that cylinder axis makes angle θ with 

direction of motion of trolley as shown in Figure 

starts to move forward with constant acceleration a. 

If initial distance of mid point of cylinder axis from 

rear edge of trolley bed is d, calculate the distance s 

which the trolley goes before the cylinder rolls off the 

edge of its horizontal bed. Assume dimensions of 

cylinder to be very small in comparison to other 

dimensions. Neglect slipping. 

θ 

d 

Calculate also, frictional force acting on the cylinder. 

Sol. Since, axis of cylinder is inclined at angle θ with the 

direction of motion of trolley, therefore components 

of acceleration a of trolley are acosθ along axis of 

cylinder and asinθ normal to axis of the cylinder. 

Cylinder rolls backward due to this normal 

component asinθ. 

Let mass and radius of cylinder be m and r 

respectively and let angular acceleration of cylinder 

be α. 

Due to angular acceleration, cylinder axis has 

acceleration relative to trolley bed, which will be 

equal to rα normal to cylinder axis. But component 

of acceleration of trolley normal to cylinder axis is 

asinθ. Therefore, net acceleration of cylinder axis is 

(asinθ – rα) normal to axis. 

Consider free body diagram of the cylinder as shown 

figure 

Note : There are two components of friction (i) F 1 

(normal to cylinder axis) and 

(ii) F 2 (along cylinder axis). F 2 prevents cylinder 

from sliding along axis or acosθ component of 

acceleration of cylinder along axis is due to F 2 . 

mg 

∴ F 2 = ma cos θ 

F 2 is not shown in the free body diagram because in 

this diagram forces action normal to cylinder axis are 

shown. 

For horizontal forces, 

F 1 = m (a sin θ – rα) 

…(1) 

2 

mr 

F 1 r = I α where I = 

2 

∴ F 1 = 2 

1 mrα …(2) 

Form equation (1) and (2), 

rα = 3 

2 a sin θ 

The cylinder will roll off the edge of trolley bed 

when its centre of mass reaches the edge. Since. 

cylinder axis is inclined at an angle 'θ' with direction 

of motion of trolley, therefore, its centre of mass 

follows a straight line path relative to the trolley bed, 

and that straight line is normal to cylinder axis. 

Hence, displacement of centre of mass of the cylinde, 

relative to trolley is equal to (d. cosec θ). 

considereing motion of cylinder relative to the 

trolley, 

u = 0, acceleration = rα = 3 

2 a sin θ, s = d cosec θ, 

t = ? 

Using, s = ut + 2 

1 at 2 , or t = 

3d 

asin 

2 

θ 

Now considering motion of trolley during this 

interval ofd time, 

u = 0, acceleration a , t = 

3d 

asin 

2 

θ 

, s = ? 

Using, s = ut + 2 

1 at 2 , s = 2 

3 d cosec 2 θ 

F 1 = 2 

1 m. rα = 3 

1 ma sin θ 

Ans. 

Total frictional force acting on the cylinder is 

l.α 

F = 

2 2 

1 F2 

F + 

N 

F 1 

m(a.sin θ –rα) 

2 θ 

2 

1 

= ma sin + 9cos θ 

Ans. 

3 


2. A particle of mass m is placed on centre of curvature 

of a fixed, uniform semi-circular ring of radius R and 

M as shown in Figure. Claculate 

M 

R 

(i) interaction force between the ring and the particle 

and 

(ii) work required to displace the particle from centre 

of curvature to infinity. 

Sol. To calculate, interaction force, consider two equal are 

lengths R dθ each of the semi-circular ring as shown 

in figure 

Rdθ 

Rdθ 

θ 

θ 

dθ 

dθ 

m 

M 

Mass of each arc, dM = 

πR 

Rdθ = mdθ 

π 

Gravitational force exerted by each arc on the 

particle, 

GmdM GMm 

dF = 

2 = dθ 

2 

R πR 

Since, force exerted by each arc is directly 

towards the arc, therefore, resultant of these two 

forces is along negative x-axis and the resultant force 

= dF 1 cos θ 

2GMm 

= cos θ dθ 

2 

πR 

Total force on the particle is 

θ=π/ 

2 

2GMm 

F = 

2 

πR 

∫ 

cosθd 

θ 

θ= 0 

2GMm 

or F = 

Ans. (i) 

2 

πR 

Work done during displacement of particle from 

centre of the curvature to infinity is used to increase 

gravitational potential energy of the system. 

Initial gravitational potential energy of particle with 

each arc is 

Gm.dM GMm 

dU = – = – dθ 

R πR 

∴ Total initial potential energy, 

U 1 = – 

or U 1 = – 

GMm 

πR 

GMm 

R 

π/ 

2 

∫ 

d θ 

θ= – π/ 

2 

x 

When separation between particle and semicircular 

ring becomes large, potential energy becomes U 2 = 0 

GMm 

∴ Work done = U 2 – U 1 = 

Ans.(ii) 

R 

3. A long round conductor of radius a is made of a 

material whose thermal conductivity depends on 

distance r from axis of the conductor as K = cr 2 , 

where c is a constant. Calculate 

(i) thermal resistance per unit length of such a 

conductor and 

(ii) temperature gradient if rate of heat flow through 

the rod is H. 

Sol. Since, thermal conductivity of material of the 

conductor depends upon distance from its axis, 

therefore, conductivity at every point of a co-axial 

cylindrical surface will be the same. To calculate 

thermal resistance of the given conductor, it may be 

assumed to be composed of thin co-axial cylindrical 

shells which are in parallel combination with each 

other. 

Consider a thin co-axial cylindrical shell of radius x, 

radial thickness dx and of unit length as shown in 

figure 

Its cross sectional area, A = 2πx.dx 

Thermal conductivity K = cx 2 and length l = 1 m 

l l 

∴ Its thermal resistance, dR = = KA 

2 

(cx )2πx.dx 

1 

or dR = 

3 

2πcx .dx 

Since, such cylindrical shells are in parallel with each 

other, therefore, equivalent resistance R per unit 

length is given by 

1 = 

∫ dR 

1 = 

∫ 

R 

x= 

a 

x= 

0 

2 πcx 

2 

or R = 

Ans.(i) 

4 

πca 

Since, temperature gradient is temperature difference 

per unit length, therefore, temperature gradient = rate 

of heat flow × resistance per unit length 

or 

dθ 2H 

= H × R = 

4 

dt 

πca 

3 . 

dx 

Ans. (ii) 

4. Switch S of circuit shown in Figure is in position 1 

for a long time. At instant t = 0, it is thrown from 

position 1 to 2. Calculate thermal power P 1 (t) and 

P 2 (t) generated across resistance R 1 and R 2 

respectively. 


S 

1 

2 

C 

+ – 

E 

Sol. Since, initially the switch was in position 1 for a long 

time, therefore, initially the capacitor was fully 

charged or potential difference across capacitor at 

t = 0 was equal to emf E fo the battery. 

∴ Initial charge on capacitor, q 0 = CE 

When switch is thrown to position 2, capacitor starts 

to discharge through resistance R 1 and R 2 . To 

calculate thermal power P 1 (t) and P 2 (t) generated 

across R 1 and R 2 respectively, current I at time t 

through the circuit must be known. 

Let at instant t, charge remaining on the capacitor be 

q and let current through the circuit be I. 

Applying Kirchhoff's voltage law on the mesh in the 

circuit of figure 

C 

I 

+ – 

q 

R 1 

R 1 

q – IR2 – IR 1 = 0 

C 

or 

q 

I = 

( R1 + R2) 

C 

...(1) 

Since, the capacitor is discharging, therfore, 

dq 

I = – dt 

∴ From equation (1), 

dq dt = – 

q ( R1 + R2) 

C 

...(2) 

Knowing that at t = 0, q = q 0 = CE, integrating 

equation (2), 

q= 

? 

∫ 

q= 

CE 

dq = – 

q ∫ 

t 

t= 

0 

dt 

( R + R 

R 2 

1 2) 

q t 

∴ log = – CE ( R1 + R2) 

C 

– t /( R R ) C 

or q = CEe 

1+ 2 

dq 

But I = – , dt 

E t /( R 

therefore, I = e 

1+ 

( R 1 + R2) 

Hence, thermal power across R 1 is 

R 2 

I 

C 

– R2 

) C 

...(3) 

P 1 = I 2 R 1 

2 

E R1 

– 2t 

/( R R C 

or P 1 = 

e 

1+ 2 ) 

( R1 

+ R2) 

C 

Similarly, thermal power across R 2 , P 2 = I 2 R 2 

or P 2 = 

( R 

2 

E R2 

–2t 

/( R R ) C 

e 

1+ 

2 

2 

1 + R2) 

Ans. 

Ans. 

5. Two plane mirrors, a source S of light, emitting 

mono-chromatic rays of wavelength λ and a screen 

are arranged as shown in figure. If angle θ is very 

small, calculate fringe width of interference pattern 

formed on screen by reflected rays. 

θ 

θ 

a 

S 

b 

Screen 

Sol. Since, interference is due to reflected rays, therefore, 

images S 1 and S 2 of the source S behave like two 

coherent sources as shown in figure 

d 

R 

M 

θ 

θ 

N 

S 

a 

D 

Distance of source S from each mirror = a cos θ 

∴ SS 1 = SS 2 = 2 × a cos θ 

Distance between S 1 and S 2 , d = SS 1 sin θ + SS 2 sin θ 

= 4a cos θ sin θ 

But θ is very small, therefore cos θ ≈ 1 and sin θ ≈ θ 

∴ 

Distance 

b 

d = 4aθ 

RS = SS 1 cos θ = 2a.cos 2 θ ≈ 2a 

∴ Distance of screen from two coherent sources S 1 

and S 2 is 

D = RO = RS + SO 

or D = (2a + b) 

Now the arrangement is similar to Young's double 

slit arrangement. 

∴ Fringe width, ω = 

Dλ = 

d 

O 

(2a 

+ b) 

λ 

4aθ 

Ans. 


PHYSICS FUNDAMENTAL FOR IIT-JEE 

Matter Waves, Photo-electric Effect 

KEY CONCEPTS & PROBLEM SOLVING STRATEGY 

Matter Waves : 

Planck's quantum theory : Wave-particle duality - 

Planck gave quantum theory while explaining the 

radiation spectrum of a black body. According to 

Planck's theory, energy is always exchanged in 

integral multiples of a quanta of light or photon. 

Each photon has an energy E that depends only 

on the frequency ν of electromagnetic radiation 

and is given by : 

E = hν .....(1) 

where h = 6.6 × 10 –34 joule-sec, is Planck's 

constant. In any interaction, the photon either 

gives up all of its energy or none of it. 

From Einstein's mass-energy equivalence 

principle, we have 

E = mc 2 .....(2) 

Using equations (1) and (2), we get ; 

mc 2 hν 

= hν or m = .....(3) 

2 

c 

where m represents the mass of a photon in 

motion. The velocity v of a photon is equal to 

that of light, i.e., v = c. 

According to theory of relativity, the rest mass m 0 

of a photon is given by : 

m 0 = 

m 1− 

v 

c 

2 

2 

hν 

Here, m = 

2 and v = c 

c 

Hence, m 0 = 0 ....(4) 

i.e., rest mass of photon is zero, i.e., energy of 

photon is totally kinetic. 

The momentum p of each photon is given by : 

p = mc = 

hν hν 

2 × c = = 

c c 

h 

c / ν 

= λ 

h 

......(5) 

The left hand side of the above equation involves 

the particle aspect of photons (momentum) while 

the right hand side involves the wave aspect 

(wavelength) and the Planck's constant is the 

bridge between the two sides. This shows that 

electromagnetic radiation exhibits a waveparticle 

duality. In certain circumstances, it 

behaves like a wave, while in other circumstances 

it behaves like a particle. 

The wave-particle is not the sole monopoly of 

e.m. waves. Even a material particle in motion 

according to de Broglie will have a wavelength. 

The de Broglie wavelength λ of the matter waves 

is also given by : 

h h h 

λ = = = mv p 2mK 

where K is the kinetic energy of the particle. 

If a particle of mass m kg and charge q coulomb 

is accelerated from rest through a potential 

difference of V volt. Then 

1 mv 2 = qV or mv = 2 mqV 

2 

Hence, λ = 

h 

2mqV 

12.34 

= Å 

V 

Photoelectric effect : 

When light of suitable frequency (electromagnetic 

radiation) is allowed to fall on a metal surface, 

electrons are emitted from the surface. These 

electrons are known as photoelectrons and the effect 

is known as photoelectric effect. Photoelectric 

effect, light energy is converted into electrical 

energy. 

Laws of photolectric effect : 

The kinetic energy of the emitted electron is 

independent of intensity of incident radiation. 

But the photoelectric current increases with the 

increase of intensity of incident radiation. 

The kinetic energy of the emitted electron 

depends on the frequency of the incident 

radiation. It increases with the increase of 

frequency of incident radiation. 

If the frequency of the incident radiation is less 

than a certain value, then photoelectric emission 

is not possible. This frequency is known as 

threshold frequency. This threshold frequency 

varies from emitter to emitter, i.e., depends on 

the material. 

There is no time lag between the arrival of light 

and the emission of photoelectrons, i.e., it is an 

instantaneous phenomenon. 


Failure of wave theory : 

Wave theory of light could not explain the laws of 

photoelectric effect. 

According to wave theory, the kinetic energy of 

the emitted electrons should increase with the 

increase of intensity of incident radiation. 

Kinetic energy of the emitted electron does not 

depend on the frequency of incident radiation 

according to wave theory. 

Wave theory failed to explain the existence of 

threshold frequency. 

According to wave theory there must be a time 

lag between the arrival of light and emission of 

photoelectrons. 

Einstein's theory of photoelectric effect : 

Einstein explained the laws of photoelectric effect 

on the basis of Planck's quantum theory of 

radiation. 

Einstein treated photoelectric effect as a collision 

between a photon and an atom in which photon is 

absorbed by the atom and an electron is emitted. 

According to law of conservation of energy, 

hν = hν 0 + 2 

1 mv 

2 

where hν is the energy of the incident photon; hv 0 

is the minimum energy required to detach the 

electron from the atom (work function or 

ionisation energy) and (1/2) mv 2 is the kinetic 

energy of the emitted electron. 

The above equation is known as Einstein's 

photoelectric equation. Kinetic energy of the 

emitted electron, 

= 2 

1 mv 2 = h(ν – ν 0 ) = hν – W 

Explanation of laws of photoelectric effect : 

(a) The KE of the emitted electron increases with the 

increase of frequency of incident radiation since 

W (work function) is constant for a given emitter. 

KE is directly proportional to (ν – ν 0 ) 

(b) Keeping the frequency of incident radiation 

constant if the intensity of incident light is 

increased, more photons collide with more atoms 

and more photoelectrons are emitted. The KE of 

the emitted electron remains constant since the 

same photon collides with the same atom (i.e., the 

nature of the collision does not change). With the 

increase in the intensity of incident light 

photoelectric current increases. 

(c) According to Einstein's equation, if the frequency 

of incident radiation is less than certain minimum 

value, the photoelectric emission is not possible. 

This frequency is known as threshold frequency. 

Hence, the frequency of incident radiation below 

which photoelectric emission is not possible is 

known as threshold frequency or cut-off 

frequency. It is given by : 

hν − (1/ 2)mv 

ν 0 = 

h 

On the other hand, if the wavelength of the 

incident radiation is more than certain critical 

value, then photoelectric emission is not possible. 

This wavelength is known as threshold 

wavelength of cut-off wavelength. It is given by : 

hc 

λ 0 = 

2 

[hν − (1/ 2)mv ] 

(d) Since Einstein treated photoelectric effect as a 

collision between a photon and an atom, he 

explained the instantaneous nature of 

photoelectric effect. 

Some other important points : 

Stopping potential : The negative potential 

applied to the collector in order to prevent the 

electron from reaching the collector (i.e., to 

reduce the photoelectric current to zero) is known 

as stopping potential. 

1 2 

eV 0 = mv max. = hν – W = h(ν – ν 0 ) 

2 

Millikan measured K.E. of emitted electrons or 

stopping potentials for different frequencies of 

incident radiation for a given emitter. He plotted a 

graph with the frequency on x-axis and stopping 

potential on y-axis. The graph so obtained was a 

straight line as shown in figure. 

V0(stopping potential) 

ν 0 

Frequency of incident light 

2 

Millikan measured the slope of the straight line 

(=h/e) and calculated the value of Planck's constant. 

I 

Full intensity 

75% intensity 

50% intensity 

25% intensity 

– V 0 

+ 

Potential difference 


The intercept of V 0 versus ν graph on frequency 

axis is equal to threshold frequency (ν 0 ). From 

this, the work function (hν 0 ) can be calculated. 

Graphs in photoelectric effect : 

(a) Photoelectric current versus potential difference 

graphs for varying intensity (keeping same metal 

plate and same frequency of incident light) : 

These graphs indicate that stopping potential is 

independent of the intensity and saturation current 

is directly proportional to the intensity of light. 

ν 2 >ν 1 

I 

ν 2 

ν 1 

– (V 0 ) 2 (V 0 ) 1 

+ 

Potential difference 

(b) Photoelectric current versus potential difference 

graphs for varying frequency (keeping same 

metal plate and same intensity of incident light) : 

These graphs indicate that the stopping potential 

is constant for a given frequency. The stopping 

potential increases with increase of frequency. 

The KE of the emitted electrons is proportional to 

the frequency of incident light. 

Stopping potential 

B 1 

B 2 

B 3 

ν 0 

A 1 A 2 A 3 Frequency 

(c) Stopping potential versus frequency graphs for 

different metals : These graphs indicate that the 

stops is same for all metal, since they are parallel 

straight lines. The slope is a universal constant 

(=h/e). Further, the threshold frequency varies 

with emitter since the intercepts on frequency axis 

are different for different metals. 

Solved Examples 

1. (i) A stopping potential of 0.82 V is required to stop 

the emission of photoelectrons from the surface 

of a metal by light of wavelength 4000 Å. For 

light of wavelength 3000 Å, the stopping 

potential is 1.85 V. Find the value of Planck's 

constant. 

(ii) At stopping potential, if the wavelength of the 

incident light is kept at 4000 Å but the intensity 

of light is increased two times, will photoelectric 

current be obtained? Give reasons for your 

answer. 

hc 

Sol. (i) We have = eV1 + W 

λ 1 

and 

⇒ 

or h = 

hc = eV2 + W 

λ 2 

⎛ 1 1 ⎞ 

hc ⎜ − 

⎟ = e(V 2 – V 1 ) 

⎝ λ2 

λ1 

⎠ 

e( 

V2 

⎛ 1 

e 

⎜ 

⎝ λ2 

− V1 

) 

= 

1 ⎞ 

− 

⎟ 

λ1 

⎠ 

−19 

1.6× 

10 (1.85 − 0.82) 

8⎛ 

1 1 

3× 

10 ⎜ − 

−7 

−7 

⎝ 3× 

10 4× 

10 

= 6.592 × 10 –34 Js 

(ii) No, because the stopping potential depends only 

on the wavelength of light and not on its intensity. 

2. A small plate of a metal (work function = 1.17 eV) is 

plated at a distance of 2m from a monochromatic 

light source of wavelength 4.8 × 10 –7 m and power 

1.0 watt. The light falls normally on the plate. Find 

the number of photons striking the metal plate per 

square metre per second. If a constant magnetic field 

of strength 10 –4 tesla is parallel to the metal surface, 

find the radius of the largest circular path followed by 

the emitted photoelectrons. 

−34 

8 

hc 6.6× 

10 × 3× 

10 

Sol. Energy of one photon = = λ −7 

4.8× 

10 

= 4.125 × 10 –19 J 

Number of photons emitted per second 

= 

1.0 

−19 

4.125× 

10 

= 2.424 × 10 18 

Number of photons striking the plate per square 

metre per second 

= 

18 

2.424× 

10 

2 

4× 

3.14× 

(2) 

= 4.82 × 10 16 

Maximum kinetic energy of photoelectrons emitted 

from the plate 

E max = λ 

hc – W 

= 4.125 × 10 –19 – 1.17 × 1.6 × 10 –19 

= 2.253 × 10 –19 J 

⎞ 

⎟ 

⎠ 


3. A monochromatic light source of frequency 

ν illuminates a metallic surface and ejects 

photoelectrons. The photoelectrons having maximum 

energy are just able to ionize the hydrogen atom in 

ground state. When the whole experiment is repeated 

with an incident radiation of frequency (5/6) ν, the 

photoelectrons so emitted are able to excite the 

hydrogen atom beam which then emits a radiation of 

wavelength 1215 Å. Find the work function of the 

metal and the frequency ν. 

Sol. In the first case, 

E max = Ionization energy = 13.6 eV 

= 21.76 × 10 –19 J 

So, hν = 21.76 × 10 –19 J ....(1) 

In the second case, 

E' max = λ 

hc 

−34 

8 

6.6× 

10 × 3× 

10 

= 

−10 

1215× 

10 

=16.3×10 –19 J 

5νh 

So, = 16.3 × 10 –19 + W ...(2) 

6 

Dividing Eq.(1) by Eq.(2) 

−19 

6 21.76× 

10 + W 

= 

5 

−19 

16.3× 

10 + W 

Solving, we get 

W = 11.0 × 10 – 19 J = 6.875 eV 

From Eq.(1) ν = 

−19 

−19 

21.76× 

10 + 11.0× 

10 

−34 

6.6× 

10 

= 5 × 10 15 Hz 

4. The radiation, emitted when an electron jumps from 

n = 3 to n = 2 orbit in a hydrogen atom, falls on a 

metal to produce photoelectrons. The electrons from 

the metal surface with maximum kinetic energy are 

made to move perpendicular to a magnetic field of 

1/320 T in a radius of 10 –3 m. Find (i) the kinetic 

energy of electrons, (ii) wavelength of radiation and 

(iii) the work function of metal. 

Sol. (i) Speed of an electron in the magnetic field, 

Ber 

v = 

m 

Kinetic energy of electrons 

E max = 2 

1 mv 2 = 

2 

2 

2 

B e r 

2m 

⎛ 1 ⎞ (1.6 × 10 ) × (10 

= ⎜ ⎟⎠ × 

⎝ 320 

−31 

2× 

9.1× 

10 

= 1.374 × 10 –19 J 

= 0.8588 eV 

2 

−19 

2 

−3 

) 

2 

(ii) Energy of the photon emitted from a hydrogen 

atom 

hc ⎡ 1 1 ⎤ 

hν = = λ 

⎢ − ⎥ 

⎣ 2 2 

3 2 ⎦ 

= 1.888 eV 

Wavelength of radiation, 

λ = 

6.62× 

10 

−34 

× 3× 

10 

1.888× 

1.6× 

10 

= 6.572 × 10 –7 m 

= 6572 Å 

−19 

(iii) Work function of metal W = hν – E max 

= 1.8888 – 0.8588 

= 1.03 eV 

5. X-rays are produced in an X-ray tube by electrons 

accelerated through a potential difference of 50.0 kV. 

An electron makes three collisions in the target 

before coming to rest and loses half of its kinetic 

energy in each of the first two collisions. Determine 

the wavelengths of the resulting photons. Neglect the 

recoil of the heavy target atoms. 

Sol. Initial kinetic energy of the electron = 50.0 keV 

Kinetic energy after first collision = 25.0 keV 

Energy of the photon produced in the first collision, 

E 1 = 50.0 – 25.0 = 25.0 keV 

Wavelength of this photon 

−34 

hc 6.6× 

10 × 3× 

10 

λ 1 = = 

E −19 

3 

1 1.6× 

10 × 25.0× 

10 

= 0.495 × 10 –10 m = 0.495 Å 

Kinetic energy of the electron after second collision 

= 12.5 eV 

Energy of the photon produced in the second 

collision, E 2 = 25.0 – 12.5 = 12.5 keV 

Wavelength of this photon 

−34 

hc 6.6 × 10 × 3× 

10 

λ 2 = = 

−19 

3 

E 2 1.6 × 10 × 12.5 × 10 

= 0.99 × 10 –10 m 

= 0.99 Å 

Kinetic energy of the electron after third collision = 0 

Energy of the photon produced in the third collision, 

E 3 = 12.5 – 0 = 12.5 keV 

This is same as E 2 . Therefore, wavelength of this 

photon, λ 3 = λ 2 = 0.99 Å. 

8 

8 

8 



PHYSICS FUNDAMENTAL FOR IIT-JEE 

Thermal Expansion, Thermodynamics 

KEY CONCEPTS & PROBLEM SOLVING STRATEGY 

Thermal Expansion : 

.(a) When the temperature of a substance is increased, 

it expands. The heat energy which is supplied to 

the substance is gained by the constituent 

particles of the substance as its kinetic energy. 

Because of this the collisions between the 

constituents particles are accompanied with 

greater force which increase the distance between 

the constituent particles. 

∆l = lα∆T ; ∆A = Aβ∆T ; ∆V = Vγ∆T 

or l' = l (1 + α∆T) ; A' = A(1 + β∆T) ; 

V' = V(1 + γ∆T) 

(b) Also ρ = ρ'(1 + γ∆T) where ρ' is the density at 

higher temperature clearly ρ' < ρ for substances 

which have positive value of γ 

* β = 2α and γ = 3α 

Water has negative value of γ for certain temperature 

range (0º to 4ºC). This means that for that 

temperature range the volume decreases with 

increase in temperature. In other words the density 

increases with increase in temperature. 

30 ml 

25 ml 

20 ml 

15 ml 

10 ml 

5 ml 

0 ml 

If a liquid is kept in a container and the temperature 

of the system is increased then the volume of the 

liquid as well as the container increases. The 

apparent change in volume of the liquid as shown by 

the scale is 

∆V app = V(γ – 3α) ∆T 

Where V is the volume of liquid at lower temperature 

∆V app is the apparent change in volume 

γ is the coefficient of cubical expansion of liquid 

α is the coefficients of linear expansion of the 

container. 

Loss or gain in time by a pendulum clock with 

change in temperature is ∆t = 2 

1 α(∆T) × t 

Where ∆t is the loss or gain in time in a time interval t 

∆T is change in temperature and d is coefficient of 

linear expansion. 

If a rod is heated or cooled but not allowed to expand 

or contract then the thermal stresses developed 

F = γα∆T. 

A 

If a scale is calibrated at a temperature T 1 but used at 

a temperature T 2 , then the observed reading will be 

wrong. In this case the actual reading is given by 

R = R 0 (1 + α∆T) 

Where R 0 is the observed reading, R is the actual 

reading. 

For difference between two rods to the same at all 

temperatures l 1α 1 = l 2 α 2 . 

Thermodynamics 

According to first law of thermodynamics 

q = ∆U + W 

For an isothermal process (for a gaseous system) 

(a) The pressure volume relationship is ρV = constt. 

(b) ∆U = 0 

(c) q = W 

(d) W = 2.303 nRT log 10 

V f 

p 

= 2.303 nRT log10 

i 

Vi 

pf 

(e) Graphs T 2 > T 1 

P 

P 

V 

T 2 

T 1 

V 

T 

T 

These lines are called isotherms (parameters at 

constant temperature) 

For an adiabatic process (for a gaseous system) 

(a) The pressure-volume relationship is PV γ = constt. 

(b) The pressure-volume-temperature relationship is 

PV = constt. 

T 

(c) From (a) and (b) TV γ–I = constt. 

(d) q = 0 

(e) W = –∆U 


(f) ∆U = nc v ∆T where c v = 

(g) W = 

(h) Graphs 

P 

p V − p V 

i 

i 

f 

γ −1 

P 

f 

= 

nR( 

T i − T 

γ −1 

R 

γ −1 

V 

T 

Please note that P-V graph line (isotherm) is 

steeper. 

For isochoric process 

(a) P ∝ T 

(b) W = 0 

(c) q = ∆U 

(d) ∆U = nC v ∆T 

R 

where C v = 

γ −1 

(e) Graphs 

P 

P 

V 

V 

T 

For isobaric process 

(a) V ∝ T 

(b) W = P∆V = P(V f – V i ) = nR(T f – T i ) 

(c) ∆U = nC v ∆T 

(d) q = nC p ∆T 

(e) Graphs 

P 

P 

V 

V 

T 

T 

For a cyclic process 

(a) ∆U = 0 ⇒ q = W 

(b) Work done is the area enclosed in p-V graph. 

For any process depicted by P-V diagram, area under 

the graph represents the word done. 

Kirchoff's law states that good absorbers are good 

emitters also. 

Problem solving Strategy : Thermal Expansion 

Step 1: Identify the relevant concepts: Decide 

whether the problem involves changes in length 

(linear thermal expansion) or in volume (volume 

thermal expansion) 

Step 2: Set up the problem using the following steps: 

Eq. ∆L = αL 0 ∆T for linear expansion and 

Eq. ∆V = βV 0 ∆T for volume expansion. 

Identify which quantities in Eq. ∆L = αL 0 ∆T or 

∆V = βV 0 ∆T are known and which are the 

unknown target variables. 

f 

) 

V 

T 

T 

Step 3: Execute the solution as follows: 

Solve for the target variables. Often you will be 

given two temperatures and asked to compute ∆T. 

Or you may be given an initial temperature T 0 and 

asked to find a final temperature corresponding to 

a given length or volume change. In this case, 

plan to find ∆T first; then the final temperature is 

T 0 + ∆T. 

Unit consistency is crucial, as always. L 0 and ∆L 

(or V 0 ∆V) must have the same units, and if you 

use a value or α or β in K –1 or (Cº) –1 , then ∆T 

must be in kelvins or Celsius degrees (Cº). But 

you can use K and Cº interchangeably. 

Step 4: Evaluate your answer: Check whether your 

results make sense. Remember that the sizes of holes 

in a material expand with temperature just as the 

same way as any other linear dimension, and the 

volume of a hole (such as the volume of a container) 

expands the same way as the corresponding solid 

shape. 

Problem solving strategy : Thermodynamics I st Law 

Step 1: Identify the relevant concepts : The first law 

of thermodynamics is the statement of the law of 

conservation of energy in its most general form. You 

can apply it to any situation in which you are 

concerned with changes in the internal energy of a 

system, with heat flow into or out of a system, and/or 

with work done by or on a system. 

Step 2: Set up the problem using the following steps 

Carefully define what the thermodynamics system is. 

The first law of thermodynamics focuses on 

systems that go through thermodynamic 

processes. Some problems involve processes 

with more than one step. so make sure that you 

identify the initial and final state for each step. 

Identify the known quantities and the target 

variables. 

Check whether you have enough equations. The 

first law, ∆U = Q – W, can be applied just once to 

each step in a thermodynamic process, so you will 

often need additional equations. These often 

include Eq. 

V2 

∫ 

W = pdV for the work done in a 

V1 

volume change and the equation of state of the 

material that makes up the thermodynamic system 

(for an ideal gas, pV = nRT). 

Step 3: Execute the solution as follows : 

You shouldn't be surprised to be told that 

consistent units are essential. If p is a Pa and V in 

m 3 , then W is in joules. Otherwise, you may want 

to convert the pressure and volume units into 

units of Pa and m 3 . If a heat capacity is given in 

terms of calories, usually the simplest procedure 

is to convert it to joules. Be especially careful 

with moles. When you use n = m tot /M to convert 


etween total mass and number of moles, 

remember that if m tot is in kilograms, M must be 

in kilograms per mole. The usual units for M are 

grams per mole; be careful ! 

The internal energy change ∆U in any 

thermodynamic process or series of processes in 

independent of the path, whether the substance is 

an ideal gas or not. This point is of the utmost 

importance in the problems in this topic. 

Sometimes you will be given enough information 

about one path between the given initial and final 

states to calculate ∆U for that path. Since ∆U is 

the same for every possible path between the 

same two states, you can then relate the various 

energy quantities for other paths. 

When a process consists of several distinct steps, 

it often helps to make a chart showing Q, W, and 

∆U for each step. Put these quantities for each 

step on a different line, and arrange them so the 

Q's, W's, and ∆U's form columns. Then you can 

apply the first law to each line ; in addition, you 

can add each column and apply the first law to the 

sums. Do you see why ? 

Using above steps, solve for the target variables. 

Step 4: Evaluate your answer : Check your results for 

reasonableness. In particular, make sure that each of 

your answers has the correct algebraic sign. 

Remember that a positive Q means that heat flows 

into the system, and that a negative Q means that heat 

flows into the system, and that a negative Q means 

that heat flows out of the system. A positive W 

means that work is done by the system on its 

environment, while a negative W means that work is 

done on the system by its environment. 

Solved Examples 

1. A metallic bob weighs 50 g in air. It it is immersed 

in a liquid at a temperature of 25ºC, it weighs 45 g. 

When the temperature of the liquid is raised to 100ºC, 

it weighs 45.1 g. Calculate the coefficient of cubical 

expansion of the liquid given that the coefficient of 

linear expansion of the metal is 2 × 10 –6 (ºC) –1 . 

Sol. Loss in weight in liquid at 25ºC = (50 – 45) = 5 gm 

Weight of liquid displaced at 25ºC = V 25 ρ 25 g 

∴ 5 = V 25 ρ 25 g ...(1) 

Similarly, V 100 ρ 100 g = 50 – 45.1 = 4.9 gm ...(2) 

From eq.(1) & (2) we get, 

5 V = 25 ρ25 

. 

4.9 V100 

ρ 100 

Now, V 100 = V 25 (1 + γ metal × 75)= V 25 (1 + 3α metal × 75) 

= V 25 (1 + 3 × 12 × 10 –6 × 75) 

or V 100 = V 25 (1 + 0.0027) = V 25 × 1.0027 

Also, ρ 25 = ρ 100 (1 + γ × 75) 

where, γ = Required coefficient of expansion of the liquid 

5 V ρ 

= 

100(1 

+ 75γ 

× 

4.9 V25 

× 1.0027 

ρ100 

or γ = 3.1 × 10 –4 (ºC) –1 

25 ) 

1+ 

75γ 

= 

1.0027 

2. A one litre flask contains some mercury. It is found 

that at different temperature the volume of air inside 

the flask remains the same. What is the volume of 

mercury in flask ? Given that the coefficient of linear 

expansion of glass = 9 × 10 –6 (ºC) –1 and coefficient of 

volume expansion of mercury = 1.8 × 10 –4 (ºC –1 ). 

Sol. Let V = Volume of the vessel 

V' = Volume of mercury 

For unoccupied volume to remain constant increase 

in volume of mercury should be equal to increase in 

volume of vessel. 

V × γ g 

∴ V' γ m ∆T = Vγ g ∆T or V' = 

γ 

1000× 

27× 

10 

∴ V' = 

−4 

1.8× 

10 

−6 

m 

= 150 cm 3 

3. A clock with a metallic pendulum gains 6 seconds 

each day when the temperature is 20ºC and loses 12 

seconds each day when the temperature is 40ºC. Find 

the coefficient of linear expansion of the metal. 

Sol. Time taken for one oscillation of the pendulum is 

L 

T = 2 π or T 2 = 4π 2 L 

× .....(1) 

g 

g 

Partially differentiating, we get 

2T∆t = 4π 2 ∆L 

× 

.....(2) 

g 

Dividing (2) by (1), we get 

∆T ∆ L α L∆t 

1 

= = = α∆t 

T 2L 

2L 

2 

where ∆t is the change in temperature. Now, 

One day = 24 hours = 86400 sec 

Let t be the temperature at which the clock keeps 

correct time. 

At 20ºC, the gain in time is 

6 = 2 

1 α × (t – 20) × 86400 ....(3) 

At 40ºC, the loss in time is 

12 = 2 

1 α× (40 – t) × 86400 ...(4) 

Dividing (4) by (3), we have 

12 40 − t = 

6 t − 20 

80 

which gives t = ºC. 

3 

Using the value in equation(3), we have 

1 ⎛ 80 ⎞ 

6 = × α × ⎜ − 20 ⎟⎠ × 86400 

2 ⎝ 3 

which gives α = 2.1 × 10 –5 perºC 


4. A piston can freely move inside a horizontal cylinder 

closed from both ends. Initially, the piston separates 

the inside space of the cylinder into two equal parts 

each of volume V 0 , in which an ideal gas is contained 

under the same pressure p 0 and at the same 

temperature. What work has to be performed in order 

to increase isothermally the volume of one part of gas 

η times compared to that of the other by slowly 

moving the piston ? 

Sol. Let volume of chambers changes by ∆V. According 

to the problem, the final volume of left chamber is η 

times final volume of right chamber. 

∴ V 0 + ∆V = η(V 0 – ∆V) 

⎛ η −1⎞ 

or ∆V = ⎜ ⎟ V 0 

⎝ η + 1⎠ 

Sol. Let A 1 = Cross section of upper piston 

A 2 = Cross section of lower piston 

T = Tension in the string 

P = Gas pressure 

m 1 = Mass of upper piston 

m 2 = Mass of lower piston 

Now, consider FBD of upper piston 

P 0 

P 0 

P 0 ,v 0 ,T 0 

P 0 ,v 0 ,T 0 

P 0 A 1 

As piston is moved slowly therefore, change in 

kinetic energy is zero. By work-energy theorem, we 

can write 

ext 

W gas in right chamber + W gas in left chamber + W Agent = ∆KE 

ext 

W Agent = (W gas(R) + W gas(L) ) 

We know that in isothermal process, work done is 

given by 

⎛V 

f ⎞ 

W = nRT ln ⎜ ⎟ 

⎝ Vi 

⎠ 

∴ Work done by gas in left chamber (W L ) 

⎛V 

⎞ 

= P 0 V 0 ln 

⎜ 

0 + ∆V 

⎛ 2η 

⎞ 

⎟ = P 0 V 0 ln ⎜ ⎟ 

⎝ V0 

⎠ ⎝ η + 1⎠ 

Similarly, work done by gas in right chamber (W R ) 

⎛V 

⎞ 

= P 0 V 0 ln 

⎜ 

0 − ∆V 

⎛ 2η 

⎞ 

⎟ = P 0 V 0 ln ⎜ ⎟ 

⎝ V0 

⎠ ⎝ η + 1⎠ 

ext 

⎛ 2η 

⎞ ⎛ 2η 

⎞ 

W Agent = –P 0 V 0 ln ⎜ ⎟ – P 0 V 0 ln ⎜ ⎟ 

⎝ η + 1⎠ 

⎝ η + 1⎠ 

⎛ η + 1⎞ 

= P 0 V 0 ln ⎜ ⎟ 

⎝ 4η 

⎠ 

5. A smooth vertical tube having two different sections 

is open from both ends equipped with two pistons of 

different areas figure. Each piston slides within a 

respective tube section. One mole of ideal gas is 

enclosed between the pistons tied with a nonstretchable 

thread. The cross-sectional area of the 

upper piston is ∆S greater than that of the lower one. 

The combined mass of the two pistons is equal to m. 

The outside air pressure is P 0 . By how many kelvins 

must the gas between the pistons be heated to shift 

the pistons through l. 

2 

PA 1 

m 1 g 

From equilibrium consideration of upper piston 

we get, P 0 A 1 + T + m 1 g = PA 1 

Similarly, consider FBD of lower piston 

T 

PA 2 

P 0 A 2 

m 2 g 

∴ P 0 A 2 + T = m 2 g + PA 2 

Eliminating T, we get 

( m1 + m2) 

g 

P = P 0 + 

A1 

− A2 

According to problem 

m = m 1 + m 2 

and ∆S = A 1 – A 2 

mg 

∴ P = P 0 + 

∆S 

Now, PV = RT 

or P∆V = R∆T or ∆T = 

But ∆V = (A 1 – A 2 )l = ∆S. l 

⎛ mg ⎞ 

∴ ∆T = ⎜ P 0 + ⎟ ∆S.l 

⎝ ∆ S ⎠ 

l 

l 

l 

P∆V 

R 


KEY CONCEPT 

Organic 

Chemistry 

Fundamentals 

CARBONYL 

COMPOUNDS 

Reduction of Aldehydes and Ketones by Hydride 

Transfer : 

R δ+ δ– 

H 3 B – H + C = O 

R´ 

R 

R 

– H – OH 

H – C – O H – C – O – H 

Hydride transfer Alkoxide ion Alcohol 

R 

R´ 

These steps are repeated until all hydrogen atoms 

attached to boron have been transferred. 

Sodium borohydride is a less powerful reducing 

agent than lithium aluminum hydride. Lithium 

aluminum hydride reduces acids, aldehydes, and 

ketones but sodium borohydride reduces only 

aldehydes and ketones : 

O 

C 

Reduced by LiAlH 4 

O 

< C < C < 

O– R OR´ R R´ R 

O 

Ease of reduction 

R´ 

Reduced by NaBH 4 

Lithium aluminum hydride reacts violently with 

water, and therefore reductions with lithium 

aluminum hydride must be carried out in anhydrous 

solutions, usually in anhydrous ether. (Ethyl acetate 

is added cautiously after the reaction is over to 

decompose excess LiAlH 4 ; then water is added to 

decompose the aluminum complex.) Sodium 

borohydride reductions, by contrast, can be carried 

out in water or alcohol solutions. 

The Addition of Ylides : The Wittig reaction : 

Aldehydes and ketones react with phosphorus ylides 

to yield alkenes and triphenylphosphine oxide. (An 

ylide is a neutral molecule having a negative carbon 

adjacent to a positive heteroatom.) Phosphorus ylides 

are also called phosphoranes : 

O 

C 

H 

R 

R 

+ .. R´´ 

C = O + (C 6 H 5 ) 3 P – C 

R´´´ 

Aldehyde or 

ketone 

Phosphorus ylide 

or phosphorane 

R 

R´ 

C = C 

Alkene 

[(E) and(Z) isomers] 

R´´ 

+ O =P(C 6 H 5 ) 3 

R´´´ 

Triphenyl phosphine 

oxide 

This reaction, known as the Wittig reaction, has 

proved to be a valuable method for synthesizing 

alkenes. The Wittig reaction is applicable to a wide 

variety of compounds, and although a mixture of (E) 

and (Z) isomers may result, the Wittig reaction offers 

a great advantage over most other alkene syntheses in 

that no ambiguity exists as to the location of the 

double bond in the product. (This is in contrast to E1 

eliminations, which may yield multiple alkene 

products by rearrangement to more stable carbocation 

intermediates, and both E1 and E2 elimination 

reactions, which may produce multiple products 

when different β hydrogens are available for 

removal.) 

Phosphorus ylides are easily prepared from 

triphenylphosphine and primary or secondary alkyl 

halides. Their preparation involves two reactions : 

General Reaction 

Reaction 1 

R´´ 

R´´ 

+ 

(C 6 H 5 ) 3 P : + CH – X → (C 6 H 5 ) 3 P – CH X – 

R´´´ 

R´´´ 

Triphenylphosphine An alkyltriphenylphosphonium 

halide 

Reaction 2 

R´´ 

R´´ 

+ 

(C 6 H 5 ) 3 P – C – H : B – + 

⎯→ (C 6 H 5 ) 3 P – C : – + H:B 

R´´´ 

R´´´ 

A phosphorus ylide 

Specific Example 

Reaction 1 

+ 

(C 6 H 5 ) 3 P : + CH 3 Br C6H6 ⎯→ (C 6 H 5 ) 3 P – CH 3 Br – 

Methyltriphenylphosphonium 

bromide (89%) 


Reaction 2 

+ 

(C 6 H 5 ) 3 P – CH 3 + C 6 H 5 Li ⎯→ 

Br – 

+ 

(C 6 H 5 ) 3 P – CH 2 : – + C 6 H 6 + LiBr 

The first reaction is a nucleophilic substitution 

reaction. Triphenylphosphine is an excellent 

nucleophile and a weak base. It reacts readily with 1º 

and 2º alkyl halide by an S N 2 mechanism to displace 

a halide ion from the alkyl halide to give an 

alkyltriphenylphosphonium salt. The second reaction 

is an acid-base reaction. A strong base (usually an 

alkyllithium or phenyllithium) removes a proton from 

the carbon that is attached to phosphorus to give the 

ylide. 

Phosphorus ylides can be represented as a hybrid of 

the two resonance structures shown here. Quantum 

mechanical calculations indicate that the contribution 

made by the first structure is relatively unimportant. 

R´´ 

+ 

–R´´ 

(C 6 H 5 ) 3 P = C 

(C 6 H 5 ) 3 P – C : 

R´´´ 

R´´´ 

The mechanism of the Wittig reaction has been the 

subject of considerable study. An early mechanistic 

proposal suggested that the ylide, acting as a 

carbanion, attacks the carbonyl carbon of the 

aldehyde or ketone to form an unstable intermediate 

with separated charges called a betaine. In the next 

step, the betaine is envisioned as becoming an 

unstable four-membered cyclic system called an 

oxaphosphetane, which then spontaneously loses 

triphenylphosphine oxide to become an alkene. 

However, studies by E. Vedejs and others suggest 

that the betaine is not an intermediate and that the 

oxaphosphetane is formed directly by a cycloaddition 

reaction. The driving force for the Wittig reaction is 

the formation of the very strong (∆Hº = 540 kJ mol –1 ) 

phosphorus –oxygen bond in triphenylphosphine 

oxide. 

R ´ R ´´ 

R–C + – :C–R ´ 

:O: 

Aldehyde 

or ketone 

P(C 6 H 5 ) 3 

+ 

Ylide 

R ´ R ´´ 

R – C – C – R´´´ 

– :O: .. P(C 6 H 5 ) 3 

+ 

Betaine 

(may not be formed) 

R´ 

R 

Alkene 

(+diastereomer) 

R ´ R ´´ 

R – C – C – R´´´ 

:O.. 

– P(C 6 H 5 ) 3 

Oxaphosphetane 

R´´ 

C = C + O = P(C 6 H 5 )3 

R´´´ 

Triphenylphosphine 

oxide 

Specific Example : 

Methylenecyclohexane 

(86%) 

– + 

O + :CH 2 – P(C 6 H 5 ) 3 CH 2 

CH 2 + O=P(C 6 H 5 ) 3 

O P(C 6 H 5 ) 

– 

3 

+ 

CH 2 

O –P(C 6 H 5 ) 3 

Michael Additions : 

Conjugate additions of enolate anions to 

α-β-unsaturated carbonyl compound are known 

generally as Michael additions. An example is the 

addition of cyclohexanone to C 6 H 5 CH=CHCOC 6 H 5 : 

C 6 H 5 

O O 

O O 

– 

CH 

OH – 

C 6H 5CH=CH–CC 6H 5 

CH δ– 

C—O δ– 

O – +H 3 O + 

C 6 H 5 

O 

C 6 H 5 

CH H 

C 

H 

C = O 

C 6 H 5 

The sequence that follows illustrates how a conjugate 

aldol addition (Michael addition) followed by a 

simple aldol condensation may be used to build one 

ring onto another. This procedure is known as the 

Robinson anulation (ring-forming) reaction (after the 

English chemist, Sir Robert Robinon, who won the 

Nobel Prize in chemistry in 1947 for his research on 

naturally occurring compounds) : 

O 

O 

CH 3 

+ CH 2 = CHCCH 3 

O 

2-Methylcyclohexane-1, 

3-dione 

Methyl vinyl 

ketone 

OH – 

CH 3OH 

(conjugate 

addition) 

O 

aldol 

condensation 

CH 3 

CH 2 

CH 2 

O 

H 3 C 

O 

CH 3 

(65%) 

C 

base 

(–H 2O) 

O 

O 


KEY CONCEPT 

Inorganic 

Chemistry 

Fundamentals 

CO-ORDINATION COMPOUND 

& 

METALLURGY 

Tetragonal distortion of octahedral complexes (Jahn- 

Teller distortion) : 

The shape of transition metal complexes are affected 

by whether the d orbitals are symmetrically or 

asymmetrically filled. 

Repulsion by six ligands in an octahedral complex 

splits the d orbitals on the central metal into t 2g and e g 

levels. It follows that there is a corresponding 

repulsion between the d electrons and the ligands. If 

the d electrons are symmetrically arranged, they will 

repel all six ligands equally. Thus the structure will 

be a completely regular octahedron. The symmetrical 

arrangements of d electrons are shown in Table. 

Symmetrical electronic arrangements : 

Electronic 

configuration 

d 5 

d 6 

d 8 

d 10 

t 2g 

All other arrangements have an asymmetrical 

arrangement of d electrons. If the d electrons are 

asymmetrically arranged, they will repel some 

ligands in the complex more than others. Thus the 

structure is distorted because some ligands are 

prevented from approaching the metal. 

as closely as others. The e g orbitals point directly at 

the ligands. Thus asymmetric filling of the e g orbitals 

in some ligands being repelled more than others. This 

causes a significant distortion of the octahedral 

shape. In contrast the t 2g orbitals do not point directly 

at the ligands, but point in between the ligand 

directions. Thus asymmetric filling of the t 2g orbitals 

has only a very small effect on the stereochemistry. 

Distortion caused by asymmetric filling of the t 2g 

orbitals is usually too small to measure. The 

electronic arrangements which will produce a large 

distortion are shown in Table. 

The two e g orbitals d x 

2 − y 

2 and d 

z 2 are normally 

degenerate. However, if they are asymmetrically 

filled then this degeneracy is destroyed, and the two 

e g 

orbitals are no longer equal in energy. If the d 2 

z 

orbital contains one. 

Asymmetrical electronic arrangements : 

Electronic 

configuration 

d 4 

d 7 

d 9 

t 2g 

more electron than the d x 

2 − y 

2 orbital then the ligands 

approaching along +z and –z will encounter greater 

repulsion than the other four ligands. The repulsion 

and distortion result in elongation of the octahedron 

along the z axis. This is called tetragonal distortion. 

Strictly it should be called tetragonal elongation. This 

form of distortion is commonly obsered. 

If the d orbital contains the extra electron, then 

2 2 

x − y 

elongation will occur along the x and y axes. This 

means that the ligands approach more closely along 

the z-axis. Thus there will be four long bonds and 

two short bonds. This is equivalent to compressing 

the octahedron along the z axis, and is called 

tetragonal compression, and it is not possible to 

predict which will occur. 

For example, the crystal structure of CrF 2 is a 

distorted rutile (TiO 2 ) structure. Cr 2+ is octahedrally 

surrounded by six F – , and there are four Cr–F bonds 

of length 1.98 – 2.01 Å, and two longer bonds of 

length 2.43 Å. The octahedron is said to be 

tetragonally distorted. The electronic arrangement in 

Cr 2+ is d 4 . F – is a weak field ligand, and so the t 2g 

level contains three electrons and the e g level contains 

one electron. The d x 

2 − y 

2 orbital has four lobes whilst 

the d 2 orbital has only two lobes pointing at the 

z 

ligands. To minimize repulsion with the ligands, the 

single e g electron will occupy the d 2 orbital. This is 

z 

equivalent to splitting the degeneracy of the e g level 

so that d 2 is of lower energy, i.e. more stable, and 

z 

d − 

is of higher energy, i.e. less stable. Thus the 

2 y 

2 x 

e g 


two ligands approaching along the +z and –z 

directions are subjected to greater repulsion than the 

four ligands along +x, –x, +y and –y. This causes 

tetragonal distortion with four short bonds and two 

long bonds. In the same way MnF 3 contains Mn 3+ 

with a d 4 configuration, and forms a tetragonally 

distorted octahedral structure. 

Many Cu(+II) salts and complexes also show 

tetragonally distorted octahedral structures. Cu 2+ has 

a d 9 configuration : 

t 2g 

e g 

To minimize repulsion with the ligands, two 

electrons occupy the d 2 orbital and one electron 

z 

occupies the d x 

2 − y 

2 orbital. Thus the two ligands 

along –z and –z are repelled more strongly than are 

the other four ligands. 

The examples above show that whenever the d 2 and 

z 

d − 

orbitals are unequally occupied, distortion 

2 y 

2 x 

occurs. This is know as Jahn–Teller distortion. 

Leaching : 

It involves the treatment of the ore with a suitable 

reagents as to make it soluble while impurities 

remain insoluble. The ore is recovered from the 

solution by suitable chemical method. For example, 

bauxite ore contains ferric oxide, titanium oxide and 

silica as impurities. When the powdered ore is 

digested with an aqueous solution of sodium 

hydroxide at about 150ºC under pressure, the alumina 

(Al 2 O 3 ) dissolves forming soluble sodium metaaluminate 

while ferric oxide (Fe 2 O 3 ), TiO 2 and silica 

remain as insoluble part. 

Al 2 O 3 + 2NaOH → 2NaAlO 2 + H 2 O 

Pure alumina is recovered from the filtrate 

NaAlO 2 + 2H 2 O ⎯→ Al(OH) 3 + NaOH 

2Al(OH) 3 

Ignited ⎯⎯ 

(autoclave) 

⎯ → Al 2 O 3 + 3H 2 O 

Gold and silver are also extracted from their native 

ores by Leaching (Mac-Arthur Forrest cyanide 

process). Both silver and gold particles dissolve in 

dilute solution of sodium cyanide in presence of 

oxygen of the air forming complex cyanides. 

4Ag + 8NaCN + 2H 2 O + O 2 

⎯→ 4NaAg(CN) 2 + 4NaOH 

Sod. argentocyanide 

4Au + 8NaCN + 2H 2 O + O 2 

⎯→ 4NaAu(CN) 2 + 4NaOH 

Sod. aurocyanide 

Ag or Au is recovered from the solution by the 

addition of electropositive metal like zinc. 

2NaAg(CN) 2 + Zn ⎯→ Na 2 Zn(CN) 4 + 2Ag ↓ 

2NaAu(CN) 2 + Zn ⎯→ Na 2 Zn(CN) 4 + 2Au ↓ 

Soluble complex 

Special Methods : 

Mond's process : Nickel is purified by this method. 

Impure nickel is treated with carbon monoxide at 60– 

80º C when volatile compound, nickel carbonyl, is 

formed. Nickel carbonyl decomposes at 180ºC to 

form pure nickel and carbon monoxide which can 

again be used. 

Impure nickel + CO 60–80ºC NI(CO) 4 

Gaseous compound 

180ºC 

Ni + 4CO 

Zone refining or Fractional crystallisation : 

Elements such as Si, Ge, Ga, etc., which are used as 

semiconductors are refined by this method. Highly 

pure metals are obtained. The method is based on the 

difference in solubility of impurities in molten and 

solid state of the metal. A movable heater is fitted 

around a rod of the impure metal. The heater is 

slowly moved across the rod. The metal melts at the 

point of heating and as the heater moves on from one 

end of the rod to the other end, the pure metal 

crystallises while the impurities pass on the adjacent 

melted zone. 

Molten zone 

containing 

impurity 

Pure metal 

Moving circular 

heater 

Impure 

zone 

Different metallurgical processes can be broadly 

divided into three main types. 

Pyrometallurgy : Extraction is done using heat 

energy. The metals like Cu, Fe, Zn, Pb, Sn, Ni, Cr, 

Hg, etc., which are found in nature in the form of 

oxides, carbonates, sulphides are extracted by this 

process. 

Hydrometallurgy : Extraction of metals involving 

aqueous solution is known as hydrometallurgy. 

Silver, gold, etc., are extracted by this process. 

Electrometallurgy : Extraction of highly reactive 

metals such as Na, K, Ca, Mg, Al, etc., by 

carrying electrolysis of one of the suitable 

compound in fused or molten state. 


UNDERSTANDING 

Physical Chemistry 

1. The critical temperature and pressure for NO are 177 

K and 6.485 MPa, respectively, and for CCl 4 these 

are 550 K and 4.56 MPa, respectively. Which gas (i) 

has smaller value for the van der Walls constant b; 

(ii) has smaller value of constant a; (iii) has larger 

critical volume; and (iv) is most nearly ideal in 

behaviour at 300 K and 1.013 MPa. 

Sol. We have T c (NO) = 177 K T c (CCl 4 ) = 550 K 

p c (NO) = 6.485 MPa p c (CCl 4 ) = 4.56 MPa 

(i) Since 

Thus, 

p 

T 

c 

c 

= 

2 

a / 27b 

8a / 27Rb 

(177 K)(8.314 MPa cm K 

b(NO) = 

(8)(6.485 MPa) 

= 28.36 cm 3 mol –1 

and 

= 

R therefore, b = 

8b 

3 –1 −1 

mol 

) 

3 −1 

−1 

T R 

550 K)(8.314 MPa cm K mol ) 

b(CCl 4 ) = 

(8)(4.56MPa) 

= 125.35 cm 3 mol –1 

Hence b(NO) < b(CCl 4 ) 

(ii) Since a = 27p c b 2 

therefore 

a(NO) = (27) (6.485 MPa) (28.36 cm 3 mol –1 ) 2 

= 140827 MPa cm 6 mol –2 ≡ 140.827 kPa dm 6 mol –2 

a(CCl 4 ) = (27) (4.56 MPa) (125.35 cm 3 mol –1 ) 2 

= 1934538 MPa cm 6 mol –2 ≡ 1934.538 KPa dm 6 mol –2 

Hence a(NO) < a(CCl 4 ) 

(iii) Since V c = 3b 

therefore, V c (NO) = 3 × (28.36 cm 3 mol –1 ) 

= 85.08 cm 3 mol –1 

V c (CCl 4 ) = 3 × (125.35 cm 3 mol –1 ) 

= 376.05 cm 3 mol –1 

Hence V c (NO) < V c (CCl 4 ) 

(iv) NO is more ideal in behaviour at 300 K and 

1.013 MPa, because its critical temperature is less 

than 300 K, whereas for CCl 4 the corresponding 

critical temperature is greater than 300 K. 

2. Potassium alum is KA1(SO 4 ) 2 .12H 2 O. As a strong 

electrolyte, it is considered to be 100% dissociated 

into K + , Al 3+ , and SO 2– 4 . The solution is acidic 

because of the hydrolysis of Al 3+ , but not so acidic as 

might be expected, because the SO 2– 4 can sponge up 

some of the H 3 O + by forming HSO – 4 . Given a 

solution made by dissolving 11.4 g of 

KA1(SO 4 ) 2 .12H 2 O in enough water to make 0.10 dm 3 

of solution, calculate its [H 3 O + ] : 

(a) Considering the hydrolysis 

c 

8p 

c 

Al 3+ + 2H 2 O Al(OH) 2+ + H 3 O + 

with K h = 1.4 × 10 –5 M 

(b) Allowing also for the equilibrium 

HSO – 4 + H 2 O H 3 O + 2– 

+ SO 4 

with K 2 = 1.26 × 10 –2 M 

11.4 g 

Sol. (a) Amount of alum = 

−1 

474.38 g mol 

= 0.024 mol 

0.024 mol 

Molarity of the prepared solution = 

3 

0.1dm 

= 0.24 M 

Hydrolysis of Al 3+ is 

Al 3+ + 2H 2 O Al(OH) 2+ + H 3 O + 

2+ 

+ 

[Al(OH) ][H3O 

] 

K h = 

3+ 

[Al ] 

If x is the concentration of Al 3+ that has hydrolyzed, 

we have 

(x)(x) 

K h = 

0.24M − x 

= 1.4 × 10 –5 M 

Solving for x, we get 

[H 3 O + ] = x = 1.82 × 10 –3 M 

(b) We will have to consider the following equilibria. 

Al 3+ + 2H 2 O Al(OH) 2+ + H 3 O + 

H 3 O + 2– 

+ SO 4 HSO – 4 + H 2 O 

Let z be the concentration of SO 2– 4 that combines 

with H 3 O + and y be the net concentration of H 3 O + 

that is present in the solution. Since the concentration 

z of SO 2– 

4 combines with the concentration z of 

H 3 O + , it is obvious that the net concentration of H 3 O + 

produced in the hydrolysis reaction of Al 3+ is (y + z). 

Thus, the concentration (y + z) of Al 3+ out of 0.24 M 

hydrolyzes in the solution. With these, the 

concentrations of various species in the solution are 

3+ 

2 

Al + 2H 2 O Al(OH) 

+ 3 O 0.24 M−y−z 

y+ 

z 

y 

H 3 O 2− 

+ SO 4 HSO − 

4 + H 2 O 

y 0.48 M−z 

z 

(y + z)(y) 

Thus, K h = 

= 1.4 × 10 –5 M ...(i) 

(0.24M − y − z) 

z 

1 

K 2 = 

= 

y(0.48M − z) 1.26× 

10 

From Eq. (ii), we get 

(0.48M)y 

z = 

2 

(1.26× 10 

− M) + y 

Substituting this in Eq. (i), we get 

−2 

M 

...(ii) 


⎛ 

⎞ 

⎜ 

(0.48M)y 

Step 6. 

⎟ 

y + 

y 

−2 

m 

⎝ (1.26× 

10 M) + y 

2 0 .6× 

84 

⎠ 

= 1.4 × 10 –5 

× 1000 = 0.6 or m 2 = = 0.0504 

84 

1000 

⎛ 

⎞ 

⎜ 

(0.48M)y 

⎟ 

0.24 − y − 

0 .0504× 

1000 

−2 

⎝ (1.26× 

10 M) + y 

∴ Strength of NaHCO 

⎠ 

3 solution = 

10 

= 4.24 g L –1 

Making an assumption that y

5. The freezing point of an aqueous solution of KCN 

containing 0.189 mol kg –1 was – 0.704 ºC. On adding 

0.095 mol of Hg(CN) 2 , the freezing point of the 

solution became –0.530ºC. Assuming that the 

complex is formed according to the equation 

Hg(CN) 2 + x CN – x – 

→ Hg (CN) x + 2 

Find the formula of the complex. 

Sol. Molality of the solution containing only KCN is 

(–∆Tf 

) (0.704 K) 

m = = 

= 0.379 mol kg –1 

K 

–1 

f (1.86 K kg mol ) 

This is just double of the given molality 

( = 0.189 mol kg –1 ) of KCN, indicating complete 

dissociation of KCN. Molality of the solution after 

the formation of the complex 

(–∆Tf 

) (0.530 K) 

m = = 

= 0.285 mol kg –1 

–1 

K f (1.86 K kg mol ) 

If it be assumed that the whole of Hg(CN) 2 is 

converted into complex, the amounts of various 

species in 1 kg of solvent after the formation of the 

complex will be 

n(K + ) = 0.189 mol, 

n(CN – ) = (0.189 – x) mol 

x– 

n(Hg(CN) x+ 2 ) = 0.095 mol 

Total amount of species in 1 kg solvent becomes 

n total = [0.189 + (0.189 – x) + 0.095] mol 

= (0.473 – x) mol Equating this to 0.285 mol, 

we get 

(0.473 – x) mol = 0.285 mol 

i.e. x = (0.473 – 0.285) = 0.188 

Number of CN – 0.188mol 

units combined = = 2 

0.095mol 

Thus, the formula of the complex is 

• Parsec is the unit of 

2– 

Hg (CN) 4 . 

MEMORABLE POINTS 

• Estimated radius of universe is 

Distance 

10 25 m 

• Estimated age of Sun is 

10 18 s 

• 18/5 km h –1 equal to 1 ms –1 

• 1 femtometre (1 fm) is equal to 10 –15 m 

• Dot product of force and velocity is Power 

• Moment of momentum is equal to 

Angular momentum 

• Rocket propulision is based on the principle of 

Conservation of linear momentum 

• The largest of astronomical unit, light year and 

parsec is Parsec 

TRUE OR FALSE 

1. The magnitude of charge on one gram of 

electrons is 1.60 × 10 –19 coulomb. 

2. Chromyl chloride test of Cl – radical is not given 

by HgCl 2 . 

3. The energy levels in a hydrogen atom can be 

compared with the steps of a ladder placed at 

equal distance. 

4. In S N 1 mechanism, the leaving group in the 

molecule, leaves the molecule, well before 

joining of an attacking group. 

5. Metamerism is special type of isomerism where 

isomers exist simultaneously in dynamic 

equilibrium. 

6. Addition of HCN with formaldehyde is an 

example of electrophilic addition reaction. 

7. Ligroin is essentially petroleum ether containing 

aliphatic hydrocarbons and is generally used in 

dry cleaning clothes. 

Sol. 

1. [False] Thomson through his experiment 

determined the charge to mass ratio of an 

electron and the value of e/m is equal to 1.76 × 

10 8 coulomb/gm. Hence one gm of electrons 

have charge 1.76 × 10 8 C. 

2. [True] 

3. [False] 

4. [True] S N 1 reaction mechanism takes place in 

two steps as : 

R—X 

R + + OH – 

⎯ Slow ⎯→ ⎯ R + + X – 

⎯ Fast ⎯→ 

ROH 

5. [False] In metamerism isomers differ in structure 

due to difference in distribution of carbon atoms 

about the functional group. 

For example : 

CH 3 CH 2 –O–CH 2 CH 3 and CH 3 –OCH 2 CH 2 CH 3 

Conditions mentioned in the statement are 

associated with phenomenon of trautomerism. 

6. [False] 

H 

H – C = O + H + CN – 

7. [True] 

H 

H – C = OH 

CN 



`tà{xÅtà|vtÄ V{tÄÄxÇzxá 

10 

Set 

This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety 

of possible twists and turns of problems in mathematics that would be very helpful in facing 

IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and 

we hope that this section would prove a rich resource for practicing challenging problems and 

enhancing the preparation level of IIT JEE aspirants. 

By : Shailendra Maheshwari 

Solutions will be published in next issue 

Joint Director Academics, Career Point, Kota 

1. Let f(x) = sinx and 

⎧{max 

f ( t); 

0 ≤ t ≤ x 

g(x) = ⎨ 2 

⎩ sin x / 2 

; 

; 

for 0 ≤ x ≤ π 

x > π 

Discuss the continuity and differentiability of g(x) in 

(0, ∞) 

2. Is the inequality sin 2 x < x sin(sin x) true for 

0 < x < π/2 ? Justify your answer. 

3. A shop sells 6 different flavours of ice-cream. In how 

many ways can a customer choose 4 ice-cream cones 

if 

(i) they are all of different flavours; 

(ii) they are not necessarily of different flavours; 

(iii) they contain only 3 different flavoures; 

(iv) they contain only 2 or 3 different flavoures ? 

4. Using vector method, show that the internal 

(external) bisector of any angle of a triangle divides 

the opposite side internally (externally) in the ratio of 

the other two sides containing the triangle. 

5. Prove that 

(a) cos x + n C 1 cos 2x + n C 2 cos 3x + ............ 

...... + n C n cos(n + 1)x = 2 n . cos n ⎛ n + 2 ⎞ 

x/2. cos ⎜ x ⎟ 

⎝ 2 ⎠ 

(b) sin x + n C 1 sin 2x + n C 2 sin 3x + ............... 

....... + n C n sin(n + 1)x = 2 n . cos n x/2 . sin ⎛ n + 2 ⎞ 

⎜ x ⎟ 

⎝ 2 ⎠ 

6. In a town with a population of n, a person sands two 

letters to two sperate people, each of whom is asked 

to repeat the procedure. Thus, for each letter 

received, two letters are sent to separate persons 

chosen at random (irrespective of what happened in 

the past). What is the probability that in the first k 

stages, the person who started the chain will not 

receive a letter ? 

7. Prove the identity : 

x 

x 2 

−z 

4 

2 2 

zx− 

z 

∫ 

e dz = 

x 4 

e 

0 ∫ 

e dz, deriving for the 

0 

x 2 

zx− 

z 

function f(x) = 

∫ 

e dz a differential equation 

0 

and solving it. 

8. Prove that 

∫ 

sin nθsecθ 

dθ 

2cos( n −1) 

θ 

= – 

– 

n −1 

∫sin( 

n – 2) θ secθdθ 

dθ. 

Hence or otherwise evaluate 

∫ π / 2 cos5θsin 3θ 

dθ. 

0 cos θ 

9. Find the latus rectum of parabola 

9x 2 – 24 xy + 16y 2 – 18x – 101y + 19 = 0. 

10. A circle of radius 1 unit touches positive x-axis and 

positive y-axis at A and B respectively. A variable 

line passing through origin intersects the circle in two 

points in two points D and E. Find the equation of the 

lines for which area of ∆ DEB is maximum. 

WHAT MAKES A STAR? 

So you're out one night and you look up into the 

sky. Assuming you aren't in a city with tons of 

smog or clouds, you will probably see a sky 

filled with little dots of light. Those dots (this 

should not be a surprise) are stars. Some are only 

a few hundred light years away and some are 

thousands of light years away. They all have 

some things in common. You see, stars are huge 

balls of fire. They aren't just any fire. That fire is 

from a constant number of nuclear reactions. 


MATHEMATICAL CHALLENGES 

SOLUTION FOR JANUARY ISSUE (SET # 9) 

1. as φ (a) = φ (b) = φ (c) 

so by Rolle’s theorem there must exist at least a point 

x = α & x = β each of intervals (a, c) & (c, b) such 

that φ′(α) = 0 & φ′(β) = 0. Again by Rolle’s theorem, 

there must exist at least a point x = µ such that 

α < µ < β where φ′(µ) = 0 

2 f ( a) 

2 f ( b) 

so 

+ 

( a − b)( 

a − c) 

( b − c) ( b − a) 

2 f ( c) 

+ 

– f ′′ (µ) = 0 

( c − a)( 

c − b) 

f ( a) 

f ( b) 

so 

+ 

( a − b)( 

a − c) 

( b − c) ( b − a) 

f ( c) 

1 

+ 

= f ′′ (µ) 

( c − a)( 

c − b) 

2 

where a < µ < b. 

2. Required probability 

r 2 

5 5 5 5 1 ⎛ 5 ⎞ 

− 

1 

1 . . . ........ . = ⎜ ⎟ . (r – 2) times 

6 6 6 6 6 ⎝ 6 ⎠ 6 

Note : any number in 1st loss 

same no. does not in 2nd (any other comes). 

Now 3rd is also diff. (and in same r − 2 times) 

Now (r − 1) th & r th must be same. 

3. 2s = a + b + c 

ON = − BN + BO 

Let BN = x 

2BN + 2CN + 2AR = 2s 

x + (a − x) + (b − a + x) = s 

x = s − b 

A 

∆ s ( s − a)( 

s − b)( 

s − c) 

so r = k = = s s 

r = k = 

s ( s − a)( 

s − b)( 

s − c) 

s 

2sk = s( s − a)( 

a − b + c)( 

a + b − c) 

= s ( s − a)( 

a − 2x)( 

a + 2x) 

2sk = s( 

s − a)( 

a − 4h 

) 

required locus is 

4s 2 y 2 = A(a 2 – 4x 2 ) 

2 

⇒ s 2 y 2 + Ax 2 Aa 

= 

4 

where A is = s (s – a) 

here h 2 < as so it is an ellipse 

4. f (0) = c 

f (1) = a + b + c & f (−1) = a − b + c 

solving these, 

a = 2 

1 [f (1) + f (−1) − 2 f (0)] , 

b = 2 

1 [f (1) − f (−1)] & c = f (0) 

2 

2 

x( x +1) 

so f (x) = f (1) + (1− x 2 x( x −1) 

) f (0) + f(−1) 

2 

2 

2 | f (x) | < | x | | x + 1 | + 2| 1 − x 2 | + | x | | x − 1| ; 

as | f (1) | , | f (0) |, | f (−1) | ≤ 1. 

2 | f (x) | ≤ | x | (x + 1) + 2 (1 − x 2 ) + | x | (1 − x) as 

x ∈ [−1, 1] 

so 2 | f (x) | ≤ 2 (|x| + 1 − x 2 5 

5 

) ≤ 2 . so | f (x) | ≤ 4 4 

Now as g (x) = x 2 f (1/x) = 2 

1 (1 + x) f (1) 

+ (x 2 − 1) f (0) + 2 

1 (1 − x) f (−1) 

B 

M 

I (h,k) 

N 

so h = ON = 2 

a − (s − b) 

r 

−2s + a + 2b 

= 

= 

2 

R 

O 

b − c 

2 

& r = k. 

C 

so 2 | g (x) | ≤ | x + 1 | + 2 | 1 − x 2 | + | 1 − x| 

⇒ 2 | g (x) | ≤ x + 1 + 2 (1 − x 2 ) | + 1 − x ; 

as x ∈ [−1, 1] 

⇒ 2 | g (x) | < − 2x 2 + 4 ≤ 4. 

⇒ |g (x) | ≤ 2. 

5. Oil bed is being shown by the plane A′ PQ. θ be the 

angle between the planes A′ PQ & A′ B′ C′. Let A′ B′ 

C′ be the x − y plane with x-axis along A′ C′ and 

origin at A′. The P.V.s of the various points are 

defined as follows 


A 

B 

C 

⎛ 5x 

x ⎞ 

= − 2 

∫ 

⎜sin 

sin ⎟ dx 

⎝ 2 2 ⎠ 

⎛ 6x 

4x 

⎞ 

= 

∫ 

⎜cos 

− cos ⎟ dx 

⎝ 2 2 ⎠ 

A´ 

B´ 

P 

Q 

point C′ : b î , point B′ : cos A î + c sin A ĵ , 

point Q : b î – z kˆ , point P : cos A î + c sin A ĵ – y kˆ 

normal vector to the plane A′ B′ C′ 

= n r 1 = bc sin A kˆ 

normal vector to the plane A'PQ = n r 

2 

= cz sin A î + (by – cz cos A) ĵ + bc sin A kˆ 

r r 

n1. 

n2 

so cos θ = r r 

| n || n | 

1 

1 

bc sin A 

= 

2 2 2 

2 2 2 2 

[ c z sin A + ( by − cz cos A) 

+ b c sin A] 

b c sin A 

cos θ = 

2 2 2 2 2 2 2 

[ b c sin A + ( c z + b y − 2bycz 

cos A)] 

2 

[ c z 

so tan θ = 

2 

so tan θ . sin A = 

2 

2 

C´ 

+ b y − 2bycz 

cos A] 

bc sin A 

z 

b 

2 

2 

y 

+ 

c 

2 

2 

1/ 2 

2yz 

− cos A 

bc 

6. 

cos8x 

− cos 7x 

2sin 5x 

∫ 

. 

1+ 

2cos5x 

2sin 5x 

dx 

= 

∫ 

sin13x 

−sin 3x 

−sin12x 

+ sin 2x 

dx 

2(sin 5x 

+ sin10x) 

= 

∫ 

sin13x 

+ sin 2x 

−sin 3x 

– sin12x 

dx 

2(sin 5x 

+ sin10x) 

= 

∫ 

15x 

11x 

15x 

9x 

2sin cos − 2sin cos 

2 2 2 2 dx 

15x 

5x 

2.2.sin cos 

2 2 

= 

∫ 

11x 

9x 

cos − cos 

2 2 dx 

5x 

2cos 

2 

= 

∫ − x 

2sin 5x 

sin 

2 dx 

5x 

2cos 

2 

1/ 2 

1/ 2 

7. 

= 

∫ 

(cos 3x − cos 2x) 

dx 

sin 3x sin x 

= − + C 

3 22 

2 

d y 

= 2 

2 

dx ∫ x f ( t) 

dt 

0 

integrate using by parts method 

dy 

⎡ 

x 

x 

= 2 

dx 

⎥ ⎥ ⎤ 

⎢x 

− 

⎢ ∫ 

f ( t) 

dt 

∫ 

x . f ( x) 

dx 

⎣ 0 

0 ⎦ 

⎡ 

x 

⎤ 

= 2 ⎢ 

⎥ 

⎢∫( 

x − t) 

f ( t) 

dt 

⎥ 

⎣ 0 

⎦ 

again integrating, 

⎡ x 

y = 2 ⎢ x − − 

⎢ ∫( 

x t) 

f ( t) 

dt 

⎣ 0 

⎡ 

=2 ⎢x 

⎢ 

⎣ 

x 

x 

∫ 

0 

( x − t) 

x 

f ( t) 

dt − 

2 

x 

∫ 

0 

2 x 

2 

= 

∫ 

2 ( x − xt) 

f ( t) 

dt − 

∫ 

x 

0 

x 

x 

0 

∫ 

0 

⎛ 

x⎜ 

⎜ 

⎝ 

2 

x 

∫ 

0 

⎞ ⎤ 

f ( t) 

dt − 0⎟ 

dx⎥ 

⎟ ⎥ 

⎠ ⎦ 

f ( t) 

dt + 

x 

∫ 

0 

2 

x 

2 

x 

f ( t) 

dt + 

∫ 

t 

y = 

∫ 

( x − 2xt 

+ t ) f ( t) 

dt = 

∫( 

x − t) 

8. To prove that 

0 

2 

Let b 

a = c > 0 

2 

⎛ 

⎜⎛ 

a ⎞ 

⎜ ⎟ 

⎜ 

⎝⎝ 

b ⎠ 

α 

⎞ 

+ 1⎟ 

⎟ 

⎠ 

1/ α 

x 

0 

⎛ 

< ⎜⎛ 

a ⎞ 

⎜ ⎟ 

⎜ 

⎝⎝ 

b ⎠ 

β 

2 

0 

⎤ 

f ( x) 

dx⎥ 

⎥ 

⎦ 

2 

f ( t) 

dt 

⎞ 

+ 1⎟ 

⎟ 

⎠ 

1/ β 

so (c α + 1) 1/α < (c β + 1) 1/β . 

Let f (x) = (c x + 1) 1/x ; x > 0 

f ′(x) = (c x + 1) 1/x ln (c x ⎛ 1 ⎞ 

+ 1) ⎜ − 2 

⎟ 

⎝ x ⎠ 

1 1 

+ (c x + 1) –1 

x . c x ln c 

x 

x 

1 

−1 

x 

f ( t) 

dt 

( c + 1) 

x 

x x x 

= 

[ − ( c + 1) l n ( c + 1) + c ln 

c ] < 0 

2 

x 

so f (x) is decreasing function 

so f (α) < f (β). Hence proved. 

9. Point P (x, 1/2) under the given condition are length 

PB = OB 


O 

P 

C 

(t – 1) 

A 

θ 

B (t, 1) 

1. Emeralds have been produced synthetically in 

labs since 1848 and can be virtually 

indistinguishable from the genuine article. 

rθ = t ; so θ = t 

PB θ 

from ∆PAB : = PA sin 

2 2 

t 

⇒ PB = 2 sin 2 

θ t 

Now ∠ PBC = = ; 2 2 

........(1) 

2. In the last 200 years the use of metals has 

increased as scientists have discovered new 

ones: until the 17th Century only 12 metals 

were known - there are now 86. 

3. The only person to have an element named 

after him while still alive was Glenn Seaborg, 

the most prolific of all the element hunters. 

so from ∠ PCB ; 2 

θ = 2 

t 

1/ 2 

so from ∆ PCB ; PB 

= sin 2 

t 

........(2) 

from (1) & (2) PB = 1 ; so θ = t = π/3 

thus | PB | 2 = (t − x) 2 + 4 

1 = 1. 

3 

| t − x | = ; t − x = ; as t > x 

23 

2 

π 3 

so x = − 3 2 

10. Let x n = n −1 

+ n + 1 be rational, then 

1 1 = is also rational 

n −1 

+ n + 1 

x n 

1 n + 1 − n −1 

= is also rational 

2 

x n 

n +1 − n −1 

is also rational 

as n + 1 + n −1 

& n + 1 − n −1 

are rational 

so n + 1 + n −1 

must be rational 

i.e. (n + 1) & (n – 1) are perfect squares. 

This is not possible as any two perfect squares differe 

at least by 3. Hence there is not positive integer n for 

which n −1 

+ n + 1 is a rational. 

4. Traffic lights with red and green gas lights were 

first introduced in London in 1868. 

Unfortunately, they exploded and killed a 

policeman. The first successful system was 

installed in Cleveland, Ohio in 1914. 

5. In 1998, design student Damini Kumar at South 

Bank University patented a teapot with a 

special grooved spout, which she claims 

virtually rules out dribbling. 

6. Even though most items in the home today are 

technologically up to date, most of us are still 

using the standard light bulb designed in 1928! 

7. A chest x-ray is comprised of 90,000 to 

130,000 electron volts. 

8. The strength of early lasers was measured in 

Gillettes, the number of blue razor blades a 

given beam could puncture. 

9. The first commercial radio station in the United 

States, KDKA Pittsburgh, began broadcasting 

in November 1920. 

10. A British rocket attack on US soldiers is 

celebrated in the lyrics of the US National 

Anthem. 

11. Until the late 1800s, it was forbidden for 

women in the United States to obtain a patent, 

so if a woman had invented something she 

would file for a patent under her husband or 

father's name. For this reason, the number of 

early female inventors remains a mystery. 

12. Milt Garland, a 102 year old engineer, invented 

a technology that forms ice on the exterior of a 

casing instead of inside it, which is used to 

create indoor ice rinks. 


Students' Forum 

Expert’s Solution for Question asked by IIT-JEE Aspirants 

MATHS 

1. Evaluate 

∫ 

( tx + 1− 

x) 

1 

0 

n 

dx where n ∈ I + and t is a 

parameter independent of x. Hence show that 

∫ 

1 

0 

k 

n−k 

x (1 − x) 

= [ n C k (n + 1)] –1 

+ 1 

t n 

1 

n −1 

Sol. 

∫ 

[( t −1) 

x + 1] = 

0 

( t −1)( 

n + 1) 

1 

= [1 + t + t 2 + t 3 + ..... + t k + .... + t n ] ...(1) 

n + 1 

1 

1 

n 

n 

Also 

∫ 

[ tx + (1 − x)] 

dx = 

0 ∫ 

[(1 − x) 

+ tx] 

dx 

0 

1 

1 

n 

n 

= 

∫ 

C0 ( 1 − x) 

dx + t 

n 

n−1 

0 ∫ 

C1 ( 1 − x) 

x dx 

0 

+ .... + t k n 1 

n−k 

C k 

∫ 

(1 − x) 

x k dx ...... ...(2) 

0 

As R.H.S of (1) = R.H.S. of (2) compare equ. of t k on 

both sides 

1 

⇒ = n 1 

n−k 

C k 

n + 1 ∫ 

(1 − x) 

x k dx 

0 

1 

n−k 

∫ 

(1 − x) 

x k 1 

dx = 

Hence proved. 

0 

( n + 1) 

n 

C k 

2. Let point A describes a curve C such that the 

difference between its distances from the points (0, 0) 

and (3, 4) is 5. Then find the no. of points at which 

the circle x 2 + y 2 = 4 and c intersect. 

Sol. Locus of the point A is curve C which is satisfying 

|AB – AP| = PB 

where P : (0, 0), B : (3, 4) curve C represents two 

rays BA or PA and it is clear from diagram that 

curve C and given circle are intersecting at only one 

point. 

A 

P(0, 0) 

A 

B(3, 4) 

3. Let [x] stands for the greatest integer function find 

2 

3 x + sin x 

the derivative of f(x) = ( x + [ x + 1]) , where it 

exists in (1, 1.5). Indicate the point(s) where it does 

not exist. Give reason(s) for your conclusion. 

Sol. The greatest integer [x 3 + 1] takes jump from 2 to 3 at 

3 2 and again from 3 to 4 at 3 3 in [1, 1.5] and 

therefore it is discontinuous at these two points. As a 

result the given function is discontinuous at 3 2 and 

hence not differentiable. 

To find the derivative at other points we write : 

in (1, 3 2 

x + sin x 

2 ), f(x) = ( x + 2) 

2 

x + sin x−1 

⇒ f ´(x) = ( x + 2) 

{x 2 + sin x + (x + 2) (2x + cos x) log (x + 2)} 

in ( 3 2 , 3 2 

x + sin x 

3 ), f(x) = ( x + 3) , 

2 

x + sin x−1 

f ´(x) = ( x + 3) {x 2 + sin x 

+ (2x + cos x) (x + 3) × log e (x + 3)} 

in ( 3 2 

x + sin x 

5 , 1.5), f(x) = ( x + 4) , 

2 

x + sin x−1 

f ´(x) = ( x + 4) , {x 2 + sin x + (2x + cos x) 

(x + 4) × log e (x + 4)} 

4. For three unit vectors â , bˆ and ĉ not all collinear 

given that aˆ× cˆ 

= cˆ× bˆ 

and b ˆ × aˆ 

= aˆ× 

c ˆ . Show 

that cosα + cos β + cos γ = –3/2, where α, β and γ are 

the angles between â and bˆ , bˆ and ĉ and ĉ and â 

respectively. 

Sol. â × ĉ = cˆ× bˆ 

⇒ ( â + bˆ ) × ĉ = → 0 

⇒ ĉ is collinear with â + bˆ ⇒ 

same λ ∈ R 

Similarly bˆ + ĉ = µ â for some scalar u 

â + bˆ = λ ĉ for 

Now â + bˆ = λ ĉ ⇒ â + bˆ + ĉ = (λ + 1) → c 

Similarly ⇒ â + bˆ + ĉ = (µ + 1) â 

Hence (λ + 1) ĉ = (µ + 1) â , 

either λ + 1 = µ + 1 = 0 or ĉ is collinear with â . 

But ĉ can not be collinear to â other wise cˆ × aˆ 

= 0 

⇒ cˆ × bˆ 

= 0 

⇒ bˆ is collinear to with ĉ 


⇒ â bˆ and ĉ are collinear. 

Hence ĉ is not collinear to â 

⇒ λ + 1 = µ + 1 = 0 

⇒ λ ± µ = –1 

Hence bˆ + ĉ = µ â 

⇒ â + bˆ + ĉ = → 0 

⇒ ( â + bˆ + ĉ ) . ( â + bˆ + ĉ ) = 0 

⇒ 1 + 1 + 1 + 2 ( â . bˆ + bˆ . ĉ + ĉ . â) = 0 

⇒ â . bˆ + bˆ . ĉ + ĉ . â = – 2 

3 

⇒ cos α + cos β + cos γ = – 2 

3 

5. Let S be the coefficients of x 49 in given expression 

f(x) and if P be product of roots of the equation 

f(x) = 0, then find the value of P 

S , given that : 

f(x) = (x – 1) 2 ⎛ 

⎜ 

⎝ 

x 

2 

⎞ ⎛ 1 ⎞ ⎛ 

− 2⎟ ⎜ x − ⎟ ⎜ 

⎠ ⎝ 2 ⎠ ⎝ 

x 

3 

⎞ ⎛ 1 ⎞ 

− 3⎟ ⎜ x − ⎟ , 

⎠ ⎝ 3 ⎠ 

⎛ x ⎞ ⎛ 

......... ⎜ − 25⎟ ⎜ x 

⎝ 25 ⎠ ⎝ 

Sol. Here we can write f(x) as : 

⎧ ⎛ x ⎞⎛ 

x ⎞ ⎛ x ⎞⎫ 

f(x) = ⎨( 

x −1) 

⎜ − 2⎟⎜ 

− 3⎟... 

⎜ − 25⎟⎬ 

⎩ ⎝ 2 ⎠⎝ 

3 ⎠ ⎝ 25 ⎠⎭ 

1 

− 25 

⎧ ⎛ 1 ⎞⎛ 

1 ⎞ ⎛ 1 ⎞⎫ 

× ⎨( x −1) 

⎜ x − ⎟⎜ 

x − ⎟... 

⎜ x − ⎟⎬ 

⎩ ⎝ 2 ⎠⎝ 

3 ⎠ ⎝ 25 ⎠⎭ 

Now roots of f(x) = 0 are; 

1 2 , 2 2 , 3 2 , ..... , 25 2 1 1 1 

and 1, , , ....., 2 3 25 

Now f(x) is the polynomial of degree 50, 

So coefficient of x 49 will be : 

S = – (sum of roots) 

= – (1 2 + 2 2 + ... + 25 2 ⎛ 1 1 1 ⎞ 

) – ⎜1 

+ + + .... + ⎟ 

⎝ 2 3 25 ⎠ 

⎧25× 

26× 

51 ⎫ 

1 

= – ⎨ + K⎬ 

where, K = 

⎩ 6 ⎭ ∑ 

n 

n= 

1 

⇒ S = –(K + 5525). 

Product of roots : 

1 2 . 2 2 . 3 2 .... 25 2 1 1 1 

. 1 . . .... = 1 . 2 . 3 ...25 

2 3 25 

25 

⎞ 

⎟ 

⎠ 

6. A man standing at a distance 5m in front of the base 

of a building 10m high on which a flagstaff is 

mounted observes that the top of the building and the 

top of a mountain behind the building are along the 

same straight line. When he recedes by a distance of 

48 m he observes that now the top of the flagstaff and 

the top of the mountain are along the same straight 

line. If at both the locations, the flagstaff subtends the 

same angle at the man’s eye, find the height of 

mountain. 

Sol. CD : Flagstaf 

DE : Building 

KF : Mountain (height = h say) 

The figure illustrates the situation. 

Since, ∠CBD = ∠CAD = α say, points A, B, C and 

D are concyclic. 

⇒ ∠ABD = ∠ACD = 90º – (α + β) 

⇒ ∠ABC = 90º – (α + β) + α = 90º – β = ∠KCH 

B 

α 

90º – β 

48 

β 

A 5 

90º – β 

C 

α 10 

Now, h = KH + HF 

= (CH) tan (90º – β) + (BE) tan(90º – β) 

(Q HF = CE) 

= [DG + (BA + AE) cot β 

= [KG cot β + (48 + 5)] cot β 

⇒ h = [(h – 10)cotβ + 53] cot β 

(Q KG = KF – GF) 

5 1 

Putting cot β = = , we get 

10 2 

⎛ h −10 

⎞ 

h = ⎜ + 53⎟ ⎝ 2 ⎠ 

1 

2 

⇒ 4h = h – 10 + 106 

⇒ 3h = 96 

⇒ h = 32 m 

E 

D 

β 

K 

H 

G 

F 

∴ P = 25 ! 

S −( K + 5525) 

Hence = P 25! 

25 

, where K = ∑ 

n= 

1 

1 

n 


MATHS 

INTEGRATION 

Mathematics Fundamentals 

Integration : 

If d f(x) = F(x), then ∫ F(x 

) dx = f(x) + c, where c 

dx 

is an arbitrary constant called constant of integration. 

1. 

∫ 

x n dx = 

1 

2. 

∫ 

dx 

x 

+ 1 

x n (n ≠ –1) 

n + 1 

= log x 

3. 

∫ 

e x dx = e x 

4. 

∫ 

a x dx = 

x 

a 

log 

e 

a 

5. 

∫ 

sin x dx = – cos x 

6. 

∫ 

cos x dx 

2 

= sin x 

7. 

∫ 

sec x dx = tan x 

8. 

∫ 

cos ec 2 x dx = – cot x 

9. 

∫ 

sec x tan x dx = sec x 

10. 

∫ 

cosec x cot x dx = – cosec x 

11. 

∫ sec x dx = log(sec x + tan x) = log tan ⎛ x π 

⎟ ⎞ 

⎜ + 

⎝ 2 4 ⎠ 

12. 

∫ cosec x dx = – log (cosec x + cot x) = log tan ⎛ x 

⎟ ⎞ 

⎜ 

⎝ 2 ⎠ 

13. 

∫ 

tan x dx = – log cos x 

14. 

∫ 

cot x dx = log sin x 

dx 

15. 

∫ 

a 2 − x 

2 

dx 

16. 

∫ a 2 + x 

2 

dx 

17. 

∫ 

x x 2 − a 

2 

= sin –1 x = – cos 

–1 x 

a a 

1 

= tan 

–1 x 1 = – cot 

–1 ⎛ x ⎞ 

⎜ ⎟ 

a a a ⎝ a ⎠ 

1 

= sec 

–1 x 1 = – cosec 

–1 ⎛ x ⎞ 

⎜ ⎟ 

a a a ⎝ a ⎠ 

18. 

∫ x − a 

2 

2 

1 

= 

1 x − a log , when x > a 

2a x + a 

1 1 a + x 

19. 

∫ 

dx = log , when x < a 

a 

2 − x 

2 2a a − x 

dx 

20. 

∫ 

x 2 − a 

2 

dx 

21. 

∫ 

x 2 + a 

2 

= log 

⎧ 2 2 ⎫ 

⎨x + x − a ⎬ = cos h –1 ⎛ x ⎞ 

⎜ ⎟ 

⎩ 

⎭ ⎝ a ⎠ 

= log 

⎧ 2 2 ⎫ 

⎨x + x + a ⎬ = sin h –1 ⎛ x ⎞ 

⎜ ⎟ 

⎩ 

⎭ ⎝ a ⎠ 

22. 

∫ 

a 2 − x 

2 1 

dx = x 

2 2 1 

a − x + a 2 sin –1 ⎛ x ⎞ 

⎜ ⎟ 2 2 ⎝ a ⎠ 

23. 

∫ 

x 2 − a 

2 1 

dx = x 

2 2 

x − a 2 

1 

– a 2 log 

⎧ 

⎨x 

+ 2 ⎩ 

24. 

∫ 

x 2 + a 

2 1 

dx = x 

2 2 

x + a 2 

x 

2 

− a 

1 

+ a 

2 

log 

⎧ 2 2 ⎫ 

⎨x 

+ x + a ⎬ 

2 ⎩ 

⎭ 

f ´( x) 

25. 

∫ 

dx = log f(x) 

f ( x) 

f ´( x) 

26. 

∫ 

dx = 2 f (x) 

f ( x) 

Integration by Decomposition into Sum : 

1. Trigonometrical transformations : For the 

integrations of the trigonometrical products such as 

sin 2 x, cos 2 x, sin 3 x, cos 3 x, sin ax cos bx, etc., they are 

expressed as the sum or difference of the sines and 

cosines of multiples of angles. 

2. Partial fractions : If the given function is in the 

form of fractions of two polynomials, then for its 

integration, decompose it into partial fractions (if 

possible). 

Integration of some special integrals : 

dx 

(i) 

∫ 2 

ax + bx + c 

This may be reduced to one of the forms of the above 

formulae (16), (18) or (19). 

2 

⎫ 

⎬ 

⎭ 


dx 

dx 

1 

(ii) 

∫ 4. 

2 

∫ 

, at first x = and then a + ct 2 = z 2 

ax + bx + c 

2 

2 

( px + r) 

ax + c 

t 

This can be reduced to one of the forms of the above Some Important Integrals : 

formulae (15), (20) or (21). 

(iii) 

∫ 

ax 2 

dx 

⎛ x − α ⎞ 

+ bx + c dx 

1. To evaluate 

∫ 

, 

( x − α)( 

x − β) 

∫ 

⎜ ⎟ dx, 

⎝ β − x ⎠ 

This can be reduced to one of the forms of the above 

formulae (22), (23) or (24). 

∫ 

( x − α)( 

β − x) 

dx. Put x = α cos 2 θ + β sin 2 θ 

( px + q) 

dx ( px + q) 

dx 

(iv) 

∫ 

, dx dx 

2 

ax + bx + c ∫ 2 


ax + bx + c 

∫ 

, 

a + b cos x ∫ 

, 

a + b sin x 

For the evaluation of any of these integrals, put 

dx 

px + q = A {differentiation of (ax 2 + bx + c)} + B ∫ a + b cos x + c sin x 

Find A and B by comparing the coefficients of like 

powers of x on the two sides. 

⎛ x ⎞ 

⎛ ⎞ 

⎜2 tan ⎟ ⎜ − 

2 x 

1 tan ⎟ 

1. If k is a constant, then 

Replace sin x = 

⎝ 2 ⎠ 

and cos x = 

⎝ 2 ⎠ 

∫ k dx = kx and ⎛ ⎞ 

∫ 

k f ( x) 

dx = k 

∫ 

f ( x) 

dx 

⎜ + 

2 x 

⎛ ⎞ 

1 tan ⎟ 

⎜ + 

2 x 

1 tan ⎟ 

⎝ 2 ⎠ 

⎝ 2 ⎠ 

2. 

∫ 

{ f 1 ( x) 

± f2( 

x)} 

dx = 

∫ 

f 1 ( x) 

dx ± 

∫ 

f 2 ( x) 

dx 

x 

Then put tan = t. 2 

Some Proper Substitutions : 

p cos x + q sin x 

1. 

∫ 

f(ax + b) dx, ax + b = t 


∫ 

dx 

a + b cos x + c sin x 

2. 

∫ f(axn + b)x n–1 dx, ax n + b = t 

Put p cos x + q sin x = A(a + b cos x + c sin x) 

+ B. diff. of (a + b cos x + c sin x) + C 

3. 

∫ 

f{φ(x)} φ´(x) dx, φ(x) = t 

A, B and C can be calculated by equating the 

coefficients of cos x, sin x and the constant terms. 

f ´( x) 

4. 

∫ 

dx , f(x) = t 

dx 

f ( x) 


∫ 

, 

2 

2 

a cos x + 2b 

sin x cos x + c sin x 

5. 

∫ 

a 

2 − x 

2 dx, x = a sin θ or a cos θ 

dx 

dx 

∫ 

, 

2 

a cos x + b ∫ 2 

a + b sin x 

6. 

∫ 

a 

2 + x 

2 dx, x = a tan θ 

In the above type of questions divide N r and D r by 

cos 2 x. The numerator will become sec 2 x and in the 

2 2 

a − x 

7. 

∫ 

dx, x 2 = a 2 denominator we will have a quadratic equation in tan 

cos 2θ 

2 2 

a + x 

x (change sec 2 x into 1 + tan 2 x). 

Putting tan x = t the question will reduce to the form 

8. 

∫ 

a ± x dx, a ± x = t 2 

dt 

∫ 2 

a − x 

at + bt + c 

9. 

∫ 

dx, x = a cos 2θ 

a + x 

5. Integration of rational function of the given form 

10. 

∫ 

2ax − x 

2 

2 2 

2 2 

x + a 

x − a 

dx, x = a(1 – cos θ) 

(i) 

∫ 

dx, (ii) 

4 2 4 

x + kx + a ∫ 

dx, where 

4 2 4 

x + kx + a 

11. 

∫ 

x 

2 − a 

2 dx, x = a sec θ 

k is a constant, positive, negative or zero. 

These integrals can be obtained by dividing 

Substitution for Some irrational Functions : 

numerator and denominator by x 2 , then putting 

dx 

1. 

∫ 

, ax + b = t 2 

a 2 a 2 

x – = t and x + = t respectively. 

( px + q) 

ax + b 

x 

x 

dx 

1 

Integration of Product of Two Functions : 

2. 

∫ 

, px + q = 

2 

( px + q) 

ax + bx + c t 

1. 

∫ f 1(x) f 2 (x) dx = f 1 (x) 

∫ f ' 

2(x) dx – 

∫[ ( f 1 ( x) 

∫ 

f2( 

x) 

dx] 

dx 

dx 

3. 

∫ 

, ax + b = t 2 2 

Proper choice of the first and second functions : 

( px + qx + r) 

ax + b 

Integration with the help of the above rule is called 


integration by parts, In the above rule, there are two 

terms on R.H.S. and in both the terms integral of the 

second function is involve. Therefore in the product 

of two functions if one of the two functions is not 

directly integrable (e.g. log x, sin –1 x, cos –1 x, tan –1 x 

etc.) we take it as the first function and the remaining 

function is taken as the second function. If there is no 

other function, then unity is taken as the second 

function. If in the integral both the functions are 

easily integrable, then the first function is chosen in 

such a way that the derivative of the function is a 

simple functions and the function thus obtained under 

the integral sign is easily integrable than the original 

function. 

2. 

∫ 

sin( bx + c) 

dx 

e ax 

= 

= 

a 

e ax 

2 

+ b 

2 

2 

ae ax + b 

3. 

∫ 

cos( bx + c) 

dx 

e ax 

= 

= 

a 

e ax 

2 

+ b 

2 

2 

ae ax + b 

[a sin (bx + c) – b cos (bx + c)] 

2 

⎡ 

sin ⎢bx 

+ c − tan 

⎣ 

−1 

b ⎤ 

a ⎥ ⎦ 

[a cos (bx + c) + b sin(bx + c)] 

2 

⎡ 

cos ⎢bx 

+ c − tan 

⎣ 

4. 

∫ ekx {kf(x) + f '(x)} dx = e kx f(x) 

⎛ x ⎞ 

5. 

∫ 

log e x = x(log e x – 1) = x log e ⎜ ⎟ 

⎝ e ⎠ 

Integration of Trigonometric Functions : 

1. To evaluate the integrals of the form 

−1 

b ⎤ 

a ⎥ ⎦ 

I = 

∫ sinm x cos n x dx, where m and n are rational 

numbers. 

(i) Substitute sin x = t, if n is odd; 

(ii) Substitute cos x = t, if m is odd; 

(iii) Substitute tan x = t, if m + n is a negative even 

integer; and 

(iv) Substitute cot x = t, if 2 

1 (m + 1) + 2 

1 (n – 1) is an 

integer. 

2. Integrals of the form 

∫ 

R (sin x, cos x) dx, where R is 

a rational function of sin x and cos x, are transformed 

into integrals of a rational function by the substitution 

tan 2 

x = t, where –π < x < π. This is the so called 

universal substitution. Sometimes it is more 

convenient to make the substitution cot 2 

x 

= t for 

0 < x < 2π. 

The above substitution enables us to integrate any 

function of the form R (sin x, cos x). However, in 

practice, it sometimes leads to extremely complex 

rational functions. In some cases, the integral can be 

simplified by – 

(i) Substituting sin x = t, if the integral is of the form 

∫ 

R (sin x) cos x dx. 

(ii) Substituting cos x = t, if the integral is of the form 

∫ 

R (cos x) sin x dx. 

dt 

(iii) Substituting tan x = t, i.e. dx = , if the 

2 

1+ 

t 

integral is dependent only on tan x. 

Some Useful Integrals : 

dx 

1. (When a > b) 

∫ a + b cos x 

= 

2 

2 

a − b 

2 

⎡ 

tan –1 

⎢ 

⎢⎣ 

dx 

2. (When a < b) 

∫ a + b cos x 

= – 

2 

1 

b − a 

2 

log 

a − b 

⎥ ⎥ ⎤ 

tan x + c 

a + b 2⎦ 

x 

b − a tan − 

a 

x 

b − a tan + 

a 

dx 1 x 

3. (when a = b) 

∫ 

= tan + c 

a + b cos x a 2 

dx 

4. (When a > b) 

∫ a + bsin 

x 

= 

2 

2 

a − b 

dx 

5. (When a < b) 

∫ a + bsin 

x 

= 

2 

1 

b − a 

2 

2 

log 

dx 

6. (When a = b) 

∫ a + bsin 

x 

a + b 

a + b 

⎧ ⎛ x ⎞ ⎫ 

⎪a 

tan⎜ 

⎟ + b ⎪ 

tan –1 

⎝ 2 ⎠ 

⎨ 

⎪ ⎪ ⎬ + c 

2 2 

⎪ a − b 

⎪⎩ 

⎭ 

⎛ x ⎞ 

a tan⎜ 

⎟ + b − 

⎝ 2 ⎠ 

⎛ x ⎞ 

a tan⎜ 

⎟ + b + 

⎝ 2 ⎠ 

b 

b 

2 

2 

− a 

− a 

2 

2 

+ c 

= 1 [tan x – sec x] + c 

a 


MATHS 

TRIGONOMETRICAL 

EQUATION 

Mathematics Fundamentals 

Functions with their Periods : 

Function 

sin (ax + b), cos (ax + b), sec (ax + b), 

cosec (ax + b) 

tan(ax + b), cot (ax + b) 

|sin (ax + b)|, |cos (ax + b)|, |sec (ax + b)|, 

|cosec (ax + b)| 

|tan (ax + b)|, |cot (ax + b)| 

Trigonometrical Equations with their General 

Solution: 

Trgonometrical equation 

sin θ = 0 

General Solution 

θ = nπ 

cos θ = 0 θ = nπ + π/2 

tan θ = 0 

θ = nπ 

sin θ = 1 θ = 2nπ + π/2 

cos θ = 1 

sin θ = sin α 

cos θ = cos α 

tan θ = tan α 

sin 2 θ = sin 2 α 

tan 2 θ = tan 2 α 

cos 2 θ = cos 2 α 


* 

cosθ = cosα 


* 



* 

cosθ = cos α 

θ = 2nπ 

θ = nπ + (–1) n α 

θ = 2nπ ± α 

θ = nπ + α 

θ = nπ ± α 

θ = nπ ± α 

θ = nπ ± α 

θ = 2nπ + α 

θ = 2nπ + α 

θ = 2nπ + α 

Period 

2π/a 

π/a 

π/a 

π/2a 

* If α be the least positive value of θ which satisfy 

two given trigonometrical equations, then the general 

value of θ will be 2nπ + α. 

Note : 

1. If while solving an equation we have to square it, 

then the roots found after squaring must be 

checked whether they satisfy the original equation 

or not. e.g. Let x = 3. Squaring, we get x 2 = 9, 

∴ x = 3 and – 3 but x = – 3 does not satisfy the 

original equation x = 3. 

2. Any value of x which makes both R.H.S. and 

L.H.S. equal will be a root but the value of x for 

which ∞ = ∞ will not be a solution as it is an 

indeterminate form. 

3. If xy = xz, then x(y – z) = 0 ⇒ either x = 0 or 

y = z or both. But x 

y = x 

z ⇒ y = z only and not 

x = 0, as it will make ∞ = ∞. Similarly, if ay = az, 

then it will also imply y = z only as a ≠ 0 being a 

constant. 

Similarly, x + y = x + z ⇒ y = z and x – y = x – z 

⇒ y = z. Here we do not take x = 0 as in the 

above because x is an additive factor and not 

multiplicative factor. 

4. When cos θ = 0, then sin θ = 1 or –1. We have to 

verify which value of sin θ is to be chosen which 

⎛ 1 ⎞ 

satisfies the equation cos θ = 0 ⇒ θ = ⎜n 

+ ⎟ π 

⎝ 2 ⎠ 

If sin θ = 1, then obviously n = even. But if 

sin θ = –1, then n = odd. 

Similarly, when sin θ = 0, then θ = nπ and cos θ = 1 

or –1. 

If cos θ = 1, then n is even and if cos θ = –1, then 

n is odd. 

5. The equations a cos θ ± b sin θ = c are solved as 

follows : 

Put a = r cos α, b = r sin α so that r = 

and α = tan –1 b/a. 

The given equation becomes 

r[cos θ cos α ± sin θ sin α] = c ; 

cos (θ ± α) = r 

c provided r 

c ≤ 1. 

2 

a + b 

2 


Relation between the sides and the angle of a triangle: 

1. Sine formula : 

sin A 

a 

= 

sin B 

b 

= 

sin C 

c 

= 

1 

2R 

Where R is the radius of circumcircle of triangle 

ABC. 

2. Cosine formulae : 

2 2 2 

2 2 2 

b + c − a a + c − b 

cos A = 

, cos B = 

, 

2bc 

2ac 

2 

2 

2 

a + b − c 

cos C = 

2ab 

It should be remembered that, in a triangle ABC 

If ∠A = 60º, then b 2 + c 2 – a 2 = bc 

If ∠B = 60º, then a 2 + c 2 – b 2 = ac 

If ∠C = 60º, then a 2 + b 2 – c 2 = ab 

3. Projection formulae : 

a = b cos C + c cos B, b = c cos A + a cos C 

c = a cos B + b cos A 

Trigonometrical Ratios of the Half Angles of a Triangle: 

a + b + c 

If s = in triangle ABC, where a, b and c are 

2 

the lengths of sides of ∆ABC, then 

(a) cos 2 

A = 

cos 2 

C = 

(b) sin 2 

A = 

sin 2 

C = 

s ( s − a) 

B s ( s − b) 

, cos = , 

bc 2 ac 

s ( s − c) 

ab 

( s − b)( 

s − c) 

B ( s − a)( 

s − c) 

, sin = , 

bc 2 ac 

( s − a)( 

s − b) 

ab 

A ( s − b)( 

s − c) 

(c) tan = , 

2 s( 

s − a) 

B ( s − a)( 

s − c) 

C 

tan = , tan 2 s( 

s − b) 

2 

( s − a)( 

s − b) 

s( 

s − c) 

Napier's Analogy : 

B − C b − c A C − A c − a B 

tan = cot , tan = cot 

2 b + c 2 2 c + a 2 

A − B a − b C 

tan = cot 

2 a + b 2 

Area of Triangle : 

∆ = 2 

1 bc sin A= 2 

1 ca sin B = 2 

1 ab sin C 

∆ = 

1 a 

2 sin B sin C 1 b 

= 

2 sinC 

sin A 1 c 

= 

2 sin Asin 

B 

2 sin( B + C) 

2 sin( C + A) 

2 sin( A + B) 

2 

sin A = s s a bc 

2∆ 

Similarly sin B = ca 

2∆ 

( − )( s − b)( 

s − c) 

= bc 

2∆ 

& sin C = ab 

Some Important Results : 

A B s − c A B 

1. tan tan = ∴ cot cot = 

2 2 s 2 2 

2. tan 2 

A + tan 2 

B = s 

c cot 2 

C = ∆ 

c (s – c) 

A B a − b 

3. tan – tan = (s – c) 

2 2 ∆ 

A B 

tan + tan 

A B 

4. cot + cot = 2 2 = 

2 2 A B 

tan tan 

2 2 

5. Also note the following identities : 

Σ(p – q) = (p – q) + (q – r) + (r – p) = 0 

s 

s − c 

c C 

cot 

s − c 2 

Σp(q – r) = p(q – r) + q(r – p) + r(p – q) = 0 

Σ(p + a)(q – r) = Σp(q – r) + aΣ(q – r) = 0 

Solution of Triangles : 

1. Introduction : In a triangle, there are six 

elements viz. three sides and three angles. In 

plane geometry we have done that if three of the 

elements are given, at least one of which must be 

a side, then the other three elements can be 

uniquely determined. The procedure of 

determining unknown elements from the known 

elements is called solving a triangle. 

2. Solution of a right angled triangle : 

Case I. When two sides are given : Let the 

triangle be right angled at C. Then we can 

determine the remaining elements as given in the 

following table. 

Given 

(i) a, b 

(ii) a, c 

Required 

tanA = b 

a , B = 90º – A, c = 

a 

sin A 

sinA = c 

a , b = c cos A, B = 90º – A 

Case II. When a side and an acute angle are given – 

In this case, we can determine 

Given 

(i) a, A 

(ii) c, A 

Required 

B = 90º – A, b = a cot A, c = 

a 

sin A 

B = 90º – A, a = c sin A, b = c cos A 


a 

Based on New Pattern 

IIT-JEE 2011 

XtraEdge Test Series # 10 

Time : 3 Hours 

Syllabus : 

Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus 

Instructions : 

Section - I 

• Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct 

answer and -1 mark for wrong answer. 

• Question 10 to 14 are multiple choice questions with one or more than one correct asnwer. +4 marks will be 

awarded for correct answer and –1 mark for wrong answer. 

• Question 15 to 20 are passage based single correct type questions. +4 marks will be awarded for correct answer and 

-1 mark for wrong answer.. 

Section - II 

• Question 21 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly 

matched answer and No Negative marks for wrong answer. However, +1 mark will be given for a correctly 

marked answer in any row. 

PHYSICS 

Questions 1 to 9 are multiple choice questions. Each 

question has four choices (A), (B), (C) and (D), out of 

which ONLY ONE is correct. Mark your response in 

OMR sheet against the question number of that 

question. + 3 marks will be given for each correct 

answer and – 1 mark for each wrong answer. 

1. Two men B and C are watching man A. B watches A 

to be stationary and C watches A moving. Then - 

(A) Man A may be at absolute rest 

(B) Man B may be at absolute rest 

(C) Man C may be at absolute rest 


2. A particle of mass m is placed on the centre of a fixed 

uniform semi-circular ring of radius R and mass M as 

shown. Then work required to displace the particle 

slowly from centre of ring to infinity is : (Assume 

only gravitational interaction of ring and particle) 

M 

GMm 

(A) 

R 

GMm 

(C) 

πR 

m 

R 

(B) – 

(D) – 

GMm 

R 

GMm 

πR 

3. A ideal diatomic gas occupies a volume V 1 at a 

pressure P 1 . The gas undergoes process in which the 

pressure is proportional to the volume. At the end of 

process the rms sped of the gas molecules has double 

from its initial value then the heat supplied to the gas 

in the given process is - 

(A) 7 P 1 V 1 (B) 8 P 1 V 1 

(C) 9 P 1 V 1 (D) 10 P 1 V 1 

4. An electron gum T emits electron accelerated by a 

potential difference U in a vacuum in the direction of 

the line a as shown in figure. Target M is placed at a 

distance d as shown in figure. Find the magnetic field 

perpendicular to the plane determine by line a and 

the point M in order that electron hit the target M – 

a 

(A) 

(C) 

Electron gun 

2Um e sin α 

2 (B) 

e d 

2Um e sin α 

e d 

d 

α 

M 

(D) 8 

Target 

2Um e sin α 

e 2d 

2Um e sin α 

e d 

5. When 24.8 KeV x-rays strike a material, the 

photoelectrons emitted from K shell are abserved to 

move in a circle of radius 23 mm in a magnetic field 

of 2 × 10 –2 T. The binding energy of K-shell 

electrons is - 

(A) 6.2 KeV 

(B) 5.4 KeV 

(C) 7.4 KeV 

(D) 8.6 KeV 


6. In the circuit shown the cell is ideal. The coil has an 

inductance of 4H ans zero resistance. F is a fuse of 

zero resistance and will blow when the current 

through it reaches 5A. The switch is closed at t = 0. 

The fuse will blow - 

2V 

+ 

– 

(A) after 5 sec 

(C) after 10 sec 

S 

F 

L = 4H 

(B) after 2 sec 

(D) almost at once 

7. In an insulating medium (K = 1) volumetric charge 

density varies with y-coordinates according the law 

ρ = a. y. A particle of mass m having positive charge 

q is at point A(0, y 0 ) and projected with velocity 

^ 

v r = v0 i as shown in figure. At y = 0 electric field is 

zero. Neglect the gravity and fractional resistance, 

the slope of trajectory of the particle as a function of 

y(E is only along y-axis) is – 

y 

qa 3 3 

(A) (y – y ) 

2 0 

m v 

(C) 

ε 0 

qa(y 

5m 

0 

A v 0 

(0,y 0) 

x 

3 3 

– y 0 

2 

ε0v0 

) 

qa 3 3 

(B) (y – y ) 

2 0 

3m v 

(D) 

ε 0 

qa(y 

2m 

0 

3 3 

– y 0 

2 

ε0v0 

8. If E denotes electric field in a uniform conductor, I 

corresponding current through it, v d -drift velocity of 

electrons and P denotes thermal power produced in 

the conductor, then which of the following graph is 

incorrect - 

(A) 

(C) 

v d 

P 

E 

v d 

(B) 

(D) 

9. Find the de Broglie wavelength of Earth. Mass of 

Earth is 6 × 10 24 kg. Mean orbital radius of Earth 

around Sun is 150 × 10 6 km - 

(A) 3.7 m (B) 3.7 × 10 –63 

(C) 3.7 × 10 63 m (D) 3.7 × 10 –63 cm 

P 

P 

E 

I 

) 



which MULTIPLE (ONE OR MORE THAN ONE) is 

correct. Mark your response in OMR sheet against the 

question number of that question. + 4 marks will be 

given for each correct answer and –1 mark for each 

wrong answer. 

10. A pendulum of length l is suspended on a flat car that 

is moving with a velocity u on the horizontal road. If 

the car is suddenly stopped, then : (Assume bob of 

pendulum does not collide anywhere) 

θ 

l 

(A) the maximum angle θ with the initial vertical line 

through which the pendulum swing is 

⎡ u ⎤ 

sin – 1 ⎢ ⎥ 

⎢⎣ 

2 gl ⎥⎦ 

(B) the maximum angle θ with the initial vertical line 

through which the pendulum swing is 

⎡ ⎤ 

–1 u 

2sin ⎢ ⎥ 

⎢⎣ 

2 gl ⎥⎦ 

(C) If maximum angle is 60º, l = 5 m and 

g = 9.8 m/s 2 then the initial speed of car u is 

7 m/s 

(D) If maximum angle 60º, l = 5 m and g = 9.8 m/s 2 , 

then the initial speed of car u is 6 m/s 

11. A parallel plate air capacitor is connected to a 

battery. If plates of the capacitor are pulled further 

apart, then which of the following statements are 

correct? 

(A) Strength of electric field inside the capacitor 

remain unchanged, if battery is disconnected 

before pulling the plate 

(B) During the process, work is done by an external 

force applied to pull the plates whether battery is 

disconnected or it remains connected 

(C) Potential energy in the capacitor decreases if the 

battery remains connected during pulling plates 

apart 

(D) None of the above 

12. 

12Ω 

6Ω 

2H 

6V 

2Ω 

(A) Its time constant is 4 

1 sec 

(B) Its time constant is 4 sec 

(C) In steady state current through battery will be 

equal to 0.75 A 

(D) In steady state current through inductance will be 

equal to 0.75 A 

2Ω 

u 


13. In passing through a boundary refraction will not take 

place if - 

(A) light is incident normally on the boundary 

(B) the indices of refraction of the two media are 

same 

(C) the boundary is not visible 

(D) angle of incidence is lesser than angle of 

refraction but greater then 

–1⎛ 

µ 

sin 

⎜ 

⎝ µ 

14. A body moves in a circular path of radius R with 

deceleration so that at any moment of time its 

tangential and normal acceleration are equal in 

magnitude. At the initial moment t = 0, the velocity 

of body is v 0 then the velocity of body will be - 

v0 (A) v = at time.t 

⎛ v0t 

⎞ 

1+ ⎜ ⎟ 

⎝ R ⎠ 

–S/ 

0e 

R 

(B) v = v after it has moved S meter 

–SR 

(C) v = v 0 e after it has moved S meter 


R 

D 

⎞ 

⎟ 

⎠ 

This section contains 2 paragraphs; each has 

3 multiple choice questions. (Questions 15 to 20) Each 

question has 4 choices (A), (B), (C) and (D) out of which 

ONLY ONE is correct. Mark your response in OMR 

sheet against the question number of that question. + 4 

marks will be given for each correct answer and –1 

mark for each wrong answer. 

Passage # 1 (Ques. 15 to 17) 

A narrow tube is bent in the form of circle of radius R 

as shown. Two small holes S and D are made in the 

tube at the positions right angles to each other. A 

source placed at S generates a wave of intensity I 0 

which is equally divided into two parts. One part 

travels along the longer path, while the other travels 

along the shorter path. Both the part waves meet at 

point D where a detector is placed. 

S 

15. Maximum intensity produced at D is given by - 

(A) 4I 0 (B) 2I 0 

(C) 3I 0 (D) I 0 

16. The maximum value of wavelength λ to produce a 

maximum at D its given by - 

(A) πR 

(B) 2πR 

πR 

3πR 

(C) 

(D) 

2 

2 

R 

D 

17. The maximum value of wavelength λ to produce a 

minimum at D is given by - 

(A) πR 

(B) 2πR 

πR 

3πR 

(C) 

(D) 

2 

2 


A metal sphere of radius R, carrying charge q 1 is 

surrounded by a thick concentric metal shell (inner 

radius a, outer radius b). The shell carries no net 

charge. 

b 

18. Find the surface charge density σ at r = R, r = a, r = b - 

q q q 

(A) σ R = , σ 

2 a = , σ 

2 b = 

2 

4πR 

4πa 

4πb 

q – q q 

(B) σ R = , σ 

2 a = , σ 

2 b = 

2 

4πR 

4πa 

4πb 

– q q q 

(C) σ R = , σ 

2 a = , σ 

2 b = 

2 

4πR 

4πa 

4πb 

(D) σ R = 

q 

4πR 

2 

, σ a = 

R 

q 

4πa 

2 

a 

, σ b = 

– q 

4πb 

19. The potential at the centre, using infinity at the 

reference point : (potential is zero at infinity) 

1 ⎡ q q ⎤ 

1 ⎡ q q q ⎤ 

(A) 

πε 

⎢ – 

4 

⎥ (B) 

0 ⎣ R a 

⎢ + + ⎥ ⎦ 4πε0 

⎣ R a b ⎦ 

1 ⎡ q ⎤ 

1 ⎡q q q ⎤ 

(C) 

4πε 

⎢ ⎥ (D) 

⎣ R 

⎢ – + ⎥ ⎦ 4πε 

⎣b 

a R ⎦ 

0 

20. Now the outer surface is touched to a grounding wire, 

which lowers its potential to zero. Now the potential 

at the centre : (Assume at infinity also potential is 

zero) 

1 

(A) 

4πε0 

1 

(C) 

4πε 

0 

⎡ q q ⎤ 

⎢ – ⎥ 

⎣ R a ⎦ 

⎡ q ⎤ 

⎢ ⎥ 

⎣ R ⎦ 

(B) 

(D) 

1 

4πε 

0 

0 

1 

4πε 

2 

⎡ q 

⎢ + 

⎣ R 

q 

a 

⎡q q 

⎢ – 

⎣b 

a 

q ⎤ 

+ 

b 

⎥ ⎦ 

This section contains 2 questions (Questions 21, 22). 

Each question contains statements given in two 

columns which have to be matched. Statements (A, B, 

C, D) in Column I have to be matched with statements 

(P, Q, R, S) in Column II. The answers to these 

questions have to be appropriately bubbled as 

illustrated in the following example. If the correct 

matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, 

then the correctly bubbled 4 × 4 matrix should be as 

follows : 

0 

+ 

q 

R 

⎥ ⎦ 

⎤ 


P Q R S 

A P Q R S 

B P Q R S 

C P Q R S 

D P Q R S 

Mark your response in OMR sheet against the question 

number of that question in section-II. + 6 marks will be 

given for complete correct answer and No Negative 

marks for wrong answer. However, 1 mark will be 

given for a correctly marked answer in any row. 

21. Match Column-I with Column-II in the light of 

possibility of occurrence of phenomena listed in 

Column-I using the systems in Column-II 

Column-I 

Column-II 

(A) Interference (P) Non-mechanical waves 

(B) Diffraction (Q) Electromagnetic waves 

(C) Polarisation (R) Visible light waves 

(D) Reflection (S) Sound waves 

22. A satellite is revolving around earth in a circular orbit 

of m radius r 0 with velocity v 0 . a particle of mass is 

projected from satellite in forward direction with 

⎡ 5 ⎤ 

relative velocity v = ⎢ –1⎥ 

v 0 . During subsequent 

⎢⎣ 

4 ⎥⎦ 

motion of particle match the following (assume M = 

mass of earth) 

Column-I 

Column-II 

(A) Magnitude of total energy of 

3GMm 

(P) 

8r0 

Patrticle 

(B) Minimum distance of particle (Q) r 0 

from earth 

(C) Maximum distance of particle (R) 5 

3 

r0 

from eath 

(D) Minimum kinetic energy of 

particle 

(S) 

CHEMISTRY 

5GMm 

8r 0 







1. The equilibrium constant for the reaction in aqueous 

solution – 

H 3 BO 3 + glycerin (H 3 BO 3 – glycerin) is 0.90. 

How many moles of glycerin should be added per 

litre of 0.10 M H 3 BO 3 so that 80% of the H 3 BO 3 is 

converted to the boric acid glycerin complex ? 

(A) 0.08 (B) 4.44 (C) 4.52 (D) 3.6 

2. If optical rotation produced by the compound (i) is 

–30°, then rotation produced by compound (ii) is 

CH 3 

CH 3 

H OH H OH 

(i) 

(ii) 

HO H 

H OH 

CH 3 

CH 3 

(A) + 30° (B) –30° 

(C) zero 

(D) unpredictable 

3. A mixture of CO and CO 2 having a volume of 20 ml 

is mixed with x ml of oxygen and electrically 

sparked. The volume after explosion is (16 + x) ml 

under the same conditions. What would be the 

residual volume if 30 ml of the original mixture is 

treated with aqueous NaOH ? 

• (A) 12 ml (B) 10 ml 

• (C) 9 ml (D) 8 ml 

4. Rutherford’s experiment, which estabilished the 

nuclear model of the atom, used a beam of - 

(A) β–particles, which impinged on a metal foil and 

got absorbed 

(B) γ–rays, which impinged on a metal foil and 

ejected electrons 

(C) helium atoms, which impinged on a metal foil 

and got scattered 

(D) helium nuclei, which impinged on a metal foil 

and got scattered 

5. The correct order of acidic strength is – 

(A) Cl 2 O 7 >SO 2 >P 4 O 10 

(B) CO 2 >N 2 O 5 MgO >Al 2 O 3 

(D) K 2 O >CaO >MgO 

6. A reaction follows the given concentration (C) vs 

time graph. The rate for this reaction at 20 seconds 

will be – 

0.5 

0.4 

0.3 

0.2 

0.1 

0 20 40 60 80 100 

Time/second 

(A) 4 × 10 –3 Ms –1 (B) 8 × 10 –2 Ms –1 

(C) 2 × 10 –2 Ms –1 (D) 7 × 10 –3 Ms –1 

7. The potential of the Daniell cell, 

ZnSO 

Zn 4 CuSO 4 

Cu was reported by Buckbee, 

(1M) (1M ) 

Surdzial, and Metz as 

Eº = 1.1028 – 0.641 × 10 –3 T + 0.72 × 10 –5 T 2 , where 

T is the celcius temperature. Calculate ∆Sº for the 

cell reaction at 25º C – 


(A) – 45.32 (B) – 34.52 

(C) – 25.43 (D) – 54.23 

8. In a hypothetical solid C atoms form CCP lattice with 

A atoms occupying all the Tetrahedral voids and B 

atoms occupying all the octahedral voids. A and B 

atoms are of the appropriate size such that there is no 

distortion in the CCP lattice. Now if a plane is 

cut (as shown) then the cross section would like – 

Plane 

(A) Root mean square speed of molecules 

(B) Mean translational kinetic energy of molecules 

(C) Number density of molecules 

(D) Kinetic energy of molecules 

12. Which of the following samples of reducing agents 

is/are chemically equivalent to 25 mL of 0.2 N 

KMnO 4 , to be reduced to Mn 2+ + H 2 O ? 

(A) 25 mL of 0.2 M FeSO 4 to be oxidized to Fe 3+ 

(B) 50 mL of 0.1 MH 3 AsO 3 to be oxidized to H 3 AsO 4 

(C) 25 mL of 0.2 M H 2 O 2 to be oxidized to H + and O 2 

(D) 25 mL of 0.1 M SnCl 2 to be oxidized to Sn 4+ 

(A) 

C 

B 

C 

A 

B 

B 

B 

CCP unit cell 

A 

C 

B 

C 

(B) 

C 

B 

C 

C 

B 

C 

C C 

B 

C 

13. Which of the following statement is/are correct ? 

(A) [Ni(CO) 4 ] is tetrahedral, paramagnetic, sp 3 

hybridised 

(B) [Ni(CN) 4 ] 2– is square planar, diamagnetic, dsp 2 

hybridised 

(C) [Ni(CO) 4 ] is tetrahedral, diamagnetic, sp 3 

hybridised 

(D) [NiCl 4 ] 2– is tetrahedral, paramagnetic, sp 3 

hybridised 

(C) 

C 

B 

C 

A 

A 

C 

B 

C 

A 

A 

C 

B 

C 

(D) 

C 

B 

C 

C 

B 

C 

C C 

9. The favourable conditions for a spontaneous 

reactions are- 

(A) T ∆S > ∆H, ∆H = ⊕ , ∆S = ⊕ 

(B) T ∆S > ∆H, ∆H = ⊕ , ∆S = Θ 

(C) T ∆S = ∆H, ∆H = Θ , ∆S = Θ 

(D) T ∆S = ∆H, ∆H = ⊕ , ∆S = ⊕ 







wrong answer. 

10. A sample of water has a hardness expressed as 77.5 

ppm Ca 2+ . This sample is passed through an ion 

exchange column and the Ca 2+ is replaced by H + . 

Select correct statement(s) 

(A) pH of the water after it has been so treated is 2.4 

(B) Every Ca 2+ ion is replaced by one H + ion 

(C) Every Ca 2+ ion is replaced by two H + ions 

(D) pH of the solution remains unchanged 

11. Consider a sample of He gas and Ne gas both at 300 

K and 1 atmosphere. Assuming ideal behaviour 

which of the following quantities are equal for two 

samples ? 

B 

C 

14. Consider the reaction 

O 

C – OH 

Na, 

NH 3 ( l) 

⎯⎯⎯⎯ 

→ A 

EtOH 

O3,Me2S 

⎯ ⎯⎯⎯ 

→ B + C 

CH2Cl2 

Identify the correct representation of structure of the 

products - 

(A) A is 

COOH 

(B) The intermediate formed in the conversion of B to 

D is enol 

(C) The structure of C is 

O O 

(D) A can also be formed from the reaction 








D 

∆ 







Pressure 

in atm 

218 

85 

15 

J 

C 

G 

1 

H 

A 

56 

V m (in cm 3 /mol) 

F 

647 K 

B 

673 K 

573 K 

E 

473 K 

In the given figure P-V m isotherm of H 2 O is shown. 

The line (……) represent, the vanderwaals plot for 

H 2 O at 473 K. The vanderwaals constant of H 2 O is 

represented by a and b. 

15. What is the equation of the dotted line (- - - ) 

A I C F B ? 

⎛ dP ⎞ ⎛ 2 

d P ⎞ 

(A) ⎜ ⎟ 

= 0 and ⎜ ⎟ = 0 

⎝ dV 

2 

m 

⎠T 

dV 

⎝ m ⎠ 

a ⎛ 2b 

⎞ 

(B) P = ⎜ ⎟ 

1 

2 + 

V ⎝ Vm 

⎠ 

m 

a ⎛ 2b 

⎞ 

(C) P = ⎜ ⎟ 

1 – 

2 

V ⎝ Vm 

⎠ 

m 

⎛ 2 

d P ⎞ 

(D) ⎜ ⎟ 

2 

dV ⎝ m ⎠ 

T 

= 0 

16 As per the vanderwaals line I H G F (- - - -) which of 

the following section against the behaviour of gas- 

(A) I H 

(B) H G 

(C) G F 

(D) All of the given 

17. For H 2 O which of the following is / are correct- 

(A) For H 2 O, compressibility factor (Z c ) is equal to 

0.23. 

(B) For H 2 O, compressibility factor (Z c ) is lesser than 

0.375 because of stronger intermolecular 

attraction among H 2 O molecules. 

(C) For H 2 O if reduced pressure, reduced volume and 

reduced temperature are 20, 0.6 and 2 

respectively then intermolecular force of 

repulsion predominate over intermolecular 

H-bonding among H 2 O molecules. 

(D) All of the above are correct. 

T 


A useful method to convert oxime to substituted 

amide is Beckmann rearrangement which occurs 

through following steps, 

Ph 

CH 3 

H O 

⎯ − ⎯⎯ 

2 

(II) 

C = N 

→ 

CH 3 –C = N–Ph 

OH 

OH 

⎯ H ⎯→ 

+ 

(I) 

Ph 

C = N 

+ 

Me OH 2 

CH 3 − C 

⊕ H O 

= N − Ph ⎯ ⎯⎯→ 2 

(III) 

O 

(IV) CH 3 –C–NH–Ph 

18. Rate determining step in Beckmann rearrangement is 

(A) I (B) II (C) III (D) IV 

Me 

19. The compound C = N when treated 

Ph 

OH 

with H 2 SO 4 and hydrolysed the products formed are 

(A) CH 3 COOH and PhNH 2 

(B) CH 3 NH 2 and PhCOOH 

(C) PhCH 2 NH 2 and CH 3 COOH 

(D) PhCH 2 COOH and CH 3 NH 2 

20. In the following sequence of reaction 

O 

Ph – C – –CH NH OH PCl 

3 

⎯ ⎯ 2 ⎯⎯ → I ⎯ ⎯⎯ 5 →P 

pH= 

4 − 6 

∆ 

the product P may be 

(A) PhCOOH 

(B) CH 3 – –C–NH 2 

O 

O 

(C) Ph – C – NH – – CH 3 

O 

(D) Ph – C – NH 2 










follows : 


A 

B 

C 

D 

P Q R S 

P Q R 

P Q R 

P Q R 

P Q R 






21. Column –I Column II 

(A) Decomposition (P) 10 t 1/2 

of H 2 O 2 

k308K 

(B) 

(Q) 1 st order 

k 298K 

(C) Arrhenius eq. (R) Temperature coefficient 

(D) t 99.9% 

22. Column-I 

(A) 

(B) 

(C) 

(D) 

Me 

Cl 

Me 

Me 

Cl 

Cl 

Column-II 

Me 

Me 

S 

S 

S 

S 

k 2 E 

(S) log = a 

⎛ T 

k1 

2.303R 

⎜ 

2 – T 

⎝ T1T 

2 

Me 

Me 

Me 

Me 

Me 

Me 

Me 

Me 

(P) Optically active 

(Q) Cis compound 

(R) Trans compound 

(S) Optically inactive 

Me 

Me 

Me 

Me 

Br 

Me 

Br 

Me 

Me 

Me 

Me 

Me 

Me 

Br 

Me 

Br 

Me 

Me 

Cl 

Me 

Me 

Me 

Me 

Cl 

Me 

Cl 

Cl 

Me 

Me 

Me 

1 

⎟ ⎞ 

⎠ 

MATHEMATICS 







1. If α, β are the roots of the equation; 

6x 2 + 11x + 3 =0 then : 

(A) both cos –-1 α and cos –1 β are real 

(B) both cosec –1 α and cosec –1 β are real 

(C) both cot –1 α and cot –1 β are real 


2. 

y ⎛ x 

If for the differential equation y′ = + φ x 

⎟ ⎞ 

⎜ 

⎝ y ⎠ 

the 

x 

general solution is y = 

log | Cx | 

then f (x / y) is given 

by - 

(A) – x 2 / y 2 (B) y 2 / x 2 

(C) x 2 / y 2 (D) – y 2 / x 2 

3. Two flagstaffs stand on a horizontal plane. A and B 

are two points on the line joining their feet and 

between them. the angles of elevation of the tops of 

the flagstaffs as seen from A are 30º and 60º and as 

seen from B are 60º and 45º. If AB is 30m, the 

distance between the flagstaffs in meters is 

(A) 30 + 15 3 (B) 45 + 15 3 

(C) 60 – 15 3 (D) 60 + 15 3 

4. If the probability of choosing an integer k out of 2m 

integers 1, 2, 3, ...., 2m is inversely proportional to 

k 4 (1 ≤ k ≤ 2m), then the probability that chosen 

number is odd, is 

(A) equal to 1/2 (B) less than 1/2 

(C) greater than 1/2 (D) less than 1/3 

5. The line x + y = 1 meets x-axis at A and y-axis at B.P 

is the mid-point of AB (fig.) P 1 is the foot of the 

perpendicular from P to OA; M 1 is that from P 1 to 

OP; P 2 is that from M 1 to OA; M 2 is that from P 2 to 

OP; P 3 is that from M 2 to OA and so on. If P n denotes 

the n th foot of the perpendicular on OA from M n–1 , 

then OP n = 

y 

B 

M 1 

M 2 

P 

O P 3 P 2 P 1 

A 

x 


(A) 1/2 

(B) 1/2 n 

(C) 1/2 n/2 (D) 1/ 2 

6. The sum ∑∑ ( 

0≤ 

i< 

j≤10 

10 

C ) ( j C i ) is equal to 

(A) 2 10 – 1 (B) 2 10 

(C) 3 10 – 1 (D) 3 10 

j 

7. Reflection of the line a z + a z = 0 in the real axis is 

(A) a z + az = 0 

a z 

(B) = a z 

(C) (a + a ) (z + z ) = 0 


8. If g(x) is a polynomial satisfying g(x) g(y) = g(x) + 

g(y) + g(xy) – 2 for all real x and y and g(2) = 5 then 

lim g(x) is - 

x→3 

(A) 9 (B) 25 

(C) 10 

(D) none of these 

9. The domain of definition of 

f(x) = 

⎛ x –1 ⎞ 1 

log 0. 

4 ⎜ ⎟ × 

⎝ x + 5 ⎠ 

2 

x – 36 

is 

(A) (– ∞, 0) ~ {– 6} (B) (0, ∞) ~ {1, 6} 

(C) (1, ∞) ~ {6} (D) [1, ∞)~ {6} 







wrong answer. 

8 ⎡ 1 ⎤ 

10. The lim x 

→0 

⎢ 3 ⎥ (where [x] is greatest integer 

x ⎣ x ⎦ 

function) is 

(A) a nonzero real number 

(B) a rational number 

(C) an integer 

(D) zero 

x x x 

⎛ 

4 2 

sin + cos x x x x x ⎞ 

11. If l = e ⎜ 

cos − sin + cos 

⎟ 

∫ 

dx 

2 2 

x x 

⎝ cos ⎠ 

then l equals - 

(A) e x sin x ⎛ sec x ⎞ 

+ cos x ⎜ x − ⎟ + C 

⎝ x ⎠ 

(B) e x sin x ⎛ cos x ⎞ 

+ cos x ⎜ xsin 

x − ⎟ 

⎝ x ⎠ 

(C) e x sin x ⎛ x sec x ⎞ 

+ cos x ⎜ − ⎟ + C 

⎝ tan x x ⎠ 

(D) xe x sin x+cos x – 

∫ 

e 

xsin 

x+ 

cos x 

⎛ cos x − xsin 

x ⎞ 

⎜1 − 

⎟ dx 

2 2 

⎝ x cos x ⎠ 

x–1 

(log(1 + x) 

– log 2)(3.4 – 3x) 

12. Let f(x) = 

, x ≠1 

1/3 

1/ 2 

{(7 + x) 

– (1 + 3x) 

}sin πx 

The value of f(1) so that f is continuous at x = 1 is 

(A) an algebraic number 

(B) a rational number 

(C) a trance dental number 

(D) – π 

9 log 4e 

13. The solution of y 1 (x 2 y 3 + xy) = 1 is 

(A) 1/x = 2 – y 2 2 

/ 2 

+ C e − y 

(B) the solution of an equation which is reducible to 

linear equation 

(C) 2/x = 1 – y 2 + e –y/2 

2 

/ 2⎛1− 2x 

2 ⎞ 

(D) e y ⎜ + y ⎟ = C 

⎝ x ⎠ 

14. Suppose a, b, c are positive integers and 

f(x) = ax 2 – bx + c = 0 has two distinct roots in 

(0, 1), then - 

(A) a ≥ 5 (B) b ≥ 5 

(C) abc ≥ 25 (D) abc ≥ 250 









At times the methods of co-ordinates becomes 

effective in solving problems of properties of 

triangles. We may choose one vertex of the triangle 

as origin and one side passing through this vertex as 

x-axis. Thus, without loss of generality, we can 

assume that every triangle ABC has a vertex B 

situated at B(0, 0), vertex C at (a, 0) and A as (h, k). 

15. If in ∆ABC, AC = 3, BC = 4, medians AD and BE 

are perpendicular, then area of triangle ABC must be 

equal to 

(A) 7 (B) 11 

(C) 2 2 



16. Suppose the bisector AD of the interior angle A of 

∆ABC divides side BC into segments BD = 4, 

DC = 2. Then we must have 

(A) b > 6 and c < 4 

(B) 2 

(C) 2 


17. If altitudes CD = 7, AE = 6 and E divides BC such 

BE 3 

that = , then c must be 

EC 4 

(A) 2 3 (B) 5 3 (C) 3 (D) 4 3 


Among several applications of maxima and minima 

is finding the largest term of a sequence. Let be 

a sequence. Consider f(x) obtained by replacing x by 

n 

x 

n e.g. let a n = consider f(x) = on [1, ∞] 

n +1 

x 

f ´(x) = > 0 for all x. 

2 

( x +1) 

Hence max f(x) = lim f ( x) 

= 1. 

x→∞ 

18. The largest term of a n = n 2 /(n 3 + 200) is 

(A) 29/453 (B) 49/543 

(C) 43/543 (D) 41/451 

19. The largest term of the sequence 

a n = n/(n 2 + 10) is 

(A) 3/19 (B) 2/13 

(C) 1 (D) 1/7 

x +1 

20. If f(x) is the function required to find largest term in 

Q. 14 then 

(A) f is increase for all x 

(B) f decreases for all x 

(C) f has a maximum at x = 3 400 

(D) f increases on [0, 9] 










follows : 

P Q R S 

A 

B 

C 

D 

P Q R 

P Q R 

P Q R 

P Q R 

S 

S 

S 

S 






21. Column –I Column II 

(A) The period of sin πx + (P) 2 2n – 1 

π x π x 

tan + sin 

2 2 2 + .... + 

sin 

πx 

n 

2 −1 

πx 

+ tan 

2 n 

(B) g(x) = 3 + 4x, the value of 

g n (0) = gog .... o g(0) is 

(C) f(x) = x 3 + 2 n x 2 + bx + c is 

bijection if and only if 

3b ≥ d where d is equal to 

n 

(D) (2 2n – 1) ∑ 

= 

⎛ 

⎜ 

1 

r 

⎝ 2 

r 

0 

( −1) 

r 

(Q) 2 2n 

(R) 2 n 

n C r (S) 2 n + 1 

r r 

3 7 

⎞ 

+ + + .....upto infinity⎟ 

2r 3r 

2 2 

⎠ 

22. Centre of circle 

Column-I 

(A) |z – 2| 2 + |z – 4i| 2 = 20 

(B) 

Column-II 

(P) 1 – i 

z −1 

= 2 (Q) 5/3 + 0i 

z + 1 

(C) z z – (1 + i)z 

– (1 – i) z + 7 = 0 

⎛ z + 3 + 4i 

⎞ 

(D) arg ⎜ ⎟ 

⎝ z + 5 − 2i 

⎠ 

(R) – 4 – i 

(S) 1 + 2i 



Based on New Pattern 

IIT-JEE 2012 

XtraEdge Test Series # 10 

Time : 3 Hours 

Syllabus : 

Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus 

Instructions : 

Section - I 

• Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct 

answer and -1 mark for wrong answer. 

• Question 10 to 14 are multiple choice questions with one or more than one correct asnwer. +4 marks will be 

awarded for correct answer and –1 mark for wrong answer. 

• Question 15 to 20 are passage based single correct type questions. +4 marks will be awarded for correct answer and 

-1 mark for wrong answer.. 

Section - II 

• Question 21 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly 

matched answer and No Negative marks for wrong answer. However, +1 mark will be given for a correctly 

marked answer in any row. 

PHYSICS 







1. A particle is projected towards north with speed 20 

m/s at angle 45º with horizontal. Ball get horizontal 

acceleration of 7.5 m/s 2 towards east due to wind. 

Range of ball is - 

(A) 40 m 

(B) 70 m 

(C) 50 m 

(D) 60 m 

2. A cylinder of mass m 1 is kept over a block of mass m 2 

kept over smooth inclined plane shown in figure. 

Surface between cylinder and block is rough. Friction 

on cylinder- 

m 1 

3. A over head tank of capacity 10 K liter (10000 liter) 

is kept at top of building 15 m high. Water level is at 

depth 5m from ground. Water falls in tank with 

1 

velocity 5 2 m/s. The tank has to filled in hr. If 2 

efficiency of pump is 67.5%, electric power used is 

approximately - 

(A) 4 kW 

(B) 5 kW 

(C) 2 kW 

(D) 2.5 kW 

4. There ia a sphere of radius 'R'. Let E 1 and E 2 be 

gravitational field at distance r 1 and r 2 from centre- 

(A) If r 1 < R < r 2 then E 1 must be less than E 2 

(B) If r 1 < r 2 < R then E 1 must be greater than E 2 

(C) If R < r 1 < r 2 then E 1 must be less than E 2 

(D) If r 1 = R – k and r 2 = R + k (where k < R) E 1 must 

be greater than E 2 

5. A block of mass m is placed at top of frictionless 

wedge of mass 'M' placed on frictionless surface as 

shown in figure. Velocity of block on wedge at the 

time it slips off the wedge is u. Velocity of wedge at 

that instant is - 

m 2 

θ 

(A) is in upward direction 

(B) is in downward direction 

(C) is zero 

(D) will depend on angle of inclination and 

coefficient of friction between cylinder and block 

mu 

(A) M 

(C) 

mu cosθ 

m + M 

θ 

mu 

cos θ 

(B) 

M 

mu cos θ 

(D) 

2m 

+ M 


6. A rectangular plate is kept in y-z plane. Which of the 

following is correct for this plate? 

(A) I z = I x + I y 

(B) I y = I x + I z 

(C) I x = I y + I z 

(D) All of these 

7. A glass of water is to be cooled using an ice-cube. 

For which of following position water will be cooled 

fastest - 

(A) Ice is left floating 

(B) Ice is kept just submerged in water 

(C) Ice is kept bottom of glass 

(D) Water will be cooled at same rate no matter 

where ice is kept 

8. Shape of string carrying transverse wave at t = 0 and 

1 

1 

t = 1 sec is given by y = and y = 

x 2 2 

+ 1 2x 

+ 4x + 3 

respectively, where 'x' is distance in meter. Wave 

velocity is - 

(A) 1 m/s in positive x-direction 

(B) 2 m/s in negative x-direction 

(C) 1 m/s in negative x-direction 

(D) 50 cm/ sec in negative x-direction 

9. A body of mass 200 gm is heated up. Graph shows 

change in temperature as heat is supplied to body. 

Specific heat capacity of body is (in J/kg/ºC) – 

(C) The two rods have same kinetic energy but linear 

kinetic energy of 'B' will be less than that of 'A' 

(D) The kinetic energy of 'B' will depend on the 

distance from centre where the mass hit 

12. Which of the following is true, for a sample of gas 

according to kinetic theory of gases - 

(A) Net velocity of the gas molecules is zero 

(B) Net momentum of the gas molecules is zero 

(C) Net speed of the gas molecules is zero 

(D) Net kinetic energy of gas molecules is zero 

13. A cylinder is floating in a liquid kept in container. 

Coefficient of cubical expansion of cylinder is 'γ'. 

Expansion of liquid and container are negligible. 

Upon increasing temperature - 

(A) Level of liquid in container will increase 

(B) Level of liquid in container will remain same 

(C) Volume of cylinder inside water will increase 

(D) Volume of cylinder inside water will remain 

same 

14. Length of kundt's tube is 1m. When tuning fork is 

vibrated and brought near rod of the kundt's tube, the 

powder keeps on moving. If velocity of sound is 320 

m, frequency of tuning fork cannot be - 

(A) 880 Hz 

(B) 900 Hz 

(C) 960 Hz 

(D) 1040 Hz 

∆H(in ºC) 

30º 

∆H(in kJ) 

10 3 

(A) 

(B) 5 3 × 10 3 

3 

(C) 3 × 10 3 (D) 3 















wrong answer. 

10. A particle is moving along straight line with velocity 

v = t 2 – 3t + 2 m/s. Particle will retard for time t - 

(A) t < 1 (B) 1 < t < 1.5 

(C) 1.5 < t < 2 (D) t > 2 

11. Two identical rods P and Q are placed on frictionless 

horizontal surface. Two identical mass hit two rods 

separately and comes at rest after hitting. Mass hits 

rod 'P' at its centre while rod 'Q' is hit by mass a little 

distance away from centre - 

(A) Rod P and Q will have same speed 

(B) Q will have greater kinetic energy 

2h 

h 

A cylindrical container of cross-sectional area 'A' and 

height '5h' is kept at height '2h' above ground. It 

contains a liquid of density '2ρ' till height 'h'. The 

cylinder is filled with light piston as show in figure. 

15. Where should a hole is made in the container so that 

liquid, strikes ground farthest ? 

(A) At bottom of container 

(B) At height h/3 above bottom of container 

(C) At height h/2 above bottom of container 

(D) Liquid will strike ground at same distance 

irrespective of position of hole 


16. A block of mass M is kept over piston and hole is This section contains 2 questions (Questions 21, 22). 

36 32 

(A) mg (B) mg 

(in rad/sec) 

17 

5 

40 

39 44 

(B) Time taken (in sec) (Q) 

(C) mg (D) mg 3π 

17 

17 

π 

(in wrapping meter) 20 

made at a distance 'h/2' from piston. Velocity of 

efflux is - 


columns which have to be matched. Statements 

(A) gh (B) 

⎛ M ⎞ 

(A, B, C, D) in Column I have to be matched with 

⎜ + h g 

A 

⎟ 

statements (P, Q, R, S) in Column II. The answers to 

⎝ ρ ⎠ 

these questions have to be appropriately bubbled as 

(C) 


⎛ M ⎞ 

⎛ 2M 

⎞ 

⎜ + h⎟g 

(D) ⎜ + h⎟g 


⎝ 2ρA 

⎠ 

⎝ 2ρA 

⎠ 


follows : 

17. A liquid of density 'ρ' is poured in container till 

P Q R S 

height h above container. Velocity of efflux from 

A P Q R S 

hole at distance 'h/2' below piston is - 

B P Q R S 

3 

(A) 3 gh 

(B) gh 

C P Q R S 

2 

D P Q R S 

(C) 2 gh 

(D) gh 





C 



A 

B 

21. Column-I contains molar heat capacity for certain 

l 

process for an ideal gas and column II contains 

l 

corresponding processes. α,β and a are constant and 

γ is adiabatic exponent. Match the correct one : 

Column-I 

Column-II 

D 

α 

A T- shape iron frame of mass m free to rotate in 

⎛ αT 

⎞ 

(A) C = (P) V exp ⎜ – ⎟ = const. 

vertical plane about one of its end as shown in figure. 

T ⎝ R ⎠ 

The two rods AB and CD making T-shape are 

identical. Initially the frame is in the position shown (B) C = C V + αT (Q) V – aT = const. 

in figure. The frame is left to rotate freely in vertical 

plane. 

(C) C = C V + βV (R)PV γ ⎛ α( 

γ –1) ⎞ 

exp ⎜ – ⎟ = const. 

18. Moment of inertia of frame about axis of rotation - 

⎝ PV ⎠ 

2 

2 

2ml 

17ml 

(A) 

(B) 

⎛ R ⎞ 

3 

12 

(D) C = C V + aP (S) T exp ⎜ ⎟ = const. 

⎝ βV ⎠ 

2 

2 

17ml 

5ml 

(C) 

(D) 

24 

12 

22. A horizontal plane support a vertical cylinder of 

radius 20 cm and a disk of mass 2 kg is attached to 

19. Angular acceleration of frame when rod AB is 

making angle 'θ' with vertical is - 

the cylinder by a horizontal thread of length π/5 m. 

The disk can move frictionlessly on the table. An 

18sin 

θ 24sin 

θ initial velocity 1 m/s is imparted to the disk. Consider 

(A) . g (B) . g 

17l 

17l 

π 

a situation when m length of string is wrapped 

12sin 

θ 9sin 

θ 20 

(C) . g (D) . g 

5l 

2l 

on cylinder. 

Column-I 

Column-II 

20. Force due to axis on frame when frame becomes 

2 

π 

vertical - 

(A) Angular velocity of disk (P) 10 


(C) Tension in string (in N) 

20 

(R) 

3π 

(D) Time taken (in sec) after 

7π 

2 

(S) 160 

which disk will hit cylinder 

CHEMISTRY 







1. If 0.5 mol of BaCl 2 is mixed with 0.2 mol of Na 3 PO 4 , 

the maximum number of moles of Ba 3 (PO 4 ) 2 that can 

be formed is - 

(A) 0.7 (B) 0.5 

(C) 0.30 (D) 0.10 

2. If the threshold frequency of a metal for photoelectric 

effect is v 0 then which of the following will not 

happen ? 

(A) If frequency of the incident radiation is v 0 , the 

kinetic energy of the electrons ejected is zero. 

(B) If frequency of the incident radiation is v, the 

kinetic energy of the electrons ejected will be 

hv – hv 0 

(C) If frequency is kept same at v but intensity is 

increased, the number of electrons ejected will 

increase. 

(D) If frequency of the incident radiation is further 

increased, the number of photo-electrons ejected 

will be increase. 

3. Which of the following is violation of Pauli's 

exclusion principle ? 

(A) 

(B) 

(C) 

(D) 

4. The IUPAC name of compound 

HO – C = O CH 3 

NH 2 – C ==== C ––– C – H is – 

5. If optical rotation produced by the compound (i) is 

–30°, then rotation produced by compound (ii) is 

CH 3 

CH 3 

H OH H OH 

(i) 

(ii) 

HO H H OH 

CH 3 

CH 3 

(A) + 30° (B) –30° 

(C) zero 

(D) unpredictable 

6. 16 mL of a gaseous aliphatic C n H 3n O m was mixed 

with 60 mL O 2 and sparked, the gas mixture on 

cooling occupied 44 mL. After treatment with KOH 

solution the volume of gas remaining was 12 mL. 

Formula of compound is - 

(A) C 2 H 6 O 

(B) C 3 H 8 O 

(C) CH 4 O 

(D) None of the above 

7. Most stable free radical is 

CH 3 

(A) 

(C) 

(B) 

(D) 

8. At constant pressure P, A dissociates on heating 

according to the equation 

A(g) B(g) + C(g) 

The equilibrium partial pressure of A at T K is 1/9 P, 

the equilibrium K p at TK is 

8 64 16 

(A) P (B) P (C) P (D) 9 P 

9 9 9 

9. Calculate the pH of 6.66 × 10 –3 M solution of 

Al(OH) 3 . Its first dissociation is 100% where as 

second dissociation is 50% and third dissociation is 

negligible. 

(A) 2 (B) 12 (C) 11 (D) 13 







wrong answer. 

10. The IUPAC name of the following compound is - 

OH 

NH 2 

Cl 

(A) 2, 3 diamino-4-chloro-2-pentenoic acid 

(B) 4-chloro-3, 3-diamino-2-pentenoic acid 

(C) 3, 3–diamino-3-chloro-pentenoic acid 

(D) All of the above 

Br 

CN 

(A) 3-Bromo-3-cyano phenol 

(B) 3-Bromo-5-hydroxy benzonitrile 

(C) 3-Cyano-3-hydroxybromo benzene 

(D) 5-Bromo-3-hydroxy benzonitrile 


11. Which of the following is/are correct regarding the 

periodic classification of elements ? 

(A) The properties of elements are the periodic 

function of their atomic number 

(B) Non metals are lesser in number than metals 

(C) The first ionization energies of elements in a 

period do not increase with the increase in 

atomic numbers 

(D) For transition elements the d-subshells are filled 

with electrons monotonically with the increase 

in atomic number 

12. Identify the correct statements - 

H 3 C 

CH 3 

(A) The compound 

fails to undergo 

COOH O 

decarboxylation 

(B) A Grignard reagent can be successfully made 

from the following dibromide Br 

Br 

(C) Cyclopentan –1, 2– dione exists almost 100% in 

the enol form whereas diacetyl (CH 3 COCOCH 3 ) 

can exist in the keto form as well as the enol form 

(D) Among the following resonance structure given 

below, (ii) will be the major contributor to the 

resonance hybrid. 

13. Which of the following are possible products from 

aldol condensation of 6-oxoheptanal ? 

O CH 3 

O H 

C 

C 

(A) 

(C) 

O 

CH 3 

(B) 

(D) 

CH 3 

14. Which of the metal is/are used in flash bulbs? 

(A) Be 

(B) Mg 

(C) Ca 

(D) Ba 








O 


According to molecular orbital theory all atomic 

orbital combine to form molecular orbital by LCAO 

(Linear combination of atomic orbital) method. When 

two atomic orbitals have additive (constructive) 

overlapping, they form bonding molecular orbital 

(BMO) which have lower energy than atomic orbitals 

whereas when atomic orbitals overlap subtractively, 

higher energy antibonding molecular orbitals (AMO) 

are formed. Each M.O. occupies two electrons with 

opposite spin. Distribution of electrons in M.O. 

follows Aufbau principle as well as Hund's rule. 

M.O. theory can successfully explain magnetic 

behaviour of molecules. 

15. Which of the following is/are not paramagnetic ? 

(A) NO (B) B 2 

(C) CO (D) O 2 

16. Bond strength increases when 

(A) bond order increases 

(B) bond length increases 

(C) antibonding electrons increases 

(D) bond angle increases 

17. 

2– 

O 2 will have 

(A) bond order equal to H 2 and diamagnetic 

(B) bond order equal to H 2 but paramagnetic 

(C) bond order equal to N 2 and diamagnetic 

(D) bond order higher than O 2 


C 

(Resolvable) 

HBr,Peroxide 

A(C 6 H 11 Br) 

Decolourise Br 2 water and connot 

be resolved 

alc. KOH 

a single possible 

product 

E 

HBr,R 2 O 2 

(Excess) 

G 

Resolvable 

F 

nonresolvable 

HBr 

B 

(Non-resolvable) 

Zn,Heat 

D(C 6 H 12 ) 

O 3 ;Zn,H 2 O 

O 

|| 

CH 3 –C–CH 3 


18. Organic compound 'A' is – 

(A) 

(C) 

CH 2 Br 

CH 2 Br 

(B) 

(D) 

Br 

Br 

19. The resolvable orgainc compound 'C' is – 

(A) 

Br 

(C) Br 

CH 2 Br 

Br 

(B) 

(D) 

• 

• 20. The resolvable organic compound, G is – 

• 

CH 3 

CH 2 CH 3 

Br CH 3 H Br 

(A) CH 3 Br (B) Br H 

CH 3 

CH 2 CH 3 

CH 2 Br 

CH 3 H 

(C) H CH 3 

CH 2 Br 

Br 

H 

(D) Br 

* 

Br 

Br 

CH 

Br 

CH 3 CH 3 

CH 3 

Br 

H 










follows : 

A 

B 

C 

D 

P Q R S 

P Q R 

P Q R 

P Q R 

P Q R 






S 

S 

S 

S 

21. Match the following : 

Column -I 

(A) Compound show (P) 

Geometrical 

isomerism 

(B) Compound is chiral (Q) 

(C) Compound having 

plane of symmetry 

(D) Compound having 

centre of symmetry 

(R) 

(S) 

Column-II 

Me 

Me 

H 

Me 

Me 

C = C 

H 

Me 

H 

= C 

Me 

H 

Me 

H 

22. Column-I Column-II 

(Ionic species) 

(Shapes) 

+ 

(A) XeF 5 (P) Tetrahedral 

– 

(B) SiF 5 (Q) Square planar 

+ 

(C) AsF 4 (R) Trigonal bipyramidal 

– 

(D) ICl 4 (S) Square pyramidal 

MATHEMATICS 







1. If sinx + sin 2 x + sin 3 x = 1, then 

cos 6 x– 4cos 4 x + 8 cos 2 x is equal to - 

(A) 0 (B) 2 (C) 4 (D) 8 

2. A line meets the coordinate axes in A and B. A circle 

is circumscribed about the triangle OAB. If m and n 

are the distances of the tangent to the circle at the 

origin from the points A and B respectively, the 

diameter of the circle is 

(A) m(m + n) (B) m + n 

(C) n(m + n) (D) (1/2) (m + n) 

3. The line joining A(b cos α, b sin α) and B (a cos β, 

a sin β) is produced to the point M(x, y) so that 

α + β α + β 

AM : MB = b : a, then x cos + y sin 

2 2 

= 

(A) –1 (B) 0 

(C) 1 (D) a 2 + b 2 

4. The equation 

x + 3 − 4 x −1 

+ x + 8 − 6 x −1 

= 1 has 

(A) no solution (B) only one solution 

(C) only two solution (D) more than two solutions 

H 

Me 


5. Equation of the line of shortest distance between the 

x y z x − 2 y −1 

z + 2 

lines = = and = = is - 

2 − 3 1 3 − 5 2 

(A) 3(x – 21) = 3y + 92 = 3z – 32 

(B) 

(C) 

(D) 

x − ( 62 / 3) y − 31 z + (31/ 3) 

= = 

1/ 3 1/ 3 1/ 3 

x − 21 y − (92 / 3) z + (32 / 3) 

= = 

1/ 3 1/ 3 1/ 3 

x − 2 y + 3 z −1 

= = 

1/ 3 1/ 3 1/ 3 

6. The set of all x satisfying the equation 

2 2 

log3 x + (log3 

x) 

−10 

x 

= 1/x 2 is 

(A) {1, 9} (B) {1, 9, 1/81} 

(C) {1, 4, 1/81} (D) {9, 1/81} 

7. In a certain test there are n questions. In this test 2 k 

students gave wrong answers to at least (n – k) 

questions, where k = 0, 1, 2, ...... , n. If the total 

number of wrong answers is 4095, then value of n is 

(A) 11 (B) 12 

(C) 13 (D) 15 

8. Equation of the locus of the pole with respect to the 

ellipse 

2 2 

x y 

2 + 

a b 2 

= 1, of any tangent line to the 

2 2 

x y 

auxiliary circle is the curve 

4 + 

a b 4 

= λ 2 where 

(A) λ 2 = a 2 (B) λ 2 = 1/a 2 

(C) λ 2 = b 2 (D) λ 2 = 1/b 2 

9. The number of values of x ∈[0, nπ], n ∈ I that satisfy 

log |sinx| (1 + cos x) = 2 is 

(A) 0 (B) n (C) 2n (D) none 







wrong answer. 

10. If x 2 + 2hxy + y 2 = 0 represents the equations of the 

straight lines through the origin which make an angle 

α with the straight line y + x = 0, then 

(A) sec 2α = h (B) cos α = 

(C) 2 sin α = 

1+ h 

h 

(D) cot α = 

1+ h 

2h 

1+ 

h 

h −1 

11. If the numerical value of tan (cos –1 (4/5) + tan –1 (2/3) 

is a/b then 

(A) a + b = 23 (B) a – b = 11 

(C) 3b = a + 1 (D) 2a = 3b 

⎡1 

1 ⎤ ⎡1 

2 ⎤ 

12. Let E = ⎢ + ⎥ + ⎢ + ⎥ + ... upto 50 terms, then - 

⎣3 

50⎦ 

⎣3 

50 ⎦ 

(A) E is divisible by exactly 2 primes 

(B) E is prime 

(C) E ≥ 30 

(D) E ≤ 35 

13. If PQ is a double ordinate of the hyperbola 

2 

x y 

= 1 such that OPQ is an equilateral 

2 

a b 

triangle, O being the centre of the hyperbola. Then 

the eccentricity e of the hyperbola, satisfies 

(A) 1 < e < 2/ 3 (B) e = 2/ 3 

(C) e = 3 /2 (D) e > 2/ 3 

– 2 

2 

14. If z 1 , z 2 , z 3 , z 4 are the vertices of a square in that order, 

then 

(A) z 1 + z 3 = z 2 + z 4 

(B) |z 1 – z 2 | = |z 2 – z 3 | = |z 3 – z 4 | = |z 4 – z 1 | 

(C) |z 1 – z 3 | = |z 2 – z 4 | 

(D) (z 1 – z 3 )/(z 2 – z 4 ) is purely imaginary 









In ∆ABC, a = 14, b = 15, c = 13, P be a point with in 

the triangle such that ∠PAB = ∠PBC = ∠PCA = α 

and tan α = n 

m , where m and n are relatively prime 

positive integers. Let PA = x, PB = y, PC = z 

15. The area of triangle ABC is 

(A) 2 

1 sin α (cx + ay + bz) (B) 2 

1 (x 2 + y 2 + z 2 ) tan α 

(C) 2 

1 (xy + yz + zx) 


16. tan α must be equal to 

∆ 

2∆ 

(A) 

(B) 

2 2 2 

2 2 2 

a + b + c a + b + c 

4∆ 

(C) 


2 2 2 

a + b + c 


17. m + n must be equal to 

(A) 461 (B) 463 (C) 465 (D) 365 


A(3, 7) and B(6, 5) are two points. 

C : x 2 + y 2 – 4x – 6y – 3 = 0 is a circle. 

18. The chords in which the circle C cuts the members of 

the family S of circles through A and B are 

concurrent at 

(A) (2, 3) (B) (2, 23/3) 

(C) (3, 23/2) (D) (3, 2) 

19. Equation of the member of the family S which bisects 

the circumference of C is 

(A) x 2 + y 2 – 5x – 1 = 0 

(B) x 2 + y 2 – 5x + 6y – 1 = 0 

(C) x 2 + y 2 – 5x – 6y – 1 = 0 

(D) x 2 + y 2 + 5x – 6y – 1= 0 

20. If O is the origin and P is the centre of C, then 

difference of the squares of the lengths of the 

tangents from A and B to the circle C is equal to 

(A) (AB) 2 (B) (OP) 2 

(C) |(AP) 2 – (BP) 2 | (D) None of these 










follows : 

P Q R S 

A P Q R S 

B P Q R S 

C P Q R S 

D P Q R S 






21. Value of x when 

Column-I 

Column-II 

(A) 5 2 5 4 5 6 ... 5 2x = (0.04) –28 (P) 3 log 3 5 

⎛ 1 1 1 ⎞ 

(B) x 2 log 5 

⎜ + + + ... ⎟ 

= 

⎝ 4 8 16 

( 0.2) 

⎠ 

(Q) 4 

log 

2.5 

⎛ 1 1 1 ⎞ 

⎜ + + + ... ⎟ 

3 

2 3 

3 3 

(C) x = ( 0.16) 

⎝ 

⎠ 

(R) 2 

(D) 3 x–1 + 3 x–2 + 3 x–3 + ... (S) 7 

⎛ 2 1 1 ⎞ 

= 2 ⎜5 

+ 5 + 1+ 

+ + ... ⎟ 

2 

⎝ 5 5 ⎠ 

22. If a and b are two units vectors inclined at angle α to 

each other then 

Column –I 

Column-II 

(A) |a + b| < 1 if (P) 

2π < α < π 

3 

(B) |a – b| = |a + b| if (Q) π/2 < θ ≤ π 

(C) |a + b| < 2 (R) α = π/2 

(D) |a – b| < 2 (S) 0 ≤ θ < π/2 

Interesting Science Facts 

• The dinosaurs became extinct before the Rockies 

or the Alps were formed. 

• Female black widow spiders eat their males after 

mating. 

• When a flea jumps, the rate of acceleration is 20 

times that of the space shuttle during launch. 

• The earliest wine makers lived in Egypt around 

2300 BC. 

• If our Sun were just inch in diameter, the nearest 

star would be 445 miles away. 

• The Australian billy goat plum contains 100 

times more vitamin C than an orange. 

• Astronauts cannot belch - there is no gravity to 

separate liquid from gas in their stomachs. 

• The air at the summit of Mount Everest, 29,029 

feet is only a third as thick as the air at sea level. 

• One million, million, million, million, millionth 

of a second after the Big Bang the Universe was 

the size of a …pea. 

• DNA was first discovered in 1869 by Swiss 

Friedrich Mieschler. 

• The molecular structure of DNA was first 

determined by Watson and Crick in 1953. 

• The thermometer was invented in 1607 by 

Galileo. 

• Englishman Roger Bacon invented the 

magnifying glass in 1250. 


MOCK TEST-3 

CBSE BOARD PATTERN 

CLASS # XII 

SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS 

Solutions published in same issue 

General Instructions : Physics & Chemistry 

• Time given for each subject paper is 3 hrs and Max. marks 70 for each. 

• All questions are compulsory. 

• Marks for each question are indicated against it. 

• Question numbers 1 to 8 are very short-answer questions and carrying 1 mark each. 

• Question numbers 9 to 18 are short-answer questions, and carry 2 marks each. 

• Question numbers 19 to 27 are also short-answer questions, and carry 3 marks each. 

• Question numbers 28 to 30 are long-answer questions and carry 5 marks each. 

• Use of calculators is not permitted. 

General Instructions : Mathematics 

• Time given to solve this subject paper is 3 hrs and Max. marks 100. 

• All questions are compulsory. 

• The question paper consists of 29 questions divided into three sections A, B and C. 

Section A comprises of 10 questions of one mark each. 

Section B comprises of 12 questions of four marks each. 

Section C comprises of 7 questions of six marks each. 

• All question in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. 

• There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and 

2 question of six marks each. You have to attempt only one of the alternatives in all such questions. 

• Use of calculators is not permitted. 

PHYSICS 

1. Distinguish between ‘point to point’ and ‘broadcast’ 

communication modes. Give one example of each. 

2. Two identical prisms made of the same material 

placed with their bases on opposite sides (of the 

incident white light) and faces touching (or parallel) 

neither deviate nor disperse. Can this arrangement 

produce a parallel displacement of the beam ? 

3. What is the formula for the magnifying power of a 

compound microscope ? 

4. Sketch two equipotential surfaces for a point charge. 

5. What are superconductors ? 

6. Draw a labelled diagram of Hertz' experiment for 

producing E.M. waves. 

7. What is the phase difference between voltage and 

current in a series LCR circuit at resonance 

connected with an AC source ? 

8. Explain the difference between ‘Hard’ and ‘Soft’ 

X-rays. 

9. The given graph show the variation of photo electric 

current (I) with the applied voltage (V) for two 

different materials and for two different intensities 

of the incident radiations. Identify the pairs of 

curves that correspond to different materials but 

same intensity of incident radiations. 

I 

1 

3 

2 4 

10. Four nuclei of an element fuse together to form a 

heavier nucleus. If the process is accompanied by 

release of energy, which of the two-the parent or the 

daughter nucleus would have a higher binding 

energy/nucleon ? 

11. Zener diodes have higher dopant densities as 

compared to ordinary p-n junction diodes. How does 

it effects the 

(i) Width of the depletion layer ? 

(ii) Junction field ? 

V 


12. Oil floating on water looks coloured due to 

interference of light. What should be the 

approximate thickness of the film for such effects to 

be visible ? 

13. Draw a graph showing the variation of intensity with 

angle in a single slit diffraction experiment. 

14. A parallel plate air filled capacitor has capacitance 5 

µF. If plate separation made twice and whole space 

is filled with medium, capacitance becomes 20 µF. 

Find dielectric constant of the medium. 

15. Suppose you have two bars of identical dimensions, 

one made of paramagnetic substance and the other 

of diamagnetic substance. If you place these bars 

along a uniform magnetic field, show 

diagramatically, what modifications in the field 

pattern would take place in each case. 

16. In a plane e.m. wave, the electric field oscillates 

with a frequency of 2 × 10 10 s –1 and an amplitude of 

40 Vm –1 . (i) What is the wavelength of the wave and 

(ii) What is the energy density due to the electric 

field ? 

17. A solenoid has self-inductance 2 mH and current in 

it is 5 amp. Find magnetic energy stored in it. 

18. Find power factor of the adjacent circuit. 

L = 30 mH 

R = 4 Ω 

~ 

V = 200 2 sin 100 t 

19. Experimental observations have shown that X-rays 

(i) travel in vacuum with a speed of 3 × 10 8 ms –1 , 

(ii) exhibit the phenomenon of diffraction and can 

the polarized. 

What conclusion can be drawn about the nature of 

X-rays from each of these observations ? 

20. A radioactive material is reduced to 16 

1 

of its 

23. The self inductance of a solenoid is 5 mH and 

current flowing in it depends on time t as i = t 2 . 

(where i → In Amp., t → In second). Find induced 

emf in it at t = 4 s. 

24. Derive an expression for magnetic field inside a 

long solenoid. 

25. What is Wheatstone bridge ? Deduce the condition 

for which Wheatstone bridge is balanced. 

26. Explain the differences between diamagnetic, 

paramagnetic and ferromagnetic substances. 

27. Describe the method to obtain Reverse Bias 

characterstics of a P-N junction diode. Define 

reverse resistance. Draw necessary circuit diagram 

and also the reverse characterstic curve. 

28. A student has to study the input and output 

characteristics of a n-p-n silicon transistor in the 

Common Emitter configuration. What kind of a 

circuit arrangement should she use for this purpose ? 

Draw the typical shape of input characteristics likely 

to be obtained by her. What do we understand by the 

cut off, active and saturation states of the transistor? 

In which of these states does the transistor not 

remain when being used as a switch ? 

OR 

Input signals A and B are applied to the input 

terminals of the ‘dotted box’ set-up shown here. Let 

Y be the final output signal from the box. 

Draw the wave forms of the signals labelled as 

C 1 and C 2 within the box, giving (in brief) the 

reasons for getting these wave forms, Hence draw 

the wave form of the final output signal Y. Give 

reasons for your choice. 

What can we state (in words) as the relation between 

the final output signal Y and the input signals A and 

B ? 

A u1 0 1 2 3 4 

original amount in 4 days. How much material 

should one begin with so that 4 × 10 –3 kg of the 

material is left after 6 days. 

21. Why are apertures of camera lenses so small while 

the apertures of telescopes are as large as feasible ? 

22. In adjacent circuit, if current in 5 Ω resistance is 

zero, find resistance R. 

R 

1Ω 6Ω 

10Ω 

5Ω 

20Ω 

A 

B 

0 1 2 3 4 

B 

C 1 

12V 

B 

C 2 


29. Draw a labelled ray diagram of an astronomical 

telescope. Write mathematical expression for its 

magnifying power. How does the magnifying power 

get affected on increasing the aperture of the 

objective lens and why ? 

30. Derive an expression for the energy density of a 

capacitor. 

OR 

An electric flux of –6 × 10 3 Nm 2 /C passes normally 

through a spherical Gaussian surface of radius 

10 cm, due to a point charge placed at the centre. 

(i) What is the charge enclosed by the Gaussian 

surface ? 

(ii) If the radius of the Gaussian surface is doubled, 

how much flux would pass through the surface ? 

CHEMISTRY 

1. Which of the following lattices has the highest 

packing efficiency (i) simple cubic (ii) body 

centered cubic and (iii) hexagonal close packed 

lattice ? 

2. What is meant by 'specific surface area' of a solid ? 

3. Give the IUPAC name of Li [AlH 4 ] 

4 What is formula of siderite ore ? 

5. Name the monomer units of Bakelite 

6. What are antiseptics. Give two example : 

7. What are basic amino acids. Give a example ? 

8. Arrange the following in the order of their 

increasing reactivity in nucleophilic substitution 

reactions : 

CH 3 F, CH 3 I, CH 3 Br, CH 3 Cl 

9. The half life for radioactive decay of 14 C is 5730 y. 

An archaeological artefact contained wood had only 

80% of the 14 C in a living tree. Estimate the age of 

the sample. 

10. What is the effect of temperature on the rate 

constant of reaction ? How can this temperature 

effect on rate constant ? Explain using collision 

theory ? 

11. Explain the following :- 

(i) S.H.E.; (ii) Kohlrausch's law 

12. Why Actinides show much higher oxidation states 

as compared to Lanthanides ? 

13. How group I radicals like Ag + and Hg 2 2+ are 

seperated by complex formation with NH 4 OH ? 

14. What is Roasting ? 

15. Give the structure and monomer units of 

biodegradable polymer PHBV ? 

16. Name the purine and pyrimidine bases in DNA and 

RNA . 

17. What are detergents. Give a example of Cationic & 

Anionic detergents 

18. A sweet smelling organic compound 'A' is slowly 

oxidised by air in the presence of light to a highly 

poisonous gas. On warming with silver powder, it 

forms gaseous substance 'B' which is also formed by 

the action of calcium carbide on water. Identify 'A' 

and 'B' and write the equations of the reactions 

involved . 

19. (a) Gold (atomic radius = 0.144 nm) crystallises in a 

face centred unit cell. What is the length of a side 

of the cell ? 

(b) Classify each of the following as being either a 

p-type or an n-type semiconductor 

(i) Ge doped with In 

(ii) Si doped with As. 

20. Two liquids A and B form ideal solution at 323 K. A 

liquid mixture containing one mole of A and two 

moles of B has a vapour pressure of 250 bar. If one 

more mole of A is added to the solution, the vapour 

pressure increases to 300 bar. Calculate the vapour 

pressures of liquids A and B at 323 K. 

21. (a) In which of the following does adsorption take 

place and why ? 

(i) Silica gel placed in the atmosphere saturated 

with water. 

(ii) Anhydrous CaCl 2 placed in the atmosphere 

saturated with water. 

(c) Give an example of shape-selective catalysis 

22. Explain the order of basic character in hydrides of 

nitrogen family ? 

23. Give structure of Cr 2 O 7 2– ? 

24. How will you convert ? 

(i) Phenol to p-hydroxyazobenzene 

(ii) Ethyl alcohol to methyl alcohol. 

25. Write the IUPAC name of the following : 

CH 3 – O – C (CH 3 ) 3 

26. Give a suitable colour reaction test to distinguish 

between 

(i) 2-Pentanone and 3-Pentanone 

(ii) Acetone and acetaldehyde ? 

27. An organic compound A(C 3 H 6 O) is resistant to 

oxidation but forms compound B(C 3 H 8 O) on 

reduction which reacts with HBr to form the 

bromide (C). C forms a Grignard reagent which 

reacts with A to give D (C 6 H 14 O). Give the 

structures of A, B, C and D and explain the reactions 

involved. 


28. (a) What are ideal and non-ideal solutions ? What 

⎡ a + b 2 ⎤ ⎡6 

2⎤ 

type of non-idealities are exhibited by 8. If ⎢ ⎥ = 

cyclohexane-ethanol and acetone-chloroform 

⎣ 5 ab 

⎢ ⎥ find a, b. 

⎦ ⎣5 

8 ⎦ 

x y z + + = 3. 2 

p q r x + 4 

18. Evaluate 

∫ 

dx 

4 

x + 16 

mixtures ? Give reasons for your answers. 

9. Find matrix X and Y if 

(b) A solution containing 30 g of a nonvolatile 

⎡5 

2⎤ 

⎡3 

6 ⎤ 

solute exactly in 90g water has a vapour pressure X + Y = ⎢ ⎥ , X –Y= 

of 2.8 kPa at 298 K. Further 18 g of water is then 

⎣0 

9 

⎢ ⎥ ⎦ ⎣0 

−1 ⎦ 

added to solution, the new vapour pressure 

10. Using determinant, find k so that points 

becomes 2.9 kPa at 298 K. Calculate, 

(k, 2 –2k), (–k + 1, 2k) and (– 4 – k, 6 –2k) are 

(i0 Molecular mass of the solute; 

collinear. 

(ii) Vapour pressure of water at 298 K. 

29. Give structure of :- 

(a) Hypophosphorus acid (b) Pyrophosphoric acid 

Section B 

(c) Dithionic acid (d) Marshall acid 

(e) Hypophosphoric acid 

11. In two successive throws of a pair of dice, determine 

the probability of getting a total of 8, each time. 

30. (a) An optically inactive compound (A) having 

molecular formula C 4 H 11 N on treatment with HNO 2 

gave an alcohol (B). (B) on heating at 440 K gave an 

alkene (C). (C) on treatment with HBr gave an 

optically active compound (D) having the molecular 

12. If f : R → R is given by 

f(x) = sin 2 x + sin 2 (π/3 + x) + cos x . cos (π/3 + x) ∀ 

x ∈ R. g : R → R be such that g(5/4) = 1 then prove 

that gof is constant function. 

formula C 4 H 9 Br. IdentifyA, B, C and D and write 

⎧ 

2 

down their structural formulae. Also write equations 

x + ax + b , 0 ≤ x < 2 

⎪ 

involved. 

13. f(x) = ⎨ 3x 

+ 2 , 2 ≤ x ≤ 4 

(b) Explain why Alkyl amines are stronger bases 

⎪ 2ax 

+ 5b 

, 4 < x ≤ 8 

than arylamines. 

⎩ 

f(x) is continuous on [0, 8] then find a, b. 

MATHEMATICS 

Section A 

14. y = tan –1 dy 

(sec x + tan x) then find ; dx 

π π 

where – < x < . 2 2 

1. Show that relation R on the set A = {1, 2, 3} given 

⎡ 2 

2 

by R = {(1, 2), (2, 1)} is symmetric but neither 

15. Differentiate tan –1 1+ 

x − 1− 

x 

reflexive nor transitive. 

⎥ ⎥ ⎤ 

⎢ 

w.r.t,cos –1 x 2 

⎢ 2 

2 

⎣ 1+ 

x + 1− 

x ⎦ 

2. If x 2/3 + y 2/3 = a 2/3 dy 

then find . 

16. The two equal sides of an isosceles triangle with 

dx 

fixed base b are decreasing at the rate of 

3. 

3 logsin x 

Evaluate 

∫ 

cos x e dx . 

3 cm/sec. How fast is the area decreasing when the 

two sides are equal to the base. 

4. Solve : (x + y) 2 dy = a 

2 

dx 

OR 

Use lagrange's Mean Value theorem to determine a 

point P on the curve y = x − 2 where the tangent is 

5. The projection of a vector on the coordinate axes are parallel to the chord joining (2, 0) and (3, 1). 

6, –3, 2. Find its length and direction cosines. 

6. Find the values of x for which the angle between the 

vectors a r = 2x 2 î + 4x ĵ + kˆ and b r 2x 

17. Evaluate 

= 7 î –2 ĵ + x kˆ is ∫ 

dx 

2 4 

1− 

x − x 

obtuse. 

OR 

7. A plane meets the coordinate axes in A, B, C such 

2 + sin x x / 2 

Evaluate : 

that the centroid of triangle ABC is the point ∫ 

.e .dx 

1 + cos x 

(p, q, r). Show that the equation of the plane is 


dy 

19. Solve : –2y = cos 3x. 

dx 

20. For any two vectors a r and b r , show that 

(1 + | a r | 2 ) (1 + | b r | 2 ) = {(1 – a r . b r )} 2 

+ | a r + b r + ( a r × b r )| 2 

21. Find the foot of the perpendicular from the point 

x + 3 y −1 z + 4 

(0, 2, 3) on the line = = . Also, 

5 2 3 

find the length of the perpendicular. 

OR 

Find the particular solution of the differential 

dx 

equation + y cot x = 2x + x 2 cot x, x ≠ 0 given 

dy 

π 

that y = 0, when x = 

2 

22. Show that 

1 

1 

1 

x 

y 

z 

yz 1 

zx = 1 

xy 1 

and hence factorize. 

OR 

If a, b and c are real numbers and 

b + c c + a a + b 

c + a a + b b + c = 0 

a + b b + c c + a 

Show that either a + b + c = 0 or a = b = c. 

Section C 

23. Two persons A and B throw a die alternately till one 

of them gets a 'six' and wins the game. Find their 

respectively probabilities of winning. 

24. Find the area bounded by the curves y = x and 

y = x 3 . 

25. Find the shortest distance between the lines 

x −1 y − 2 z − 3 x − 2 y − 4 z − 5 

= = and = = . 

2 3 4 3 4 5 

⎡1 

− 2 0⎤ 

26. If A = 

⎢ ⎥ 

⎢ 

2 1 3 

⎥ 

, find A –1 . Using A –1 , solve the 

⎢⎣ 

0 − 2 1⎥⎦ 

system of linear equations 

x – 2y = 10 

2x + y + 3z = 8 

–2y + z = 7 

27. Show that sin p θ cos q θ attains a maximum value 

when θ = tan –1 p / q .(where p, q > 0) 

x 

y 

z 

x 

y 

z 

2 

2 

2 

3 

28. Evaluate 

∫ 

( x + 5x) 

dx , by first principle method 

1 

Evaluate 

∫ 

2 

1 

e x 

−1 

OR 

dx by first principle method 

29. There is a factory located at each of two places P 

and Q. From these locations, a certain commodity is 

delivered to each of the three depots situated at A, B 

and C. The weekly requirements of the depots are 

respectively 5, 5 and 4 units of the commodity while 

the production capacity of the factories at P and Q 

are 8 and 6 units respectively. The cost of 

transportation per unit is give below. 

To 

From 

COST (In Rs.) 

A B C 

P 16 10 15 

Q 10 12 10 

How many units should be transported from each 

factory to each depot in order that the transportation 

cost is minimum. Formulate the above as a linear 

programming problem. 

OR 

A brick manufacturer has two depots, A and B, with 

stocks of 30,000 and 20,000 bricks respectively. He 

receives orders from three builders P, Q and R for 

15, 000, 20,000 and 15000 bricks respectively. The 

cost in Rs. transporting 1000 bricks to the builders 

from the depots are given below. 

To 

From 

P Q R 

A 40 20 30 

B 20 60 40 

How should the manufacturer fulfill the orders so as 

to keep the costs of transportation minimum? 

• Pluto lies at the outer edge of the planetary 

system of our sun, and at the inner edge of the 

Kuiper Belt, a belt of icy comets that are the 

remnants of the formation of the solar system. 

• Gamma ray bursts - mysterious explosions at 

the edge of the Universe - were first detected in 

1969 by military satellites monitoring the Test 

Ban Treaty. 

• Titan is the largest moon of Saturn and the 

second largest moon in the entire solar system. 


MOCK TEST-2 (SOLUTION) 

MOCK TEST– 2 (PAPER) PUBLISHED IN JANUARY ISSUE 

1. p = q × 2l 

It's a vector quantity 

2. Sensitivity ∝ 

PHYSICS 

1 

Potential gradient 

3. Gamma rays, X-rays, ultraviolet, Infrared 

h h 

4. λ = = p 2mk 

∴ The proton will have a higher K.E. 

(mass of proton is slightly less than that of the 

neutron) 

5. The ionization energy of silicon gets (considerably) 

reduced compared to that of carbon. Silicon 

(a semi-conductor), therefore, becomes a (much) 

better conductor of electricity than carbon (an 

insulator) 

6. (0 to t 1 ), (t 3 to t 4 ) 

7. No, when the refractive index of prism material is 

same as that of the surrounding, then there is no 

dispersion. 

8. As, P ∝ f 

1 , so lens of smaller focal length is more 

powerful and more magnifying power. 

9. F = qvB sinθ 

10. S = 

(i) θ = 90°, F max = q v B 

(ii) θ = 0°, 180°, F = 0 

i 

g 

× G 

i – i 

11. V 1 = 2 V V 2 

g 

6 µF 12 µF 

V 

12 

V 1 = 2 = × V 

6 + 12 

V = 3 volt 

OR 

∈ 0 A 

= 8 

d 

∈0 

(5) A ∈ 

C' = = 10 × 0 A 

d 

d 

2 

= 10 × 8 = 80 

12. The region containing the uncompensated acceptor 

and donor ions is called depletion region there is a 

barrier at the junction which opposes the movement 

of majority charge carriers. 

P 

– ° – ° – ° – – + + + + + 

– ° – ° – ° 

– ° – ° – ° 

– ° – ° – ° 

– – 

– – 

– – 

+ + 

+ + 

+ + 

Depletion region 

N 

+ + + 

+ + + 

+ + + 

Formation of depletion region in PN junction 

diode 

The physical distance from one side of the barrier 

to the other is called the width of the barrier. The 

width of the depletion region or barrier depends 

upon the nature of the material. Its typical value is 

nearly 10 –6 m. The difference of potential from one 

side of the barrier to the other side is called 

potential barrier or height of the barrier. Its value is 

nearly 0.7 V for a silicon PN junction and 0.3 V for 

a germanium diode. 

13. Reasons : 

(i) Size of antenna 

(ii) Effective power radiated by the antenna 

14. The activity of a radioactive element at any instant, 

equals its rate of decay at that instant. Its SI unit is 

Becquerel (Bq) (= 1 decay per second) 

dN 

Activity R = – log = λN = dt 

Te 2 N 

∴ 

R 1 = 

1 

R2 

T1 

N N + 

2 N = 

1T2 

T2 

N 2T1 


15. 

Information 

Source 

Message 

Signal 

Transmitter 

Transmitted 

Signal 

Transmission 

channel 

Received 

Signal 

Receiver 

16. The permitted stationary orbits for the electron in a 

hydrogen atom are those for which the angular 

momentum of the electron is an integral multiple of 

h/2π 

h 

m v n r n = n 

2π 

h 

∴ 2πr n = n mv 

h 

But = λ n the associated de Broglie 

mv n 

wavelength for electron in its n th orbit 

Hence 2πr n = n λ n 

or circumference of n th permitted orbit 

= n × de Broglie wavelength 

associated with the electron in the n th orbit. 

17. For telescope, focal length & aperture of objective 

has to be maximum i.e., lens A and eyepiece has to 

smaller focal length and smaller aperture i.e., lens 

D. Then magnifying power will be maximum, i.e. 

f 

M.P. = 0 100 = = 20 

f e 5 

and l = f 0 + f e = 105 cm. 

18. The interference pattern due to different component 

colours of white light makes interference pattern of 

different colours and overlap on each other, so the 

central fringe is white. As violet wavelength is 

minimum and red has maximum so violet fringes 

are closer & red are farther. And there are different 

lines of different colours. 

19. 

= 

µ 0 

4π 

idy cosθ 

2 

(r') 

∴ y = r tanθ 

∴ dy = r sec 2 θdθ 

r 

and cosθ = ∴ r ' = 

r ' 

∴ dB = 

∴ B = 

µ 0 

4π 

µ 0i 

4πr 

2 

r 

cosθ 

= r secθ 

i( 

r sec θdθ)cosθ 

= 

2 

( r secθ) 

θ=π/ 

2 

∫ 

θ= – π / 2 

20. (a) τ = NiAB sin θ 

here θ = 0 

cos θdθ 

= 

∴ τ = 0 

i 

(b) F = ev d B , v d = neA 

µ 0 i 

2πr 

V 

21. (a) K = 0 R ε 

= × L R + r + R e xt L 

20 5 

= × (r = 0) 

20 + 480 10 

= 2 × 10 –2 V/m 

(b) ∴ V = Kl = 2 × 10 –2 × 6 

= 12 × 10 –2 volt 

ρl 

22. ∴ R = A 

∴ If l = 1m, A = 1 m 2 

∴ 

R = ρ 

S.I. unit, ρ = R l 

A = ohm-metre 

2 

ne τ 

ρ ' = 

m 

µ 0 i 

cosθ dθ 

4πr 

i 

r 

θ 

y 

r ' 

90°–θ 

dy 

At point P 

µ 0 id ysin (90° 

– θ) 

dB = 

2 

4π 

( r') 

P 

23. (i) A → Capacitive circuit 

B → Inductive only 

V 

(ii) For device A ; i = 

X C 

∴ I ∝ ω 

For device B : 

i = 

X L 

∴ I ∝ ω 

1 

V V = 

ωL 

= V ω C 


24. Output not symmetric for A,B = (0,1) and 

(1,0) Not gate in one input 

(i) has three zeroes NOR gate 

Thus 

A 

Y 

B 

(ii) has three one's ⇒ OR gate 

Thus 

A 

B 

h h 

25. λ = = p mv 

∴ λ e = 

9× 

10 

6.6× 

10 

–31 

–34 

× 3× 

10 

–34 

6 

Y 

= 2.44 × 10 –10 m 

6.6× 

10 

λ ball = 

= 2.2 × 10 –34 m 

–2 

3× 

10 × 100 

λ e = size of atom, λ ball

1100 1 

∴ i s = = Amp. 

22000 20 

E 

29. 

(v) Power in secondary V s i p = V p i s = 1100 W 

+ – 

+ – 

E i 

K 

E 0 

A 

B 

F 0 

f 0 

O 

v 0 

B'' 

P 

F e 

B' 

α 

β 

Due to dielectric, electric field between plates 

decreased, so p.d. decreased, consequently 

capacitance increased 

Net Electric field between plates 

E = E + + E – 

= 

= 

E = 

∴ 

∴ 

σ 

2ε 0 K 

σ 

ε 0 K 

+ 

q 

ε 0 AK 

∆U 

d 

q 

∆U 

= 

= 

σ 

2ε 0 K 

q 

∈ 0 AK 

∈ 0 KA 

d 

30. A 

⇒ C = 

∈ 0 KA 

d 

u 0 

A'' 

L 

Magnifying power, 

tanβ 

M = = 

tan α 

A'' 

B'' 

D = 

PB'' 

/ D 

⎛ D ⎞ v 

= m e m 0 = 

⎜1 + 

⎟ . 

⎝ fe 

⎠ u 

0 

0 

A' 

D 

A '' B' 

' 

= 

AB 

= – 

L 

f 0 

⎛ ⎜1 

+ 

⎝ 

A'' 

B'' 

. 

A' 

B' 

D 

fe 

CHEMISTRY 

⎞ 

⎟ 

⎠ 

A ' B' 

AB 

1. NCl 5 is not found because nitrogen do not have 

vacant d-orbitals 

2. Ores of aluminium are : 

(i) Bauxite ore : Al 2 O 3 . 2H 2 O. 

(ii) Diaspore : Al 2 O 3 . H 2 O 

i 

r 1 r 2 

δ 1 

e 

δ 2 

3. Enzymes are protein natured specific biocatalyst 

which increases the rates of reaction by lowering 

the energy of activation. 

When light passes through a prism, it gets refracted 

twice from its two non-parallel refracting surfaces 

such that net deviation is given by sum of 

deviations produced by each surface, 

δ = δ 1 + δ 2 

= (i – r 1 ) + (e – r 2 ) 

= i + e – (r 1 + r 2 ) 

δ = i + e – A 

At minimum deviation, 

r 1 = r 2 = r (say) & i = e 

so, A = r 1 + r 2 = 2r ⇒ r = A/2 

δmin 

+ A 

δ min = 2i – A ⇒ i = 

2 

As, µ = 

sin i 

sin r 

= 

⎛ δm 

+ A ⎞ 

sin⎜ 

⎟ 

⎝ 2 ⎠ 

sin A/ 

2 

4. Coordination No. of 

B.C.C. = 8 

H.C.P. = 12 

C.C.P = 12 

S.C. = 6 

M 

5. K 0 = 0.25 hr 

x = K 0 t t = 30 min 

(a 0 – a t ) = K 0 t 

a 0 = 0.25 × 0.5 + 0.075 = 0.20 M 

6. Ca +2 + 2e – → Ca 

2mol 1 mol 

1F → ½ mol = 20 g 

7. In absorption association over the surface takes 

place 

∴ It is an exothermic process. 


8. CH 3 Cl, Na, Dry ether; Wurtz Fittig reaction 

14. log 

9. Carbon monoxide is poisonous as it combines with 

haemoglobin of blood forming 

carboxyhaemoglobin due to which deficiency of 

oxygen occurs in blood. 

10. The principle oxidation state of lanthanides is +3. 

However, some lanthanides also show oxidation 

state of +2 and +4. 

For example, Eu shows oxidation state of +2 and 

Cerium shows oxidation state of + 4 . 

11. Bayer's process is used when bauxite ore contains 

0 

15. 

ferric oxide as chief impurity. 

The powdered ore is first roasted at low 

temperature to convert ferrous oxide into ferric 

oxide. It is then digested with a concentrated 

solution of sodium hydroxide. 

The aluminium oxide dissolves in caustic soda 

(NaOH) forming soluble sodium metal aluminate 

(NaAlO 2 ) while ferric oxide and silica remains 

insoluble and settle down. These are removed by 16. 

filtration. 

Al 2 O 3 . 2H 2 O + 2NaOH → 2NaAlO 2 + 3H 2 O 

(soluble) 

The sodium metaaluminate solution is agitated and 

it undergoes hydrolysis with formation of Al(OH) 3 

as precipitate. 

NaAlO 2 + 2H 2 O → NaOH + Al(OH) 3 

(Precipitate) 

The precipitate is washed and dried 

12. (i)Colligative Properties : The properties which 

depends upon no. of particles but do not depends 

upon nature of particles are called colligative 

properties. 

(ii) Reverse Osmosis : When we apply pressure 

greater than osmotic pressure on the conc. side of 

the two solution which are separated by 

semipermeable membrane. This results in 

movement of solvent molecules from high conc. to 

low conc. 

use : purification of water 

17. C 

13. The minimum additional energy which is required 

by the reactant molecule to participate in the 18. (i) 

chemical rx n is called activation energy. Catalyst 

reduces the activation energy as it offer an 

additional path to the chemical rx n with rise in C 2 H 5 Br 

temperature the activation energy is not affected but 

more and more molecules will have that minimum 

energy which is required to participate in chemical 

rx n 

ln 2 = 

⎛ 1 1 

2 ⎟ ⎞ 

⎜ − 

⎝ T1 

T2 

⎠ 

K 2 E = a 

K1 

.303 R 

E a 

2 .303× 

8.30 

E a 

⎛ 

⎜ 

⎝ 

1 

300 

1 ⎞ 

− ⎟ 310 ⎠ 

⎛ 10 ⎞ 

0.3010 = 

⎜ ⎟ 

2 .303× 

8.314 ⎝ 310×300 ⎠ 

E a = 2.303 × 8.314 × 0.3010 × 31 × 300 

= 53.598 kJ 

E + = 0.80 V 

Ag / Ag 

0 

Ag 

E 0 cell = E – 

+ 

/ Ag 

0 

Zn 2 / Zn 

E + = 0.76 V 

0 

Zn 2 / Zn 

E + 

= 0.8 – (–0.76) = 1.56 V 

∆Gº r = –nF Eº cell 

= – 2 × 96500 × 1.56; = – 301.08 kJ 

Impregnated 

with Pt or Ni 

H 2 

(5atm) 

Anode 

Porous C-electrode 

Chemical rx n 

At anode 

H 2 (g) + 2OH¯ → 2H 2 O(g) + 2e¯ 

At cathode 

NaOH + KOH 

⊕ Cathode 

1 

O2 (g) + H 2 O(l) + 2e – → 2OH¯ (aq) 

2 

overall rx n 

H 2 (g) + 2 

1 

O2 (g) → H 2 O(l) 

+ – 

6H5NH2 

→ C6H5N2 

Cl → C6H5I 

(X) 

(Y) 

(Z) 

Mg 

ether 

C 2 H 5 MgBr 

O 

O 2 

(5atm) 

C 2 H 5 – CH 2 – CH 2 MgBr 

H 2 O / H + 

CH 3 CH 2 CH 2 CH 2 OH 

Butan-1-ol 


(ii) 

CH 3 

CH 3 MgBr + CH 3 – C – CH 3 

CH 3 – C – CH 3 

(ii) Bacteriostatic – These stop the growth of 

bacteria 

Ex – Chloremphenicol, Erythromycin 

Methyl magnesium 

bromide 

O 

Propanone 

CH 3 

CH 3 – C – CH 3 + Mg 

OH 

2-Methylpropan-2-ol 

H 2O / H + 

Br 

OH 

OMgBr 

19. The preparation of K 2 Cr 2 O 7 from chromite ore is 

given in following steps : 

Step – I : Preparation of sodium chromite : 

4FeCr 2 O 4 + 16NaOH + 7O 2 

→ 8Na 2CrO 4 + 2Fe 2 O 3 + 8H 2 O 

Step – II : Conversion of sodium chromate into sodium 

dichromate : 

2Na 2 CrO 4 + H 2 SO 4 → Na 2 Cr 2 O 7 + Na 2 SO 4 + H 2 O 

Step – III : Conversion of sodium dichromate into 

potassium dichromate. 

Na 2 Cr 2 O 7 + 2KCl → K 2 Cr 2 O 7 + 2NaCl 

on increasing pH value, dichromate ions (Cr 2 O 2– 7 ) 

get converted into chromate ions (CrO 2– 4 ). 

20. (i) [CoCl 2 (en) 2 ] Cl 

Dichloridobis (ethane –1, 2–diamine) cobalt (III) 

chloride. 

(ii) Potassium tetrahydroxozincate (II) 

(iii) Tetraammine aqua chloridocobalt (III) chloride 

21. Polymers are macro molecules with number of 

repeating units called monomers 

O 

O 

( ) n 

Terylene – O – CH 2 – CH 2 – O – C – – C – 

Nylon 6, 6 ( HN – (CH 2 ) 6 – NH – CO – (CH 2 ) 4 – 

CO ) n 

22. Dissaccharide are sugar containing two 

monosaccharide unit 

(i) maltose = αD Glucose + αD Glucose 

(ii) Sucrose = αD Glucose + βD Fructose 

(iii) Lactose = βD Galactose + βD glucose 

23. Antibiotics are naturally produced chemical 

substances which kill or arrest the growth of 

bacteria 

(i) Bactericidal – These kill bacteria 

Ex – Penicillin, Ofloxacin 

24. Edge of unit cell a = 288 × 10 –10 cm. 

V = a 3 = 23.9 × 10 –24 cm 3 

d = 

Z 

3 × 

a 

7.2 = 

M 

N A 

2 × 52 

23.9 × 10 

−24 

× N A 

2× 

52 

N A = 

= 6.04 × 10 23 

–24 

23.9× 

10 × 7.2 

0 

25. (a) p H 2 O = 12.3 kPa 

In 1 molal solution 

m = 

n 

w 

B 

Kg 

n B = 1 

n H 2 O = 

55.5 

x H 2 O = 

55.5 + 1 

= 0.982 

p = 0.982 × 12.3 = 12.08 kPa 

0 

v.p. of solution p A = 0.8 p A 

(b) let mass of solute be W g 

W 

moles of solute = 40 

114 

moles of octane n 0 = = 1 114 

x B = 

∆ p 

0 

p A 

W 

40 

W 

40 

= xB 

+ 1 

p 

0 

A 

– 0.8p 

p 

0 

A 

0 

A 

0 .2× 

40 

⇒ W = = 10 g 

0.8 

= 

1000 = 55.5 

18 

W 

40 

W 

40 

+ 1 

26. (a) The associative colloide is that colloide which is 

form due association of large no. of particles. These 

particles form true solution at lower conc. but as 

conc. became greater than C.M.C. (critical micelles 

conc.) their association results in the formation of 

colloidal solution Ex. Soap, detergents 

(b) Hardy-Schulz rule states that for the coagulation 

of colloidal solution active ions are required and 

active ions are those ions which are having opposite 

charge more is the charge greater will be the 

coagulation tendency. 


(c) 

H 2O 

⊕ 

Cellophane bag 

H 2O 

30. Formic acid H – C – OH 

O 

contains both an 

H 

Dialysis is a phenomenon in which removal of 

dissolved impurities from the colloidal solution by 

means of diffussion through a suitable 

semipermeable membrane. 

In this process colloidal solution is placed in 

cellophane bag and impurities get removed through 

small pores of the bag. To enhance the removel of 

electrolyte it is placed in electric field which is 

called electro – dialysis. 

27. (a) The reactivity of aldehydes and ketones toward 

nucleophilic addition depends upon (i) + I effect (ii) 

steric hinderance. Hence the order is Di-tert butyl 

ketone < methyl-tert-butyl ketone < Acetone < 

Acetaldehyde 

(b) The acidic strength depends upon (i) nature of + 

I effect (ii) nature of atom / group attached (iii) 

position of substituent on the chain. Hence, 

(CH 3 ) 2 CHCOOH < CH 3 CH 2 CH 2 COOH < 

CH 3 CH(Br)CH 2 COOH < CH 3 CH 2 CH(Br)COOH 

(c) 4-Methoxy benzoic acid < Benzoic acid < 4- 

Nitrobenzoic acid < 3, 4-Dinitrobenzoic acid 

28. Hydrolysis of trichlorosilanes gives cross – linked 

silicones. 

R 

R 

Cl – Si – Cl + 3H 2 O –3HCl 

R 

Cl 

n HO – Si – OH 

–(n–1) H 2O 

OH 

HO – Si – OH 

R 

OH 

R 

-- O – Si – O – Si – O -- 

O 

O 

-- O – Si – O – Si – O -- 

R R 

Cross linked silicon 

29. A is C 6 H 5 CONH 2 ; B is C 6 H 5 CN; C is C 6 H 5 CH 2 OH 

the sequence of reactions is 

C 6 H 5 CONH 2 

P 2O 5 

Red 

C 6 H 5 CN 

–H 2O 

C 6 H 5 CH 2 NH 2 

Benzamide Benzonitrile Benzylamine 

C 6 H 5 COOH 

Benzoic acid 

–N 2 – H 2O HNO 2 

Oxide 

C 6 H 5 CH 2 OH 

[O] 

Benzyl alcohol 

aldehyde 

group 

– C = O 

– C – OH 

O 

as well as carboxyl 

but acetic acid contain 

only a carboxyl group. Formic acid behaves as 

reducing agent whereas acetic acid does not. 

(a) Formic acid reduces Tollen's reagent to 

metallic silver but acetic acid does not. 

HCOOH + 2[Ag(NH 3 ) 2 ] + + 2(OH)¯ 

Tollen's reagent 

2Ag ↓ + CO 2 ↑ + 2H 2 O + 4NH 3 

Silver mirror 

No silver mirror is formed with acetic acid. 

(b) Formic acid reduced Fehling solution to red ppt. 

of Cu 2 O but acetic acid does not. 

HCOOH+2Cu 2+ + 4(OH)¯ → Cu 2 O ↓ + CO 2 ↑ + 3H 2 O 

Fehling solution 

Red ppt (cuprous oxide) 

MATHEMATICS 

Section A 

1. t r (A) = a 11 +a 22 +a 33 

= 14 + (–5) + (–2) ⇒ = 14 – 7 = 7 

2. x + 10 = 3x + 4 and y 2 + 2y = 3 and y 2 –5y = –4 

⇒ 2x = 6 ⇒ (y +3) (y – 1) = 0 y 2 – 5y + 4 = 0 

⇒ (y – 4) (y – 1) = 0 

⇒ x = 3 ⇒ y = –3, 1 y = 1, 4 

⇒ x = 3, y = 1 

3. sin 10° cos 80° – sin 80° (–cos 10°) 

= sin 10° cos 80° + sin 80° cos 10° 

= sin (10 + 80)° = sin 90° = 1 

4. Diff. w.r. to x 

e x + e y dy = e 

x+y ⎞ 

⎜ 

dy 

1 + ⎟ 

dx ⎝ ⎠ 

⇒ (e y – e x+y dy 

) = e x+y – e x 

dx 

x 

dy e ( e −1) 

= 

y x 

dx e (1 − e ) 

y 


5. f(f(x)) = 

= 

6. = 

∫ 

⎛ 3x 

− 2 ⎞ 

3⎜ 

⎟ − 2 

3 f ( x) 

− 2 

= 

⎝ 2x 

− 3 ⎠ 

2 f ( x) 

− 3 ⎛ 3x 

− 2 ⎞ 

2⎜ 

⎟ − 3 

⎝ 2x 

− 3 ⎠ 

9x 

− 6 − 4x 

+ 6 

= x 

6x 

− 4 − 6x 

+ 9 

e x 

⎛ x x ⎞ 

⎜1 

− 2sin cos ⎟ 

⎜ 2 2 ⎟ dx 

⎜ 

2 x ⎟ 

2sin 

⎝ 2 ⎠ 

x ⎛ 1 2 x x ⎞ 

= 

∫ 

e ⎜ cosec − cot ⎟ dx 

⎝ 2 2 2 ⎠ 

= – 

∫ 

x x 1 x 2 x 

e cot dx + e cosec dx 

II I 2 2 ∫ 2 

⎧ x 

= – ⎨cot 

.e 

⎩ 2 

x 

− 

∫ 

− 

cosec 

2 

x 1 x ⎫ 

. e dx⎬ 

2 2 ⎭ 

1 x 2 x 

+ 2 ∫ 

e cosec dx 

2 

= –e x x 1 

cot – 

x 2 x 

2 2 ∫ 

e cosec dx + 

2 

= – e x cot 2 

x + C 

1 x 2 x 

2 ∫ 

e cosec dx + C 

2 

7. We are given that cos 2 dy 

x + y = tan x 

dx 

dy 

⇒ + (sec 2 x)y = tan x . sec 2 x ….(i) 

dx 

This is a linear differential equation of the form 

dy + Py = Q, where P = sec 2 x and Q = tan x sec 2 x 

dx 

∴ I = I.F. = e ∫ 

sec 2 

x dx 

= e tanx 

Multiplying both sides of (i) by I.F. = 

e tan x dy + sec 2 x e tanx . y = e tanx . tan x sec 2 x 

dx 

Integrating both sides w.r.t. x, we get 

ye tan x tan x 

= ∫ e . tan x sec 2 x dx + C 

[Using : y (I.F.)= ∫ Q (I.F.) dx + C] 

⇒ ye tan x = 

∫ tet dt + C, where t = tan x 

I II 

x 

e tan , we get 

⇒ ye tan x = te t – 

∫ 

e t dt + C [Integrating by parts] 

⇒ ye tan x = te t – e t + C 

⇒ ye tan x = e tanx (tan x –1) + C, 

which is the required solution. 

8. We have, 

( a r × b r ) 2 = | a r × b r | 2 

⇒ ( a r × b r ) 2 = {| a r | | b r | sin θ} 2 

⇒ ( a r × b r ) 2 = | a r | 2 | b r | 2 sin 2 θ 

⇒ ( a r × b r ) 2 = {| a r | 2 | b r | 2 } (1 – cos 2 θ) 

⇒ ( a r × b r ) 2 = | a r | 2 | b r | 2 – | a r | 2 | b r | 2 cos 2 θ 

⇒ ( a r × b r ) 2 = ( a r . a r ) ( b r . b r ) – ( a r . b r ) ( a r . b r ) 

[Q a r . b r = | a r | | b r |cos θ] 

⇒ ( a r × b r r r r r 

) 2 a. 

a a. 

b 

= r r r r 

a. 

b b. 

b 

9. Let a r = î +2 ĵ + 3 kˆ and b r = 3 î –2 ĵ + kˆ . The vector 

area of the parallelogram whose adjacent sides are 

represented by the vectors a r and b r is a r × b r . 

Now, a r × b r iˆ 

ˆj 

kˆ 

= 1 2 3 

3 − 2 1 

= (2 + 6) î – (1 – 9) ĵ + (–2 – 6) kˆ = 8 î +8 ĵ –8 kˆ 

So, area of the parallelogram = | a r × b r | 

= 

2 

2 

8 + 8 + ( −8) 

= 8 3 square units 

10. We know that the angle θ between the line 

r = a r + λ b r and the plane r . n r = d is given by 

r r 

b . n 

sin θ = r r 

| b || n | 

Here, b r = î – ĵ + kˆ and n r = 2 î – ĵ + kˆ 

∴ sin θ = 

= 

2 + 1+ 

1 

= 

3 6 

2 

⎛ 

⇒ θ = sin –1 ⎜ 

2 

⎝ 

2 

1 + ( −1) 

2 

(ˆ i − ˆj 

+ kˆ).(2ˆ 

i − ˆj 

+ kˆ) 

2 

+ 1 

4 2 2 = 

3 2 3 

2 

3 

⎟ ⎞ 

⎠ 

2 

2 

+ ( −1) 

2 

2 

+ 1 


Section B 

11. R 3 → R 3 + R 2 , R 1 → R 1 + R 2 

∆ = 

a + b 

− c 

− ( b + c) 

= (a + b) (b + c) 

R 1 → R 1 + R 3 

= (a + b) (b + c) . 

a + b 

a + b + c 

b + c 

1 

− c 

−1 

0 

− c 

−1 

= (a + b) (b + c) (–2) 

1 

a + b + c 

1 

− ( a + b) 

2 

1 

− a 

b + c 

a + b + c 

− c 

−1 

− a 

1 

−1 

− a 

1 

0 

− a 

= (a + b) (b + c). 2 (c + a) 

12. A → Integer chosen is divisible by 6 

B → integer chosen is divisible by 8 

n (A) = 33, n (B) = 25, n (A ∩ B) = 8, n(S) = 200 

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

33 25 8 50 1 

= + – = = 200 200 200 200 4 

OR 

Let E : Candidate Reaches late 

A 1 = Candidate travels by bus 

A 2 : Candidate travels by scooter 

A 3 : Candidate travels by other modes of 

transport 

3 1 3 

P(A 1 ) = , P(A2 ) = , P(A3 ) = 10 10 5 

P(E/A 1 ) = 4 

1 , P(E/A2 ) = 3 

1 , P(E/A3 ) = 0 

∴ By Baye's Theorem 

P(A 1 /E) = 

P(A1)P(E / A1) 

P(A1)P(E / A1) 

+ P(A 2 )P(E / A 2 ) + P(A 

3 1 

× 

= 

10 4 9 

= 

3 1 

+ + 0 13 

40 30 

⎡ 2 

13. Let y = tan –1 1+ 

x −1 

⎥ ⎥ ⎤ 

⎢ 

⎢ x 

⎣ ⎦ 

put x = tan θ 

y = tan –1 ⎡1− 

cosθ⎤ 

⎢ ⎥ 

⎣ sin θ ⎦ 

1 

3 

z = tan –1 x 

)P(E / A 

dz 1 = 

dx 

2 

1+ x 

3 

) 

14. 

⎡ 2 θ ⎤ 

⎢ 2sin 

y = tan –1 2 

⎢ 

⎢ 

θ θ 

2sin cos 

⎣ 2 2 ⎥ ⎥⎥⎥ ⎦ 

y = 2 

1 tan –1 x 

dy 1 1 = . 

dx 2 

2 

1+ x 

dy 1/ 2× 

1/1+ 

x 

∴ = dz 

2 

1/1+ 

x 

2 

= 2 

1 

dx ⎛ 1 ⎞ dy ⎛ 1 ⎞ 

= a ⎜1 − ⎟⎠ 

dt ⎝ t 2 = a ⎜1 + ⎟⎠ 

dt ⎝ t 2 

⎛ 1 ⎞ 

⎡ 1⎤ 

a⎜1 

+ ⎟ 

dy 

2 2 

at 

= 

⎝ t ⎠ a ( t + 1) ⎢t 

+ ⎥ 

= = 

⎣ t ⎦ 

dx ⎛ 1 

2 

⎞ a ( t −1) 

⎡ 1⎤ 

a⎜1 

− ⎟ at 

2 

⎢t 

− ⎥ 

⎝ t ⎠ ⎣ t ⎦ 

dy x = 

dx y 

OR 

x p y p = (x + y) p + q 

Take log on both sides 

p log x + q log y = (p + q) log (x + y) 

p q dy p + q ⎛ dy ⎞ 

+ . = ⎜1 

+ ⎟⎠ 

x y dx x + y ⎝ dx 

or 

or 

or 

or 

p – 

x 

p + q 

x + y 

dy 

= dx 

⎛ p + q q ⎞ 

⎜ − 

⎟ 

⎝ x + y y ⎠ 

px + py − px − qx dy ⎛ py + qy − qx − qy 

= 

x( x + y) 

⎟ ⎞ 

⎜ 

dx ⎝ y( x + y) ⎠ 

py − qx 

x 

y dy = 

x dx 

dy ⎛ py − qx 

= ⎟ ⎞ 

⎜ 

dx ⎝ y ⎠ 

15. Let f(x) = log e x, x ∈ [a, b] 

↓ 

continuous & differentiable 

f ( b) 

− f ( a) 

∴ f ′(c) = 

b − a 

1 logb 

− log a 

= 

c b − a 

Q a < c < b 

1 1 1 

⇒ < 

1 logb 

− log a 1 

⇒ < 



b − a 

b − a 

⇒ < log b – log a < 

b 

a 


16. Q differentiable at x = c 

⇒ continuous at x = c 

⇒ f(c) = f(c + ) 

⇒ c 2 = 

17. f(1) = 

lim 

→0 h 

⇒ c 2 = ac + b 

Now 

f ′(c + ) = f ′ (c – ) 

a(c + h) + b 

[ a( 

c + h) 

+ b] 

− c 

⇒ lim 

h→0 

h 

2 

[( c − h) 2 ] − c 

= lim 

h→0 

− h 

( ac + b) 

+ ah − c 

⇒ lim 

h→0 

h 

2 

= 

2 

lim 

→0 h 

2 

2 

2 

…(1) 

h − 2ch 

− h 

c + ah − c 

⇒ lim = lim (2c – h) [From (1)] 

h→0 

h h →0 

a = 2c 

…(2) 

from (1), (2) 

a = 2c, b = – c 2 

1+ 1 = 1 

2 

f(2) = 2 

2 = 1 

many-one function 

If n → odd natural number then 2n –1 is also odd 

number 

2 n −1+ 

1 

f(2n –1) = = n 

2 

If n → even natural number then 2n is also an even 

natural number 

2n 

f(2n) = = n 

2 

⇒ f is onto function. 

18. We have, 

sin x 

sin x 

I = 

∫ 

dx = 

sin 4x 

∫ 

dx 

2sin 2x 

cos 2x 

sin x 

= 

∫ 

dx 

4sin x cos x cos 2x 

1 1 1 cos x 

⇒ I = 4 ∫ 

dx = 

cos x cos 2x 

4 ∫ 

dx 

2 

cos x cos 2x 

1 cos x 

⇒ I = 4 ∫ 

dx 

2 

2 

(1 − sin x)(1 

− 2sin x) 

Putting sin x = t and cos x dx = dt, we get 

1 dt 

I = 4 ∫ 2 2 

(1 − t )(1 − 2t 

) 

Let t 2 = y. Then, 

1 

2 

1 

1 

1 

= 

2 2 

(1 − t )(1 − 2t 

) (1 − y) (1 − 2y) 

1 A B 

Let 

= + . Then, 

(1 − y) (1 − 2y) 

1− y 1− 2y 

1 = A (1 –2y) + B (1 – y) ….(i) 

Putting y = 1 and y = 2 

1 respectively in (i), we get A 

= –1 and B = 2 

1 −1 2 

∴ 

= + 

(1 − y)(1 

− 2y) 

1− 

y 1− 

2y 

1 

1 2 

⇒ 

= – + 

2 2 

2 

(1 − t )(1 − 2t 

) 1− t 1− 

2t 

1 dt 

⇒ I = 4 ∫ 2 2 

(1 − t )(1 − 2t 

) 

1 

= 4 ∫ ⎜ ⎛ 1 2 ⎞ 

− + ⎟ dt 

2 

⎝ 1− 

t 1− 2t 

2 ⎠ 

1 1 2 1 

⇒ I = – 4 ∫ 

dt + 

− 

2 

1 t 4 ∫ 

dt 

− 

2 

1 ( 2t) 

⇒ I = – 4 

1 . 2 

1 log 

1+ 

t 

1− 

t 

1 1 1+ 

2t 

+ . log + C 

2 2 2 1− 

2t 

1 1+ 

t 

⇒ I = – log + 

8 1− 

t 

1 1+ 

sin x 

⇒ I = – log + 

8 1− 

sin x 

19. We have, 

∫ 

I = { tan θ + cot θ} 

1 1+ 

2t 

log + C 

4 2 1− 

2t 

1 1+ 

2 sin x 

log + C 

4 2 1− 

2 sin x 

dθ 

⎧ 1 ⎫ 

⇒ I =∫ ⎨ tan θ + ⎬ dθ 

⎩ tan θ ⎭ 

tan θ + 1 

⇒ I = ∫ 

dθ 

tan θ 

Let tan θ = x 2 . Then, 

d (tan θ) = d(x 2 ) 

⇒ sec 2 θ dθ = 2x dx 

2x 

dx 2x 

dx 

⇒ dθ = = = 

2 

2 

sec θ 1+ 

tan θ 

x 

∴ I = 

∫ 

2 

1+ 

1/ x 

= 2 

∫ 2 

x + 1/ x 

+ 1 

. 

2 

x 

2 

2 

dx 

1+ 

1/ x 

⇒ I = 2 

∫ 

dx 

2 

( x −1/ 

x) 

+ 2 

2 

2x 

dx 

1+ 

x 

2x 

dx x + 1 

= 

4 2 

1+ 

x 

∫ 

dx 

4 

x + 1 

2 

2 

4 


1+ 

1/ x 

= 2 

∫ 

dx 

2 2 

( x −1/ 

x) 

+ ( 2) 

du 

⇒ I = 2 

∫ 2 

u + ( 2) 

where x – x 

1 = u 

2 

2 

= 

2 tan 

–1 ⎛ 

⎜ 

2 ⎝ 

u ⎞ 

⎟ + C, 

2 ⎠ 

⇒ I = 2 tan –1 ⎛ x −1/ 

x 

⎟ ⎞ 

⎜ + C 

⎝ 2 ⎠ 

⎛ 2 

⇒ I = 2 tan –1 ⎟ ⎞ 

⎜ 

x −1 

+ C 

⎝ 2 x ⎠ 

⇒ I = 2tan –1 ⎛ tan θ −1 

⎟ ⎞ 

⎜ + C 

⎝ 2 tan θ ⎠ 

OR 

2 2 

( x + 1)( x + 4) 

∫ 

dx 

2 2 

( x + 3)( x − 5) 

Consider 

2 2 

( x + 1)( x + 4) ( t + 1)( t + 4) 

= 

where t = x 2 

2 2 

( x + 3)( x − 5) ( t + 3)( t − 5) 

7t 

+ 19 

= 1 + 

( t + 3)( t − 5) 

Consider 

7t 

+ 19 A B 

= + 

( t + 3)( t − 5) t + 3 t − 5 

1 27 

A = , B = 4 4 

( x 

∴ 

∫ 

( x 

2 

2 

+ 1)( x 

2 

+ 3)( x 

2 

+ 4) 

dx 

− 5) 

1 

dx 27 dx 

+ 

+ 3 

5 

= 

∫ dx + 4 ∫ 2 

4 ∫ 2 

x 

x − 

1 

= x + tan 

–1 

⎛ 

⎜ 

4 3 ⎝ 

x ⎞ 

⎟ + 

3 ⎠ 

x − log 

827 5 

5 x + 5 

+ c 

20. Let P(x, y) be any point on the curve. The equation of 

the normal at P (x, y) to the given curve is 

1 

Y – y = – (X –x) … (i) 

dy / dx 

It is given that the normal at each point passes through 

(2, 0). Therefore, (i) also passes through (2, 0). Putting 

Y = 0 and x = 2 in (i), we get 

1 

0 – y = – (2 –x) 

dy / dx 

dy 

⇒ y = 2 – x dx 

⇒ ydy = (2 – x) dx [On integrating both sides] 

2 

2 

y (2 − x) 

⇒ = – + C 

2 2 

⇒ y 2 = – (2 – x) 2 + 2C 

…(ii) 

This passes through (2, 3). Therefore, 

9 

9 = 0 + 2C ⇒ C = 2 

Putting C = 2 

9 in (ii), we get 

y 2 = – (2 – x) 2 + 9 

This is the equation of required curve. 

21. We have, 

(2 î + 6 ĵ + 27 kˆ ) × ( î + λ ĵ + µ kˆ ) = 0 r 

⇒ 

î 

2 

1 

ĵ 

6 

λ 

kˆ 

27 

µ 

= 0 r 

⇒ (6µ –27λ) î – (2µ –27) ĵ + (2λ – 6) kˆ = 0 r 

⇒ 6µ –27λ = 0, 2µ –27 = 0 and 2λ – 6 = 0 

27 

⇒ λ = 3 and µ = 

2 

22. The equation of a plane passing through the 

intersection of the given planes is 

(4x – y + z –10) + λ(x + y – z –4) = 0 

⇒ x(4 + λ) + y (λ –1) + z (1 –λ) –10 – 4λ = 0 

This plane is parallel to the line with direction ratios 

proportional to 2, 1,1 

∴ 2(4 + λ) + 1(λ –1) + 1(1 –λ) = 0 ⇒ λ = – 4 

Putting λ = – 4 in (i), we obtain 

5y – 5z – 6 = 0 

This is the equation of the required plane. 

Now, length of the perpendicular from (1, 1, 1) on (ii) 

is given by 

d = 

5× 

1− 

5× 

1− 

6 

= 

2 2 

5 + ( −5) 

Given line 

or, 

x −1 

= 

5 

x −1 

= 

5 

3 − y 

2 

y − 3 

− 2 

= 

= 

3 2 

5 

OR 

z +1 

4 

z − (−1) 

4 

....(i) 

is passing through (1, 3, –1) and has D.R. 5, –2, 4. 

Equations of line passing through (3, 0, –4) and 

parallel to given line is 

x − 3 y − 0 z + 4 

= = 

...(ii) 

5 − 2 4 

Vector equations of line (i) & (ii) 

→ 

r = î + 3 ĵ – kˆ + λ(5 î – 2 ĵ + 4 kˆ ) 


→ 

r = 3 î – 4 ĵ + µ (5 î – 2 ĵ + 4 kˆ ) 

∴ 

→ 

a 2 – → 1 

a = 2 î – 3 ĵ – 3 kˆ 

→ 

b = 

2 2 2 

( 5) + ( −2) 

+ (4) 

Also 

∴ 

= 45 = 3 5 

iˆ 

→ 

⎛ ⎞ 

b × ⎜a → − → 

2 a1 

⎟ = 5 

⎝ ⎠ 

2 

= 18 î + 23 ĵ – 11 kˆ 

→ 

⎛ 

→ → 

⎞ 

b × ⎜a 2 − a 1 ⎟ = 

⎝ ⎠ 

2 

ˆj 

− 2 

− 3 

kˆ 

4 

− 3 

2 

( 18) + (23) + (11) = 974 

∴ Distance between two parallel lines. 

= 

⎛ 

→ → → 

b× 

⎜a2 

− a1 

⎝ 

→ 

23. We have, 

b 

∫ 

a 

f ( x) 

b 

dx 

⎞ 

⎟ 

⎠ 

= 

974 units 

45 

Section C 

= lim h[f(a) + f(a + h) + f(a+2h) +….+ 

h→0 

f (a + (n –1)h)] 

⎡ ⎛ h ⎞ nh ⎤ 

⎢sin⎜a 

+ ( n −1) 

⎟sin 

⎥ 

= lim h ⎢ ⎝ 2 ⎠ 2 

⎥ 

h→0 

⎢ 

h ⎥ 

⎢ 

sin 

⎥ 

⎣ 

2 ⎦ 

⎡ ⎛ nh h ⎞ nh ⎤ 

⎢sin⎜a 

+ − ⎟sin 

⎥ 

= lim h ⎢ ⎝ 2 2 ⎠ 2 

⎥ 

h→0 

⎢ 

h ⎥ 

⎢ 

sin 

⎥ 

⎣ 

2 ⎦ 

⎡ ⎛ b − a h ⎞ ⎛ b − a ⎞⎤ 

⎢sin⎜a 

+ − ⎟sin⎜ 

⎟⎥ 

= lim h ⎢ ⎝ 2 2 ⎠ ⎝ 2 ⎠⎥ 

h→0 

⎢ 

h 

⎥ 

⎢ 

sin 

⎥ 

⎣ 

2 

⎦ 

[Q nh = b –a] 

⎡ h 

⎤ 

⎢ 

= lim 

2 ⎛ a + b h ⎞ ⎛ b − a ⎞ 

⎢ × 2sin ⎜ − ⎟sin 

⎜ ⎟ 

h→0 

⎢ 

h ⎝ 2 2 ⎠ ⎝ 2 

sin 

⎠ 

⎢ 

⎥ ⎥⎥⎥ ⎣ 2 

⎦ 

2 

⎡ h ⎤ 

⎢ 

= lim 

2 

⎡ a + b h ⎤ ⎡b 

⎢ . lim 2 sin 

h→0 

⎢ 

h 

sin 

⎢ ⎥ ⎥⎥⎥ h→0 

⎢ − ⎥ sin 

⎣ 2 2 

⎢ ⎦ ⎣ 

⎣ 2 ⎦ 

⎡ a + b ⎤ ⎡b 

− a ⎤ 

= 2 sin ⎢ ⎥ sin 

⎣ 2 

⎢ ⎥ ⎦ ⎣ 2 ⎦ 

= cos a –cos b 

[Q 2 sin A sin B = cos (A –B) 

– cos (A + B)] 

24. A → getting a white ball from 1st bag. 

B → getting a black ball from 1st bag 

C → getting a white ball from 2nd bag 

D → getting a black ball from 2nd bag 

P(A) = 6 

4 , P(B) = 6 

2 , P(C) = 8 

3 , P(D) = 8 

5 

(A) P (both are white) = P(A). P(C) = 4 

1 

(B) P (one is white and one is black) 

13 

= P(A). P(D) + P(B). P(D) = 24 

25. Given curves are x 2 + y 2 = 16 and x 2 = 6y 

Solving these two equations 

. 

(–2 3 , 2)C 

2 

x 2 x 

+ = 16 36 

y 

D(0, 2) B(0, 4) 

x 4 + 36x 2 –576 = 0 

(x 2 + 48) (x 2 –12) = 0 

x 2 –12 = 0 

x = ± 2 3 

O(0, 0) 

A(2 3 , 2) 

∴ y = 2 

∴ Required area = 2(area of shaded portion) 

Reqd. area = 2[area of OADO + area of DABD] 

⎡ 

2 

4 

⎤ 

2 

= 2 ⎢ + − ⎥ 

⎢∫ 

6y 

dy 

∫ 

16 y dy 

⎥ 

⎣ 0 

2 ⎦ 

x 

− 

2 

a ⎤ 

⎥ ⎦ 

⎡ 

⎤ 

= 2 ⎢ 

2 4 

3/ 2 2 ⎧1 

2 16 −1 

⎫ 

6( y ) 0 + ⎨ y 16 − y + sin 

y ⎬ ⎥ 

⎢⎣ 

3 

⎩2 

2 4 ⎭2 

⎥⎦ 

⎛ 

= ⎜ 

4 

⎝ 

3 

3 

16π 

+ 3 

⎞ 

⎟ sq. units 

⎠ 


26. The given lines are 

r = ( î + 2 ĵ + 3 kˆ ) + λ (2 î + 3 ĵ + 4 kˆ ) 

…(i) 

Points (x, y) 

Value of the objective function 

Z = 400x + 300y 

and, r =(2 î + 4 ĵ + 5 kˆ ) + 2µ (2 î + 3 ĵ + 4 kˆ ) …(ii) 

Equation (ii) can re-written as 

r = (2 î + 4 ĵ + 5 kˆ )+ µ′(2 î + 3 ĵ + 4 kˆ ) …(iii) 

where µ′ = 2µ 

These two lines passes through the points having 

position vectors a r 1 = î + 2 ĵ + 3 kˆ & a r 2 = 2 î +4 ĵ +5 kˆ 

respectively and both are parallel to the vector 

b r =2 î + 3 ĵ + 4 kˆ . 

r r r 

| ( a2 

− a1 

) × b | 

∴ shortest distance = r 

…(iv) 

| b | 

We have, 

( a r 2 – a r 1 ) × b r = ( î + 2 ĵ + 2 kˆ ) × (2 î + 3 ĵ + 4 kˆ ) 

⇒ ( a r 2 – a r 1 ) × b r = 

iˆ 

1 

2 

ˆj 

2 

3 

= (8 – 6) î – (4 – 4) ĵ + (3 – 4) kˆ = 2 î – 0 ĵ – kˆ 

⇒ |( a r 2 – a r 1 )× b r | = 4 + 0 + 1 = 5 and 

|b r | = 4 + 9 + 16 = 29 

Substituting the values of |( a r 2 – a r 1 ) × b r | and | b r | in 

5 

(iv), we get shortest distance = 

29 

27. The given data may be put in the following tabular 

form: 

Refinery High grade Medium grade Low grade Cost per day 

A 100 300 200 Rs.400 

B 200 400 100 Rs.300 

Minimum 

requirement 

12,000 20,000 15,000 

kˆ 

Suppose refineries A and B should run for x and y 

days respectively to minimize the total cost. 

The mathematical form of the above LPP is 

Minimize Z = 400x + 300y 

Subject to 

100x + 200y ≥ 12,000 

300x + 400y ≥ 20,000 

200x + 100y ≥ 15,000 

and, x, y ≥ 0 

The feasible region of the above LPP is represented by 

the shaded region in fig. 

The corner points of the feasible region are 

A 2 (120, 0), P(60, 30) and B 3 (0, 150). The value of the 

objective function at these points are given in the 

following table: 

2 

4 

A 2 (120, 0) Z = 400 × 120 + 300 × 0 = 48,000 

P (60, 30) Z = 400 × 60 + 300 × 30 = 33, 000 

B 3 (0, 150) Z = 400 × 0 + 300 × 150 = 45,000 

Clearly, Z is minimum when x = 60, y = 30. Hence the 

machine A should run for 60 days and the machine B 

should run for 30 days to minimize the cost while 

satisfying the constraints. 

300x + 400y = 20000 

y 

B1(0,50) 

B3(0,150) 

22x + 100y = 15000 

B2(0,60) 

P(60,30) 

• 

O • • • A3(75,0) 

200 

A1 ⎛ ⎞ 

⎜ , 0⎟ ⎝ 3 ⎠ 

A2(120,0) 

• 

X 

100x + 200y = 12000 

−1 

1 

28. A 11 = = –3 + 2 = –1; 

− 2 3 

2 1 

A 12 = – = – (6 –1) = –5; 

1 3 

2 − 1 

A 13 = = –4 + 1 = –3; 

1 − 2 

1 1 

A 21 = – = – (3 + 2) = –5 

− 2 3 

1 1 

A 22 = = 3 – 1 = 2; 

1 3 

1 1 

A 23 = – = – (–2 –1) = 3 

1 − 2 

1 1 

A 31 = = 1 + 1 = 2; 

−1 

1 

1 1 

A 32 = – = – (1 –2) = 1 

2 1 

1 1 

A 33 = = –1 –2 = –3 

2 − 1 

⎡−1 

− 5 − 3⎤′ 

⎡−1 

− 5 2 ⎤ 

∴ adj A = 

⎢ 

⎥ 

⎢ 

− 5 2 3 

⎥ 

= 

⎢ 

⎥ 

⎢ 

− 5 2 1 

⎥ 

⎢⎣ 

2 1 − 3⎥⎦ 

⎢⎣ 

− 3 3 − 3⎥⎦ 

Also |A| = 1 (–1) + 1 (–5) + 1(–3) 

= –1 –5 – 3 = – 9 ≠ 0 ∴ A –1 exists. 

⎡−1 

− 5 2 ⎤ 

A –1 adj A 1 

= = – 

⎢ 

⎥ 

| A | 9 ⎢ 

− 5 2 1 

⎥ 

⎢⎣ 

− 3 3 − 3⎥⎦ 

The given system of equations can be written as 


⎡x⎤ 

⎡3⎤ 

AX = B where X = 

⎢ ⎥ 

⎢ 

y 

⎥ 

, B = 

⎢ ⎥ 

⎢ 

2 

⎥ 

∴ X = A –1 B 

⎢⎣ 

z⎥⎦ 

⎢⎣ 

2⎥⎦ 

⎡x⎤ 

⎡−1 

− 5 2 ⎤ ⎡3⎤ 

i.e. 

⎢ ⎥ 1 

⎢ 

y 

⎥ 

= – 

⎢ 

⎥ 

9 ⎢ 

− 5 2 1 

⎢ ⎥ 

⎥ ⎢ 

2 

⎥ 

⎢⎣ 

z⎥⎦ 

⎢⎣ 

− 3 3 − 3⎥⎦ 

⎢⎣ 

2⎥⎦ 

⎡− 

3 −10 

+ 4⎤ 

⎡1⎤ 

1 

= – 

⎢ ⎥ 

9 ⎢ 

−15 

+ 4 + 2 

⎥ 

= 

⎢ ⎥ 

⎢ 

1 

⎥ 

⎢⎣ 

− 9 + 6 − 6 ⎥⎦ 

⎢⎣ 

1⎥⎦ 

Thus x = 1, y = 1, z = 1. 

OR 

− 3 1 

A 11 = = 21 – 3 = 18 

3 − 7 

1 1 

A 12 = – = – (–7 –2) = 9 

2 − 7 

A 13 = 

A 21 = – 

A 22 = 

A 23 = – 

A 31 = 

A 32 = – 

A 33 = 

1 

2 

4 

2 

− 3 

= 3 + 6 = 9 

− 3 

− 5 

3 

−11 

= – (35 + 33) = – 68 

−7 

−11 

= –28 + 22 = – 6, 

− 7 

4 − 5 = –(12 + 10) = –22 

2 3 

− 5 

− 3 

4 

1 

−11 

= – 5 – 33 = – 38 

1 

4 −11 

= – (4 + 11) = –15 

1 1 

− 5 

= –12 + 5 = – 7 

− 3 

⎡ 18 9 9 ⎤′ 

∴ adj A = 

⎢ 

⎥ 

⎢ 

− 68 − 6 − 22 

⎥ 

= 

⎢⎣ 

− 38 −15 

− 7 ⎥⎦ 

⎡ 18 − 68 − 38 ⎤ 

⎢ 

⎥ 

⎢ 

9 − 6 −15 

⎥ 

⎢⎣ 

9 − 22 − 7 ⎥⎦ 

Also |A| = 4(18) + (–5) 9 + (–11)9 = 72 – 45 – 99 

= 72 – 144 = –72 ≠ 0 ∴ A –1 exists. 

⎡ 18 − 68 − 38 ⎤ 

∴ A –1 adj A 1 

= = – 

⎢ 

⎥ 

| A | 72 ⎢ 

9 − 6 −15 

⎥ 

⎢⎣ 

9 − 22 − 7 ⎥⎦ 

Further the given system of equations can be written as 

⎡x⎤ 

⎡12⎤ 


⎢ ⎥ 

⎢ 

y 

⎥ 

, B = 

⎢ ⎥ 

⎢ 

1 

⎥ 

⎢⎣ 

z⎥⎦ 

⎢⎣ 

2 ⎥⎦ 

∴ X = A –1 B 

⎡x⎤ 

⎡ 18 − 68 − 38 ⎤ ⎡12⎤ 

i.e. 

⎢ ⎥ 1 

⎢ 

y 

⎥ 

= – 

⎢ 

⎥ 

72 ⎢ 

9 − 6 −15 

⎢ ⎥ 

⎥ ⎢ 

1 

⎥ 

⎢⎣ 

z⎥⎦ 

⎢⎣ 

9 − 22 − 7 ⎥⎦ 

⎢⎣ 

2 ⎥⎦ 

⎡ 216 − 68 − 76 ⎤ ⎡72⎤ 

⎡−1⎤ 

1 

= – 

⎢ 

⎥ 1 

72 ⎢ 

108 − 6 − 30 

⎥ 

= – 

⎢ ⎥ 

72 ⎢ 

72 

⎥ 

= 

⎢ ⎥ 

⎢ 

−1 

⎥ 

⎢⎣ 

108 − 22 −14 

⎥⎦ 

⎢⎣ 

72⎥⎦ 

⎢⎣ 

−1⎥⎦ 

∴ x = –1, y = –1, z = –1 

29. 1st part 

OC = X 

2 

AC = R − X 

DC = R + X 

Volume 

V= 3 

1 π(R 2 –X 2 ) (R + X) 

2 

dV π = [R 2 –2RX –3X 2 ] = 0 

dX 3 

⇒ (R – 3X) (R +X) = 0 

⇒ X = R/3 (Q R + X ≠ 0) 

2 

d V 

= π/3 (–2R –6X) 

2 

dX 

2 

d V 4 

= – 

2 

πR < 0 ⇒ maximum volume 

dX 

3 

X= 

R / 3 

Put X = R/3 in V, we have 

V = 8/27 (volume of sphere) 

OR 

f ′(x) = cos x – sin 2x = 0 

= cos x [1 –2 sin x] = 0 

⇒ x = π/2, π/6 

Now 

f ′′(x) = – sin x –2 cos 2x 

f ′′(x)| x= π/2 = –1 + 2 = 1 > 0 minimum 

1 ⎛ 1 ⎞ 3 

f ′′(x)| x = π/4 = – – 2 ⎜ ⎟⎠ = – < 0 

2 ⎝ 2 2 

maximum 

1 

∴ f(0) = 2 

f(π/2) = 1 – 2 

1 = 2 

1 

f(π/6) = 2 

1 + 2 

1 × 2 

1 

= 4 

3 

A 

D 

O R • 

R R 

X 

C 

⇒ minimum value = 2 

1 

maximum value = 4 

3 

B 


ADAPTIVE TESTING 

(An initiative by a2zExam.com) 

What is Adaptive Testing? 

Adaptive testing in the tertiary environment also has a long history as it has existed since the time of the oral examination, 

and responding adaptively to students. This was totally lost in the “paper based fixed test for everyone”, because it was not 

feasible to conduct oral test of each and every student in such a short time period. But due to the development in 

technology, it is again possible to have adaptive testing for everyone. Computer Based Adaptive Testing is the way going 

forward, as it gives the benefits of Adaptive Testing and also do not require too much resource. 

Adaptive Testing Process: 

• Computer Based-adaptive test (CAT) is a form of assessment where the level of the questions administered to 

individual test-takers is dynamically tailored to their skill and knowledge levels. 

• First set of Papers are fixed for each part of syllabus, student take this test and system judges the level of the 

student based on their performance in this papers 

• Next papers onward system creates the paper especially for the student based on the student’s performance in 

previous papers. 

• Thus, each paper adapted the students level and thus is known as adaptive testing 

It is important to differentiate between online assessment tools, those readily delivered through learning management 

systems, although having the advantages of collation and immediate results are fixed rather than dynamic, as they are not 

adaptive. 


Performance Analysis by Feedback Report: 

• Analyze the bigger picture then go deep to find exact cause 

• Overall Paper’s Danger Zone Analysis: Danger zone analysis tells where the student lies out of three zone 

viz. Danger, Normal and Safe Zone represented by a white circle. Overall Paper’s Danger Zone analysis 

tells where the student lies in the distribution graph as per total marks obtained. 

 

 

 

Student or circle lies in “Red” area means student is in Danger Zone and needs to work very hard to 

achieve the goal of selection 

Student or circle lies in “Blue” area means student is in Normal Zone and should regularize and do 

systematic study to achieve the goal 

Student or circle lies in “Green” area means student is in Safe Zone but needs to keep improving to 

achieve the goal 

• Subject Wise Comparative Analysis: This analysis compares student’s subject wise marks with highest and 

average marks in the subject. It also tells percentage of question attempted by student correctly, incorrectly 

& percentage of question student did not attempted. 

 

 

 

 

Graph represents Student’s, Highest & Average Marks in each subject 

Graph implies that student is doing much better in Physics than Maths & Chemistry 

Table shows the Attempt Status & Marks of each subject in the test paper. 

Table shows that in Chemistry, student has attempted very few questions, which implies either he 

was short of time or he doesn’t knew how to answer 


Similarly, for Maths student has very high (%) of Incorrect Question, which implies either student, 

has misconception or has made lot of silly mistakes. 

• Help you know your learning gaps & how to improve or fill them 

• Skill-Wise Personal Analysis: Personal analysis is done to find out the skill wise performance. Whether student is 

able to Direct Theory or Formula based Questions, whether he has good concepts or problem solving skills, etc. 

• Inferences and Suggestions for Improvements: Based on the observation made above we deduct inferences and 

suggest methods for improvement 

Observation Inference Suggestion For Improvement 

You have got only 50 % of 

Easy and Direct Questions 

Correct. 

You are expected to grab 

all the Easy and Direct 

problems 

Your real problem lies in your unsystematic self study, Revise 

lectures on day to day basis and plan mega revision (1hrs) of 

theory on weekends. 

• Help you focus on specific weak areas 

• Topic Wise Danger Zone Analysis: For each topic the analysis is done and suggestion based on student’s zone is 

given in next column. Student should figure out the weak topics and should work according to the suggestion. 

The circle o in the graph represents student’s position 

• Question wise Detailed Analysis: Each question is analyzed based on the student’s attempt status and compared 

with overall attempt status. 

 

 

 

Question wise analysis table shows Knowledge Area & Skills which the question belongs, along with 

Students attempt status 

(%) of student attempted the question 

(%) of students among attempted which does it correctly 

Each paper has some tricky questions, which most people attempt, but does it wrong. Understanding how to solve 

those questions, improve student’s concept & learning on those topic. But the important thing is how to figure out 

those questions. 

Each paper has hard questions, which most people has left and if you have attempted you might have taken a lot of 

time; this type of question should be tried in the end. 


The above table helps you in recognizing those questions and if you keep this in mind you will be able to save time 

and marks by not attempting questions which are meant to be left. 

Benefits of Adaptive Testing 

• Adaptive test encourages student to bring out their maximum output by providing them the questions with levels 

close to their skills and knowledge level, instead of very easy or hard which most of the time de-motivates student 

from attempting. 

• Adaptive testing helps evaluator to measure the accurate skill level of students, even with small number of 

questions. 

• Better Analysis of Students Performance can be judged and provided to students using adaptive testing 

• Adaptive Testing targets the student weak areas and motivates them to improve on those areas by going to their 

level and upgrading as students improve. 

• This also brings students focus on those areas where they need improvement 

• The experience of taking an adaptive test is like participating in a high-jump event. The high-jumper, regardless of 

ability, quickly reaches a challenging level where there is about an equal chance of clearing the bar or knocking it 

down. The ‘score’ is the last height that was successfully cleared and is earned without having to jump all possible 

lower heights or trying the higher levels. 

Research shows that adaptive testing has improved the students learning by more than 22% compare to student taking 

fixed test during their preparation. 

How you can get Adaptive Testing for your preparation of Entrance Exams like 

IIT-JEE, AIEEE, BITSAT etc. 

• Visit a2zExam.com and Register. 

• Select the Course according to exam you are taking and get the Adaptive Set of papers for the preparation of the 

exam. 

• Everything will be done by them 

a2zExam - Adaptive Testing with added advantage 

First time in India, a2zExam.com brings “Adaptive Testing” for the preparation of Engineering Entrance Exams like IIT- 

JEE, AIEEE, BIT-SAT etc. a2zExam not just provide adaptive testing but also provide Feedback Report, which is used by 

the system to adapt to your level for testing, so that you can work on your weak areas before taking the next test. 

Adaptive testing is like High Jump but if every jump of high jumper (participant) is recorded and shown before the next 

jump pointing out the mistakes or good things done by them. Think how much beneficial it will be for them and the 

improvement in their records. 

Similarly, after every test you will be provided here with the feedback report pointing your strength and weakness along 

with the suggestions for improvement, so that you can work on those areas before taking the next test and improve your 

performance. 


• Adaptive Testing 

• Get Adaptive Testing for preparation of IIT-JEE, AIEEE, BITSAT etc. 

• Give edge to your preparation and improve your ranking or chance of selection 

• Don’t leave any stone unturned for such an important event of your or your child’s life. 

• Detailed Performance Analysis 

• Subject Wise, Topic Wise Personal and Comparative Analysis 

• Skill Wise Personal Analysis 

• Inferences and Suggestions for Improvement 

• Question Selection Strategy 

• Anytime Anywhere Access 

• You can take the test anytime (Day or Night), anywhere (from Home or Cyber Café or any place with Computer 

and Internet) 

• As soon as the syllabus for a test is completed take the test, no need to wait for the specific Sunday for test to be 

conducted. 


MOCK TEST-3 (SOLUTION) 

MOCK TEST– 3 PUBLISHED IN SAME ISSUE 

PHYSICS 

1. Point to point : Communication over a link between 

a single transmitter and received 

Example : Telephone 

3. (i) When final image is formed at infinity 

v0 

D 

M = – × 

u0 

u e 

(ii) When final image is formed at distance D 

v0 

⎡ D ⎤ 

M = – ⎢1 

+ ⎥ 

u0 

⎣ f ⎦ 

4. S 1 and S 2 are the two desired surfaces. 

11. (i) ‘Depletion layer’ width decreases, 

(ii) Junction field becomes very high 

12. The approximate thickness of the film should be of 

the order of wavelength of the light. 

13. 

θ 

' 

θ 

y 2 n 

θ 2 

' 

θ 1 

θ 1 

O 

' 

P 2 

P 2 

' 

P 1 

P 1 

I 

Intensity 

S 1 

S 2 

q 

5. Superconductors are those material which 

resistivity is zero below a certain temperature. 

6. 

Induction 

coil 

~ 

7. Zero 

Conducting 

Plate I 

Sphere I 

Sphere II 

Conducting 

Plate II 

Detector 

14. 

15. 

∈ 0 A 

d 

= 5 µF ………. (i) 

∈ 0 KA 

= 20 µF ……. (ii) 

2d 

K 20 

(ii) ÷ (i) = 2 5 

∴ K = 8 

8. High energy X-rays are known as hard X-rays and 

low energy X-rays are known as soft X-rays. These 

terms are relative. 

9. (1, 3) – (2, 4) 

10. The daughter element 

(release of energy is accompanied by an increase of 

B.E) 

Uniform magnetic 

field 

16. (i) λ = ν 

c 

(ii) (U av ) E = 4 

1 

∈0 

Paramagnetic 

substance 

2 

E 0 

Diamagnetic 

substance 


17. U = 2 

1 Li 2 = 2 

1 × 2 × 10 –3 (5) 2 J 

= 2.5 × 10 –2 J 

di 

23. |ε| = L dt 

= 

d 

L (t 2 ) = 2 L t 

dt 

∴ At t = 4, ε = 2 × 5 × 10 –3 × 4 = 40 × 10 –3 V 

18. X L = ωL = 100 × 30 × 10 –3 = 3 Ω 

R = 4 Ω 

V = 200 2 sin 100 t 

Z = 

2 

R + X 

2 

L 

24. 

P 

Q 

b 

B 

= 

= 5 Ω 

2 

4 + 3 

∴ cosφ = Z 

R = 5 

4 = 0.8 

2 

19. (i) X-rays are e.m. waves 

(ii) X-rays are transverse in nature 

S 

a 

Let i → current 

N → No. of turns 

l → length 

N 

n = l 

R 

20. 

1 

Reduction factor = 16 

1 = 

4 

2 

in 4 days. 

Hence life = 1 day 

1 1 

∴ For 6 days reduction factor would be 

6 = 

2 64 

∴ original amount = 4 × 10 –3 × 64kg = 0.256 kg 

21. A telescope views large objects at large distances; a 

microscope views small objects at small distances. 

Both need a small field of view. A camera views 

objects of ordinary sizes at fairly close distances. 

Here the field of view is required much more 

(compare 45° for a camera with about 1° for a 

microscope objective and something similar for a 

telescope, a moon subtends about 0.5° at the earth) 

Thus rays entering a camera lens are far from being 

paraxial and aberrations will be large and images 

will be blurred if the apertures are not very small. 

For a telescope, on the other hand, the important 

thing is its ability to resolve distant abjects (i.e., see 

them as distinct). We have seen that the resolving 

power increases with increase in aperture. 

Therefore, telescopes have as large an aperture as 

feasible. 

22. Q Current in 5 Ω is zero 

∴ bridge is balance 

∴ 

1 

6R 

6+ R 

10 

= 20 

6+ R 1 

or = 

6 R 2 

∴ R = 3 Ω 

25. 

Applying Ampere's Law along PQRS, 

∫ 

∫ 

→ 

→ 

B. dl 

= µ 0 i net 

Q 

→ 

P 

B 

R 

→ 

→ 

. dl 

+ 

∫ 

Q 

B 

S 

→ 

→ 

. dl 

+ 

∫ 

R 

B 

B.a + 0 + 0 + 0 = µ 0 n (a) i 

∴ 

I 

B = µ 0 

A 

+ 

P 

+ 

I 1 

ni 

I 2 

– 

+ 

B 

+ 

I g 

G 

R – S 

+ 

– 

D 

I 2 – I g 

Q 

– 

– 

P 

→ 

→ 

. dl 

+ 

∫ 

S 

C 

I 1 + I g 

→ 

B. dl 

= µ 0 i net 

It is an arrangement of four resistances used for 

measuring unknown resistance. 

Applying KVL in loop ADBA 

–I 1 R + I g G + I 2 P = 0 …. (i) 

and in loop DCBD, 

– (I 1 + I g ) S + (I 2 – I g ) Q – I g G = 0 

for balance bridge, I g = 0 

∴ I 1 R = I 2 P …….. (iii) 

I 1 S = I 2 Q …….. (iv) 

(iii) ÷ (iv) we get 

P R = 

Q S 


26. Diamagnetic substances are feebly magnetised in 

opposite direction to that of magnetising field 

Paramagnetic substances are feebly magnetised in 

the direction of magnetic field. 

Ferromagnetic substances are strongly magnetised 

in the direction of magnetic field. 

+ 

V 

– 

27. 

P N + – 

Reverse biasing 

In the experimental set, the P and N terminals of a 

P-N diode are connected to the negative and 

positive potential point's of a potential divider 

respectively. As the reverse bias current a feeble 

current measurable in micro amperes so a microammeter 

is used to measure it. To plot reverse bias 

characteristic, we note down reverse currents 

corresponding to various different reverse voltages 

on the diode with help of the potential divider. 

After obtaining it, the applied voltages are plotted 

along X-axis and corresponding reverse currents 

along Y-axis of a graph as shown in the fig. 

I (mA) 

– 

Characteristic of a P-N diode 

+ 

V 

Reverse bias resistance- The ratio of small change 

in reverse voltage (before break down voltage) to 

the corresponding change in reverse current for a 

P-N diode is known as its reverse bias resistance, 

i.e, Reverse bias resistance 

Smallchangein 

= 

reversevoltage 

Corresponding changein 

reversecurrent 

28. Circuit diagram for drawing the input and output 

characteristics. 

R 2 

V BB 

V BE 

I B 

B 

+ 

µA 

– 

I c 

– + 

C mA 

E V CE IB 

+ 

– 

R 1 

V cc 

29. 

Q'' 

P'' 

α 

O 

α 

Q' 

P' 

β 

Astronomical telescope 

(i) When the final image is formed on the nearest 

distance of clear vision D 

f 

M = – 0 ⎡ f ⎤ 

f 

⎢1 

+ e 

⎥ 

e ⎣ D ⎦ 

(ii) When the final image is formed at infinity 

f 

M = – 0 

f e 

On increasing the aperture of the objective lens the 

magnifying power of telescope will increase. 

30. U = 

∫ q Vdq 

0 

q 

⎛ q ⎞ 

= 

∫ 

⎜ ⎟ dq 

0 ⎝ C ⎠ 

= 

2 

q 

2C 

= 2 

1 CV 

2 

= 2 

1 

= 2 

1 

∈0 E 2 (Ad) 

⎛∈0 A ⎞ 

⎜ ⎟ (Ed) 2 

⎝ d ⎠ 

U 1 

∴ Energy density, = ∈0 E 2 

Ad 2 

OR 

(i) From Gauss' theorem 

φ = 

q in 

∈ 0 

E 

B 

+ – 

∴ q in = φ × ∈ 0 = – 6 × 10 3 × 8.85 × 10 –12 

= –5.31 × 10 –8 C 

(ii) Flux remains the same 

d 

E 

f e 

Typical shape of the input characteristics. 


CHEMISTRY 

1. H.C.P has highest 74% efficiency 

2. When in Fe(OH) 3 ppt FeCl 3 is added Fe +3 ions are 

adsorb over the surface of Fe(OH) 3 it results in the 

formation of Fe(OH) 3 solution. 

3. Lithium tetrahydrido aluminate (III) 

4. FeCO 3 is siderite ore. 

5. Phenol & Formaldehyde 

6. Antiseptics are germicides which can be applied on 

wounds 

Ex. Soframycin, Tincture iodine 

7. Amino acid in which amino group are more then 

– COOH group are called basic amino acid. 

Lysine ( R is (CH 2 ) 3 – NH 2 ) 

8. CH 3 F < CH 3 Cl < CH 3 Br < CH 3 I 

9. t 1/2 = 5730 

0.693 

λ = = 1.21 × 10 –4 yr –1 

t 

λ = 

1/ 2 

2.303 

t 

ln 

2.303 

t = 

1.21× 

10 

= 1845 yr 

−4 

a 

a 

0 

t 

10 

ln 8 

10. With the increase in temperature rate constant 

increases. It is found that with 10 K rise in 

temperature the rate of reaction become 2 – 3 times. 

With the increase in temperature 

(1) More no. of collisions occur between the 

molecules. 

(2)Only those molecules which are having 

minimum sufficient energy to participate in the 

chemical rx n , reacts with each other and form 

product 

(3) For effective collision activated molecule must 

collide in the proper orientation 

∴ Rate of rx n = P × Z . e 

P = Orientation factor 

Z = No. of collisions 

e 

−E a / RT 

−E a / RT 

= No. of activated molecules 

11. (a) Standard Hydrogen electrode - When H 2 gas at 

1 atm pr is supplied on Pt sheet dipped in the 

aqueous solution of an acid having molarity 1M 

H 2 l atm 

Pt 

1 M aq acid 

solution 

Following Chemical rx n takes place 

H 2 (g, l atm) 2H + (aq, 1M) + 2e – 

the potential of this half electrode = 0.0 V 

(b) Kohlrausch's law states conductivity of a 

solution at infinite dilution is equal to sum of 

molar conductivity of all the ions present in the 

solution. 

0 

Λ m = γ ⊕ λ ⊕ º + γ –º λ – º 

12. Actinides show much higher oxidation states than 

Lanthanides because energy difference between 5f, 

6d and 7s orbitals is less and hence electrons also 

participate from 5f orbital also. 

13. The seperation of Ag + and Hg 2+ 2 in group – I is 

carried out by dissolving the precipitate of AgCl in 

NH 4 OH, AgCl forms a soluble complex with 

NH 4 OH. 

Whereas Hg 2 Cl 2 forms a black water insoluble 

complex. 

⇒ AgCl + NH 4 OH → [ Ag(NH3 ) 2 ]Cl + 2H 2O 

(water soluble) 

⇒ Hg 2 Cl 2 + NH 4 OH → Hg(NH 2 )Cl + Hg + HCl 

black percipitate + H 2 O 

14. The concentrated ore is heated with excess of air to 

remove water and carbon dioxide to remove 

sulphur and arsenic impurities and to oxidise 

ferrous to ferric oxide for eg; 

2Fe 2 O 3 .3H 2 O → 2Fe 2 O 3 + 3H 2 O 

FeCO 3 → FeO + CO 2 ↑ 

S + O 2 → SO 2 ↑ 

4As + 3O 2 → 2As 2 O 3 

4Fe + O 2 → 2Fe 2 O 3 

15. PHBV has 3hydroxybutanoic acid & 3 

hydroxypentanoic 

acid. 

O – CH – CH 2 – CO – O – CH – CH 2 – CO n 

CH 3 CH 2 – CH 3 


16. Purine bases in DNA and RNA are Adenine & 

Guanine. 

Pyrimidine bases in DNA are Cytosine & Thymine 

while in RNA are Cytosine & Uracil 

17. Detergents are sodium or potassium salts of 

sulphonic acid. 

Cationic detergent – Cetyl trimethyl ammonium 

bromide 

Anionic – Alkyl benzene sulphonate 

18. (A) CHCl 3 (B) HC ≡ CH 

19. (a) Radius of gold r = 0.144 nm 

F.C.C. 4 r = 2a 

4r 

a = = 2 2 r 

2 

= 2 × 1.414 × 0.144 

= 0.407 nm 

Edge length a = 0.407 nm. 

(b) (i) When Ge is doped with 

2 2 

(4s 4p ) 

In 13 th gr 

2 1 

(5s 5p ) 

element all the 3e – get bonded and fourth bond 

of Ge contain only one e – and hence an e – 

deficient bond or a hole is formed and p type 

semiconductor is formed. 

(ii) Similarly when Si is doped with As 4s 2 4p 3 4e – 's 

get bonded and fifth e – remain unbonded ∴ n- 

type semiconductor is formed. 

0 0 

20. p = p A x A + p B x B 

n A = 1 

p T 1 

= 250 bar 

0 

A 

1 

250 = p × + 

0 2 

p × 3 3 

B 

n B = 2 mol. 

n A = 2 

n B = 2 mol 

0 1 

p T 2 

= 300 = p A . + 

0 1 

p B × 2 2 

solving (1) & (2) 

0 

p A = 450 bar 

..... (1) 

......(2) 

0 

p B = 150 bar 

21. (a) (i) When silica gel is placed is atmosphere 

saturated with water adsorption of moisture takes 

place 

(ii) CaCl 2 adsorbs H 2 O. 

(b) Zeolite is a shape selective catalyst which are 

metal alumino silicates M x/n (AlO 2 ) x (SiO 2 ) y .mH 2 O. 

When zeolite is heated pores are generated, these 

pores are having size 260 – 740 pm which can 

absorb molecule of definite size. Therefore it is 

called shape selective catalyst 

22. In case of nitrogen family, basic character of 

hydrides decreases from NH 3 to BiH 3 because with 

the increase in size of central element lone pair 

density about it decreases and tendency of proton 

(H + ) to coordinate with it decreases and hence basic 

character of hydrides decreases. 

23. 

24. 

O 

O 

Cr 

O¯ 

Cr 2 O 7 

2– 

O 

131º 

O 

Cr 

O¯ 

25. 2-Methoxy-2-methylpropane 

26. 

27. The compound A can be either an aldehyde or a 

ketone. Since it resists oxidation it must be a 

ketone. i.e., acetone (CH 3 COCH 3 ) 

The reactions involved are : 

O 

OH 

Reduction 

CH 3 – C – CH 3 

2[H] 

O 

O 

CH 3 – CH – CH 3 

(B) 

2-Propanol 

MgBr 

CH 3 – C – CH 3 + CH 3 – CH – CH 3 

(A) 

OMgBr 

CH 3 – C – CH 3 

OH 

CH(CH 3 ) 2 

H 2O 

CH3 – C – CH 3 

CH(CH 3 ) 2 

2, 3 – Dimethyl-2-butanol 

HBr 

–H 2O 

Br 

Mg 

CH 3 – CH – CH 3 

(C) 

2-Bromopropane 

28. (a) The ideal solution is that solution which follows 

the Raoult's law i.e for ideal solution :- 

(i) ∆H mixing = 0 

(ii) ∆V mixing = 0 

Non ideal solution is that solution which does not 

follow Raoult's law and for which : 

(i) ∆H mixing ≠ 0 

(ii) ∆V mixing = 0 

In case of cyclohexane – ethanol a solution with 

+ve deviation is obtained 

In case of a acetone – Chloroform 


–ve deviation is obtained 

Cl 

Cl 

Cl 

+δ –δ +δ 

C H O C 

(b) Moles of solute n B = M 

30 

CH 3 

CH 3 

90 

Moles of H 2 O ( n H 2 O ) = = 5 18 

5 M 

x A = = 

5 + 30 / M 6 + M 

0 

p A = p A x A 

0 M 

p A = p A . 

...... (1) 

6 + M 

When 18g of H 2 O is added to solution 

6 M 

x H 2 O = = 

6 + 30 / M 5 + M 

0 M 

p' A = p A 

...... (2) 

5 + M 

from (1) & (2) 

5 + M 2.8 

= 

6 + M 2.9 

⇒ M = 23 g/mol 

29. (a) Hypophosphorus acid 

O 

(e) Hypophosphoric acid 

O O 

HO 

P 

OH 

P 

OH 

OH 

30. (a) Since compound (A) is optically inactive and 

contains nitrogen which gives alcohol with HNO 2 , 

it is primary amine. The reactions may be given as 

CH 3 CH 2 CH 2 CH 2 NH 2 ⎯ HNO ⎯⎯ 

2 →CH 3 CH 2 CH 2 CH 2 OH+N 2 +H 2 

O 

(A) 

(B) 

1-Aminobutane 

1-Butanol 

H SO4 

,440K 

CH 3 CH 2 CH 2 CH 2 OH ⎯⎯ 

2 ⎯⎯ 

– H2O 

⎯ →CH 3 CH 2 CH = CH 2 

(B) 

(C) 

1-Butene 

CH 3 CH 2 CH=CH 2 ⎯ HBr ⎯→CH 3 CH 2 CH–CH 3 

(C) | 

Br 

(D) 

2-Bromobutane(optically active) 

MATHEMATICS 

Section A 

P 

H 

OH 

H 

(b) Pyrophosphoric acid 

O 

O 

1. Q (1, 1), (2, 2), (3, 3) ∉ R 

⇒ R is not reflexive. 

again (1, 2) ∈ R ⇒ (2, 1) ∈ R ⇒ R is symmetric 

again (1, 2) ∈ R, (2, 1) ∈ R but (1, 1) ∉ R 

⇒ R is not transitive. 

P 

HO O 

OH 

(c) Dithionic acid 

O O 

P 

OH 

OH 

2. On differentiating 

2 x –1/3 2 

+ y 

–1/3 

3 3 

dy = – 

dx 

⎡ y 

⎢ 

⎣ x 

1/3 

1/3 

⎤ 

⎥ ⎥ ⎦ 

dy = 0 

dx 

HO — S — S — OH 

O O 

(d) Marshall acid 

O 

O 

HO — S — O — O — S — OH 

O 

O 

3. We have, 

3 logsin x 

I = 

∫ 

cos x e dx = 

∫ 

cos 3 xsin x dx 

Putting cos x = t and – sin x dx = dt 

or, sin x dx = – dt, we get 

I = – 

∫ t 3 dt = 

− t 4 

+ C = – 

4 

cos 4 x + C 

4 


4. Let x + y = v. Then 

dy dv dy dv 

1 + = ⇒ = –1 

dx dx dx dx 

dy dv 

Putting x + y = v and = –1 the given 

dx dx 

differential equation, we get v 2 ⎛ dv ⎞ 

⎜ −1⎟ = a 2s 

⎝ dx ⎠ 

⇒ v 2 dv = a 2 + v 2 

dx 

⇒ v 2 dv = (a 2 + v 2 ) dx 

2 

v 

⇒ dv = dx [By separating the variables] 

2 2 

v + a 

⎛ 

2 

⎞ 

⇒ ⎜ 

a 

⎟ 

1− 

2 2 

dv = dx 

⎝ v + a ⎠ 

⇒ 

∫ 1 . dv – ∫ a2 v + a 

2 

2 

1 

dv = 

∫ 

dx + c 

[On integration] 

⇒ v – a tan –1 ⎛ v ⎞ 

⎜ ⎟ = x + c 

⎝ a ⎠ 

⇒ (x + y) –a tan –1 ⎛ x + y ⎞ 

⎜ ⎟ = x + c 

⎝ a ⎠ 

5. Let l, m, n be the direction cosines of the given 

vector r (say). Then, its projections on the coordinate 

axes are l | r |, m | r |, n | r | 

∴ l | r | = 6, m | r |= –3, n | r | = 2 … (i) 

⇒ {l | r |} 2 + {m | r |} 2 + {n| r |} 2 = 6 2 + (–3) 2 + (2) 2 

⇒ | | 2 (l 2 +m 2 + n 2 ) = 36 + 9 + 4 

⇒ | | 2 = 49 

⇒ | r | = 7 [Q l 2 + m 2 + n 2 =1] 

Putting | r | = 7 in (i), we obtain that the direction 

cosines of r 6 − 3 2 

are l = , m = , n = 7 7 7 

6. The angle θ between vectors a r and b r is given by 

r r 

a. 

b 

cos θ = r r 

| a || b | 

For the angle θ to be obtuse, we must have 

cos θ < 0 

. b 

⇒ r a r r 

r < 0 

| a || b | 

⇒ a r . b r < 0 [Q | a r |, | b r | > 0] 

⇒ 14x 2 –8x + x < 0 ⇒ 7x (2x –1) < 0 

⇒ x (2x –1) < 0 ⇒ 0 < x < 1/2 

Hence, the angle between the given vectors is obtuse 

if x ∈ (0, 1, 2) 

7. Let the equation of the required plane be 

x y z + + = 1 

…(i) 

a b c 

Then, the coordinates of A, B and C are A (a, 0, 0), 

B(0, b, 0), and C(0, 0, c) respectively. So the centroid 

of triangle ABC is (a/3, b/3, c/3). But the coordinates 

of the centroid are (p, q, r) 

a b c 

∴ p = , q = and r = 3 3 3 

⇒ a = 3p, b = 3q and c = 3r 

Substituting the values of a, b and c in (i), we get the 

required plane as 

x 

3 p 

+ y 

3 q 

+ z 

3 r 

= 1 ⇒ x y z + + = 3 

p q r 

8. a + b = 6, …(i) 

8 

ab = 8 ⇒ b = a 

Putting in (i) 

a + a 

8 = 6 ⇒ a = 4, 2 

From (ii) 

⇒ b = 4 

8 , 2 

8 ⇒ b = 2, 4 

Hence (a, b) = (4, 2) or (2, 4) 

… (ii) 

⎡5 

+ 3 2 + 6⎤ 

⎡4 

9. 2X = ⎢ ⎥ ⇒ X = 

⎣ 0 9 −1 

⎢ ⎦ ⎣0 

4⎤ 

4 

⎥ ⎦ 

⎡5 

2⎤ 

⎡4 

4⎤ 

⎡1 

and Y = ⎢ ⎥ – 

⎣0 

9 

⎢ ⎥ ⇒Y = 

⎦ ⎣0 

4 

⎢ ⎦ ⎣0 

10. ∆ = 1/2 

k 

− k + 1 

− 4 − k 

⇒ 8k 2 + 4k – 4 = 0 

⇒ k = 1/2, –1 

2 − 2k 

2k 

6 − 2k 

1 

1 

1 

Section B 

= 0 

− 2⎤ 

5 

⎥ ⎦ 

11. Let A → total of 8 in first throw 

B → total of 8 in 2nd throw 

Number of exhaustive cases when a pair of dice is 

thrown = 6 × 6 = 36 

Cases favourable to a total of 8 in each throw are 

(2, 6), (6, 2), (3, 5), (5, 3), (4, 4) 

Their number = 5 

P(A) = P(B) = 5/36 

5 5 25 

P(A and B) = P(A). P(B) = × = 36 36 1296 


12. f(x) = sin 2 x + [1 – cos 2 (π/3 + x)] 

= 1 –[cos 2 (π/3 + x) – sin 2 x] 

+ 2 

1 2 cos x cos (π/3 + x) 

+ 2 

1 [cos(π/3+2x) + cos π/3] 

= 1 – [cos (π/3+2x) –cos π/3] + 2 

1 cos (π/3 +2x) + 4 

1 

f(x) = 5/4 

g(f(x)) = g(5/4) = 1 

→ constant function 

13. Q f(x) is continuous at x = 2, 4 

∴ f(2) = f(2 – ) 

⇒ 3 × 2 + 2 = lim (2 –h) 2 + a(2 – h) + b 

h →0 

⇒ 8 = 2a + b + 4 

⇒ 2a + b = 4 

again, 

f(4) = f(4 + ) 

⇒ 3 × 4 + 2 = 

⇒ 14 = 8a + 5b 

Solve (i) , (ii) 

a = 3, b = –2 

14. y = tan –1 ⎡ 1+ 

sin x ⎤ 

⎢ ⎥ 

⎣ cos x ⎦ 

lim 2a (4 + h) + 5b 

h→0 

… (ii) 

… (i) 

⎡ 

2 

y = tan –1 (cos x / 2 + sin x / 2) 

⎥ ⎥ ⎤ 

⎢ 

2 

2 

⎢⎣ 

cos x / 2 − sin x / 2 ⎦ 

y = tan –1 ⎡cos 

x / 2 + sin x / 2⎤ 

⎢ 

⎥ 

⎣cos 

x / 2 − sin x / 2 ⎦ 

y = tan –1 [tan (π/4 + x/2)] 

y = π/4 + x/2 

dy 1 = 

dx 2 

⎡ 2 

2 

15. Let y = tan –1 1+ 

x − 1− 

x 

⎥ ⎥ ⎤ 

⎢ 

, z = cos –1 x 2 

⎢ 2 

2 

⎣ 1+ 

x + 1− 

x ⎦ 

put x 2 = cos 2θ 

y = tan –1 ⎪⎧ 

1+ 

cos 2θ 

− 1− 

cos 2θ 

⎪⎫ 

⎨ 

⎬ 

⎪⎩ 1+ 

cos 2θ 

+ 1− 

cos 2θ 

⎪ ⎭ 

y = tan –1 ⎡cosθ − sin θ⎤ 

⎢ ⎥ 

⎣cosθ + sin θ ⎦ 

y = tan –1 [tan (π/4 –θ)] 

y = 4 

π – 2 

1 cos –1 x 2 

dy − ⎛ 

= ⎜ −1 

dx 21 

⎜ 

⎝ 1− 

x 

dy x = 

dx 

4 

1− 

x 

4 

⎞ 

. 2x⎟ 

⎟ 

⎠ 

dz −1 

= .2x 

dx 

4 

1− x 

dy 

∴ = dz 

dy / dx 

= 

dz / dx 

x / 

− 2x 

/ 

16. Area A = 2 

1 × b × AD 

2 

2 b 

b 

= . 2 

x − 

4 

b 

A = 4 2 2 

4x − b 

1− 

x 

4 

1− 

x 

4 

= 

−1 

2 

dA b dx 

= 8x 

B b/2 D b/2 

dt 

2 2 

8 4x 

− b dt 

dA bx 

⎛ dx ⎞ 

= × 3 ⎜Q 

= 3⎟ dt 2 2 

4x 

− b 

⎝ dt ⎠ 

2 

⎛ dA ⎞ 3b 

⎜ ⎟ = 

⎝ dt ⎠ 

2 2 

x= 

b 4b 

− b 

× 3 = 3 b cm 2 /sec. 

OR 

(i) Since (x – 2) ≥ 0 in [2, 3] 

so f(x) = x − 2 is continuous 

1 

(ii) f ´(x) = exists for all x ∈(2, 3) 

2 x − 2 

∴ f(x) is differentiable in (2, 3) 

Thus lagrang's mean value theorem is applicable; 

∴ There exists at least one real number in (2, 3) 

such that 

f (3) − f (2) 

f ´(c) = 

3 − 2 

or 

2 

1 

= 

c − 2 

( 1) − 0 

1 

c = 2 + 4 

1 = 2.25 ∈(2, 3) 

⇒ LMV is verified and the 

req. point is (2.25, 0.5) 

2x 

17. 

∫ 

dx 

2 2 2 

1− 

x − ( x ) 

⇒ 2 c − 2 = 1 

Let x 2 = t. Then, d(x 2 ) = dt ⇒ 2x dx = dt ⇒ dx = 

∴ I = 

∫ 

= 

∫ 

⇒ Ι = 

∫ 

dt 

1− 

t − t 

dt 

1 1 

− { t 

2 + t + − −1} 

4 4 

2 

dt 

= 

∫ 

− { t 

2 + t −1} 

dt 

= 

∫ 

⎪⎧ 

⎞ ⎪⎫ 

⎨⎜ 

⎛ 1 2 

5 

− t + ⎟ − ⎬ 

⎪⎩ ⎝ 2 ⎠ 4 ⎪⎭ 

x 

5 

4 

A 

dt 

⎞ 

⎜ 

⎛ 1 

t + ⎟ 

⎝ 2 ⎠ 

− 

2 

x 

dt 

2x 

C 


= 

∫ 

2 

2 

⎛ 

⎞ 

⎜ 

5 

⎟ 

⎝ 2 ⎠ 

dt 

⎛ 1 ⎞ − ⎜t 

+ ⎟ 

⎝ 2 ⎠ 

⇒ I = sin –1 ⎛ t + 1/ 2 

⎟ ⎞ 

⎜ + C 

⎝ 5 / 2 ⎠ 

= sin –1 ⎛ 2t + 1 

⎟ ⎞ 

⎜ + C 

⎝ 5 ⎠ 

⎛ 

= sin –1 ⎟ ⎞ 

⎜ 

2x 

2 + 1 

+ C 

⎝ 5 ⎠ 

OR 

2 + sin x 

I = 

∫ 

e x/2 dx 

1 + cos x 

⎡ 2 sin x ⎤ 

= 

∫ ⎢ + ⎥ e x/2 dx 

⎣1+ cos x 1+ 

cos x ⎦ 

⎡ 

⎢ 2 

= 

∫ ⎢ 

⎢ 

2 x 

cos 

⎣ 2 

+ 

x 

2sin 

2 

2cos 

x 

cos 

2 

2 

x 

2 

⎛ x x ⎞ 

= 

∫ 

⎜sec 2 + tan ⎟ e x/2 dx 

⎝ 2 2 ⎠ 

2 tan 2 

x .e x/2 + c 

⎤ 

⎥ 

⎥ e x/2 dx 

⎥ 

⎦ 

4 

1+ 

2 

18. I = 

x 

∫ 

dx 

2 16 

x + 

2 

x 

4 

1+ 

2 

⇒ I = 

x 

∫ 

dx 

2 

2 ⎛ 4 ⎞ 

x + ⎜ ⎟ − 8 + 8 

⎝ x ⎠ 

4 

1+ 

2 

⇒ I = 

x 

∫ 

dx 

2 

⎛ 4 ⎞ 

⎜ x − ⎟ + 8 

⎝ x ⎠ 

4 ⎛ ⎞ 

Let x – = t. Then, d ⎜ x − 4 ⎟⎠ = dt 

x ⎝ x 

⎞ 

⇒ ⎜ 

⎛ 4 

1 + ⎟ dx = dt 

⎝ x 2 ⎠ 

dt 

∴ I = 

∫ 2 

2 

t + (2 2) 

⇒ I = 

1 tan 

–1 ⎛ t ⎞ 

⎜ 

⎟ + C 

2 2 ⎝ 2 2 ⎠ 

⇒ I = 

= 

2 

1 

2 

⎛ 4 ⎞ 

1 

⎜ x − ⎟ 

tan 

–1 

⎜ x ⎟ + C 

2 2 ⎜ 2 2 ⎟ 

⎝ ⎠ 

⎛ 

2 

tan –1 

⎟ ⎞ 

⎜ 

x − 4 

+ C 

⎝ 2x 

2 ⎠ 

dy 

19. We are given that + (–2)y = cos 3x 

dx 

This is a linear differential equation of the form 

dy + Py = Q, where P = –2 and Q = cos 3x 

dx 

∴ I.F. = 

∫ 

Pdx = 

∫ − 2 dx = e –2x 

e e 

Multiplying both sides of (i) by I.F. = e –2x , we get 

e –2x dy –2ye –2x = cos 3x. e –2x 

dx 

Integrating both sides w.r.t. x, we get 

ye –2x = 

∫ 

e −2x 

cos 3x dx + C 

[Using : y (I.F.) = 

∫ 

Q ( I. 

F.) 

dx + C ] 

⇒ ye –2x = I + C, where I = e –2x cos 3x 

Now, I = 

∫ 

e 

− 2 x cos3x 

dx 

I II 

1 

⇒ I = e –2x 2) 

sin 3x – e 

3 ∫ − 

3 

⇒ I = 3 

1 e –2x sin 3x + 3 

2 

⇒ I = 3 

1 e –2x sin 3x 

+ 3 

2 

⎡ − 

⎢ 

⎣ 3 

cos3x 

− 

( −2x 

∫ 

∫ 

−2 

e x sin 3x 

1 −2x 

− 2 x 

e 

⇒ I = 3 

1 e –2x sin 3x 

+ 3 

2 

⎡ 

⎢− 

e 

⎣ 3 

2 

cos3x 

− 

3 

( −2) 

e 

1 − 2x 

∫ 

e 

−2 

x 

sin 3x 

dx 

dx 

⎛ − cos3x 

⎞ ⎤ 

⎜ ⎟ dx⎥ 

⎝ 3 ⎠ ⎦ 

⎤ 

cos3x 

dx⎥ 

⎦ 

⇒ I = 3 

1 e –2x sin 3x– 9 

2 e –2x cos 3x – 9 

4 I 

⎛ 4 ⎞ 

⇒ ⎜ I + I ⎟ = 

⎝ 9 ⎠ 

2x 

e − (3 sin 3x –2 cos 3x) 

9 

…(i) 

2x 

e − 

⇒ I = (3 sin 3x – 2 cos 3x) 

13 

Substituting the value of I in (ii), we get 

2x 

ye –2x e − 

= (3 sin 3x –2 cos 3x) + C, which is the 

13 

required solution. 


20. We have, 

(1 – a r . b r ) 2 + | a r + b r +( a r × b r )| 2 

= {1 – 2( a r . b r ) + ( a r . b r ) 2 } + {( a r + b r + a r × b r ). 

(a r + b r + a r × b r )} 

= {1 –2 ( a r . b r ) + ( a r . b r ) 2 } + ( a r + b r ).( a r + b r ) 

+ ( a r + b r ). ( a r × b r )+ ( a r × b r ).( a r + b r )+ ( a r × b r ).( a r × b r ) 

= {1 –2 ( a r . b r ) +( a r . b r ) 2 } + {| a r + b r | 2 + a r . ( a r × b r ) 

+ b r . ( a r × b r ) + ( a r × b r ) . a r ( a r × b r ). b r + | a r × b r | 2 } 

= {1–2 ( a r . b r ) + ( a r . b r ) 2 } + { | a r + b r | 2 + | a r × b r | 2 } 

[Q a r ⊥( a r × b r ), b r ⊥( a r × b r ) ∴ a r ( a r × b r ) = b r .( a r × b r )= 0] 

= 1 – 2 ( a r . b r ) + ( a r . b r ) 2 + | a r | 2 + | b r | 2 +2 ( a r . b r ) 

+ |a r × b r | 2 

= 1 + | a r | 2 + | b r | 2 + ( a r . b r ) 2 + | a r × b r | 2 

= 1 + | a r | 2 + | b r | 2 + | a r | 2 | b r | 2 

[Q( a r . b r ) 2 + | a r × b r | 2 = | a r | 2 | b r | 2 ] 

= (1 + | a r | 2 ) (1+| b r | 2 ) 

Hence, (1+ | a r | 2 ) (1+ | b r | 2 ) 

= 1 – ( a r . b r ) 2 + | a r + b r + a r × b r | 2 

21. Let L be the foot of the perpendicular drawn from the 

point P (0, 2, 3) to the given line. 

The coordinates of a general point on 

x + 3 y −1 

z + 4 

x + 3 y −1 

z + 4 

= = are given by = = =λ 

5 2 3 

5 2 3 

i.e. x = 5λ –3, y = 2λ + 1, z = 3λ –4 

Let the coordinates of L be 

(5λ –3, 2λ + 1, 3λ – 4) 

So direction ratios of PL are proportional to 

5λ – 3 – 0, 2λ + 1 –2, 3λ –4 – 3 

i.e. 5λ –3, 2λ –1, 3λ – 7 

Direction ratios of the given line are proportional to 

5, 2, 3 

P(0, 2, 3) 

22. 

y.sin x = 

∫ 

(2x 

+ x cot x) 

sin x dx 

2 

= 

∫ 

2 xsin x dx + 

∫ 

x cos x dx 

2 

= 

∫ 

2 xsin x dx + x 2 sin x – 

∫ 

2 xsin x dx + c 

= x 2 sin x + c ...(1) 

Substituting y = 0 and x = π/2, we get 

0 = 

2 

π 

4 

2 

π 

+ c or c = – 4 

∴ (i) ⇒ y sin x = x 2 sin x – 

or y = x 2 – 

2 

π 

4 

cosec x 

2 

2 

π 

1 x yz x x xyz 

1 2 

1 y zx = y y xyz 

xyz 2 

1 z xy z z xyz 

C 1 ↔ C 3 

1 x 

2 

y 

⇒ 1 y 

2 

y 

1 z 

2 

z 

R 1 → R 1 – R 2 , R 2 → R 2 – R 3 

0 x − y 

2 2 

x − y 

= 0 y − z 

2 2 

y − z 

1 z 

2 

z 

0 1 x + y 

= (x – y) (y – z) 0 1 y + z 

1 z 

2 

z 

1 x + y 

= (x – y) (y – z) 

1 y + z 

= (x – y) (y – z) [(y + z) – (x + y)] 

= (x – y) (y – z) (z – x) 

4 

xyz 

= xyz 

x 

y 

z 

x 

y 

z 

2 

2 

2 

1 

1 

1 

OR 

x + 3 

= 

5 

y − 1 = z 

2 

+ 

3 

4 L (5λ –3, 2λ +1, 3λ –4) 

Since PL is perpendicular to the given line. 

∴ 5(5λ –3) + 2(2λ –1) + 3(3λ –7) = 0 ⇒ λ = 1 

Putting λ = 1 in (i), the coordinates of L are (2, 3, –1) 

∴Length of the perpendicular from P on the given 

line. 

2 

2 

2 

= PL = ( 2 − 0) + (3 − 2) + ( −1− 

3) = 21 units 

OR 

Here integrating factor = ∫cot 

xdx 

e = e log sinx = sin x 

∴ the solution of differential equation is given by 

b + c c + a a + b 

c + a a + b b + c = 0 

a + b b + c c + a 

C 1 → C 1 + C 2 + C 3 

1 c + a a + b 

⇒ 2(a + b + c) 1 a + b b + c 

1 b + c c + a 

R 2 → R 2 – R 1 , R 3 → R 3 – R 1 

1 c + a a + b 

⇒ 2(a + b + c) 0 

0 

b − c c − a 

b − a c − b 

= 0 

= 0 


⇒ 2(a + b + c) (–a 2 – b 2 – c 2 + ab + bc + ca) = 0 

⇒ –(a + b + c) [(a – b) 2 + (b – c) 2 + (c – a) 2 ] = 0 

⇒ a + b + c = 0 or a = b = c 

Section C 

23. Getting 6 → success 

1 5 

p = ⇒ q = 1 – p = 6 6 

A can win the game in 1st, 3rd, 5th …..throws. 

P(A winning) = p + qqp + qqqqp + ……… 

= p[1 + q 2 + q 4 + ……..] 

⎛ 1 

= p ⎟ ⎞ 6 

⎜ = 

⎝1− q ⎠ 11 

P(B winning) = 11 

5 

24. y = x … (i) 

y = x 3 … (ii) 

Solving (i) and (ii) 

O(0, 0), A(1, 1), B(–1, –1) 

Required area = area BCOB + Area ODAO 

C 

• 

B(–1, –1) 

Area BCOB = 

∫ 

( x − x 

3 ) dx = 

0 

−1 

y 

y = x 

• 

A(1, 1) 

D 

• 

O(0, 0) x 

⎛ 1 1 ⎞ 1 

= − ⎜ − ⎟ = 

⎝ 2 4 ⎠ 4 

1 

Area ODAO = 

∫ 

( x − x 

3 ) dx = 

0 

= 2 

1 – 4 

1 = 4 

1 

⎛ 

2 4 

⎞ 

⎜ 

x ⎟ 

− x 

2 4 

⎝ ⎠ 

⎛ 

2 4 

2 4 ⎟ ⎞ 

⎜ 

x 

− x 

⎝ ⎠ 

∴ required area = 4 

1 + 4 

1 = 2 

1 sq. units. 

25. The equations of two given lines are 

x −1 y − 2 z − 3 

= = 

… (i) 

2 3 4 

x − 2 y − 4 z − 5 

and = = 

… (ii) 

3 4 5 

Line (i) passes through (1, 2, 3) and has direction 

ratios proportional to 2, 3, 4, so its vector equation is 

1 

0 

0 

−1 

r = a r 1+ λ b r 

1 

… (iii) 

where, a r 1= î + 2 ĵ + 3 kˆ and b r 1 = 2 î + 3 ĵ + 4 kˆ 

Line (ii) passes through (2, 4, 5) and has direction 

ratio proportional to 3, 4, 5. So, its vector equation is 

r = a r 2 + µ b r 

2 

… (iv) 

where a r 2 = 2 î + 4 ĵ + 5 kˆ and b r 2 = 3 î + 4 ĵ + 5 kˆ 

The shortest distance between the lines (iii) and (iv) 

is given by 

r r r r 

( a2 

− a1).( 

b1 

× b2 

) 

S.D. = r r 

… (v) 

| b1 

× b2 

| 

We have, a r 2 – a r 1 = (2 î + 4 ĵ + 5 kˆ ) – ( î + 2 ĵ + 3 kˆ ) 

= î + 2 ĵ + 2 kˆ 

and b r 1 × b r 2 = 

iˆ 

2 

3 

ˆj 

3 

4 

kˆ 

4 

5 

= – î + 2 ĵ – kˆ 

∴ | b r 1 × b r 2 | = 1 + 4 + 1 = 16 

and,( a r 2 – a r 1 ).( b r 1 × b r 2 ) = ( î + 2 ĵ +2 kˆ ).(– î + 2 ĵ – kˆ ) 

= –1 + 4 –2 = 1 

Substituting the values of ( a r 2 – a r 1 ) . ( b r 1 × b r 2 ) and 

| b r 1 × b r 2 | in (v), we obtain S.D. = 1/ 6 

26. A 11 = 

A 13 = 

A 22 = 

A 31 = 

A 33 = 

1 

− 2 

2 

0 

1 

0 

− 2 

2 

1 

3 

2 

= 1 + 6 = 7; A 12 = – 

1 

0 

1 

= –4; A 21 = – 

− 2 

0 

1 

1 

0 

3 

1 − 2 

= 1; A 23 = – 

= –6; A 32 = – 

= 1 + 4 = 5 

− 2 

− 2 

1 

0 

1 

2 

3 

= –2 

1 

0 

1 

= 2 

− 2 

= 2 

− 2 

0 

3 

= –3 

⎡ 7 − 2 − 4⎤ 

⎡ 7 2 − 6⎤ 

∴ adj A = 

⎢ 

⎥ 

⎢ 

2 1 2 

⎥ 

= 

⎢ ⎥ 

⎢ 

− 2 1 − 3 

⎥ 

⎢⎣ 

− 6 − 3 5 ⎥⎦ 

⎢⎣ 

− 4 2 5 ⎥⎦ 

Also |A| = (1) (7) + (–2) (–2) +0(–4) = 7 + 4 = 11 

⎡ 7 2 − 6 ⎤ 

∴ A –1 adj A 1 

= = 

⎢ 

⎥ 

| A | 11 ⎢ 

− 2 1 − 3 

⎥ 

⎢⎣ 

− 4 2 5 ⎥⎦ 

Again the given system of equations can be written as 

⎡x⎤ 

⎡10⎤ 


⎢ ⎥ 

⎢ 

y 

⎥ 

, B = 

⎢ ⎥ 

⎢ 

8 

⎥ 

⇒ X = A –1 B 

⎢⎣ 

z⎥⎦ 

⎢⎣ 

7 ⎥⎦ 

′ 


⎡x⎤ 

⎡ 7 2 − 6 ⎤ ⎡10⎤ 

i.e. 

⎢ ⎥ 1 

⎢ 

y 

⎥ 

= 

⎢ 

⎥ 

11 ⎢ 

− 2 1 − 3 

⎢ ⎥ 

⎥ ⎢ 

8 

⎥ 

⎢⎣ 

z⎥⎦ 

⎢⎣ 

− 4 2 5 ⎥⎦ 

⎢⎣ 

7 ⎥⎦ 

⎡ 70 + 16 − 42 ⎤ ⎡ 44 ⎤ ⎡ 4 ⎤ 

1 

= 

⎢ 

⎥ 1 

11 ⎢ 

− 20 + 8 − 21 

⎥ 

= 

⎢ ⎥ 

11 ⎢ 

− 33 

⎥ 

= 

⎢ ⎥ 

⎢ 

− 3 

⎥ 

⎢⎣ 

− 40 + 16 + 35⎥⎦ 

⎢⎣ 

11 ⎥⎦ 

⎢⎣ 

1 ⎥⎦ 

∴ x = 4, y = –3, z = 1 

27. Let y = sin p θ cos q θ 

Let z = log y = p log sin θ + q log cos θ 

dz = p(cot θ) – q tan θ 

dθ 

dz 

For maximum = 0 

dθ 

⇒ p cot θ = q tan θ 

⇒ tan θ = 

Now 

2 

p / q 

d z 

= – p cosec 2 θ – q sec 2 θ 

2 

dθ 

= – p(1 + cot 2 θ) – q (1 + tan 2 θ) 

= – p(1 + q/p) – q(1 + p/q) (Q tan θ = p / q ) 

= – 2(p + q) 

d 

d 

b 

28. 

∫ 

a 

2 

z 

2 

θ < 0 

⇒ at θ = tan –1 

f ( x) 

dx = 

p / q 

, y attains maximum value. 

lim h[f(a) + f(a + h) + f(a + 2h) + ........ 

h→0 

b − a 

where h = 

n 

Here a = 1, b = 3, f(x) = x 2 + 5x and 

3 − 1 2 

h = = 

n n 

3 

∴ 

∫ 

( x + 5x) 

dx 

1 

2 

…….+ f(a + (n –1) h], 

= lim h [ f(1) + f(1 + h) + f(1 + 2h) + ….......... 

h→0 

….+ f(1+(n –1) h)] 

= lim h [{(1 2 + 5 × 1)} + {(1 + h) 2 + 5 (1 + h)} 

h→0 

+ {(1 + 2h) 2 + 5(1 + 2h)} + ....... 

…..+{(1 + (n –1)h) 2 + 5(1 + (n –1) h)}] 

= lim h [{(1 2 + (1 + h) 2 + (1 + 2h) 2 + ….+ 

h→0 

(1 + (n –1)h) 2 }] + 5{1 + (1 + h) + (1 + 2h) + ........ 

….+ (1 +(n –1)h)}] 

= lim h [{n + 2h (1 + 2 + 3 + …..+(n –1) + h 2 

h→0 

(1 2 + 2 2 + …..+(n –1) 2 )} + 5{n + h (1 + 2 + …... 

......+ (n –1))}] 

lim h [6n + 7h (1 + 2+ 3+ …..+(n –1) + h 2 

h→0 

(1 2 + 2 2 + ……+……+ (n –1) 2 ] 

⎡ n( 

n −1) 

2 n( 

n −1) (2n 

−1) 

⎤ 

= lim h 

h→0 

⎢6n 

+ 7h 

− + h . 

⎥ 

⎣ 

2 

6 ⎦ 

= 

= 

= 

lim 2 

n→∞ 

n 

lim 

n→∞ 

lim 

n→∞ 

⎡ 14 n( 

n −1) 

4 n( 

n −1)(2n 

−1) 

⎤ 

⎢6n 

+ . + . 

2 

⎥ 

⎣ n 2 n 6 ⎦ 

⎡ ⎛ n −1⎞ 

⎢12 

+ 14 ⎜ ⎟ + 

⎣ ⎝ n ⎠ 

8 

. 

6 

( n −1)(2n 

−1) 

⎤ 

2 ⎥ 

n ⎦ 

⎡ 

⎞ ⎛ ⎞⎛ 

⎞⎤ 

⎢ ⎜ 

⎛ 1 4 1 1 

12 + 14 1 − ⎟ + . ⎜1 

− ⎟⎜2 

− ⎟⎥ ⎣ ⎝ n ⎠ 3 ⎝ n ⎠⎝ 

n ⎠ ⎦ 

4 8 86 

= 12 + 14 + × 2 = 12 + 14 + = 3 3 3 

b 

∫ 

a 

f ( x) 

dx = 

OR 

lim h [f(a) + f(a + h) + f(a + 2h) +…..... 

h→0 

….+ f(a + (n –1) h)], 

b − a 

where h = 

n 

Here, a = –1, b = 1, f (x) = e x 1− 

( −1) 

2 

and h = = 

n n 

1 

∴ 

∫ 

e x dx 

= 

= 

= 

−1 

lim h [f(–1) + f(–1+ h) + f(–1 + 2h) + ........... 

h→0 

……..+ f (–1 + (n –1) h)] 

lim h [e –1 + e –1+h + e –1+2h +…+ e –1+(n –1)h ] 

h→0 

lim h e –1 [1 + e h + e 2h + ….+e (n –1)h ] 

h→0 

= lim h e ⎡ 

–1 ⎪⎧ 

h n 

( e ) −1⎪⎫ 

h →0 

⎥ ⎥ ⎤ 

⎢1. 

⎨ h ⎬ 

⎢⎣ 

⎪⎩ e −1 

⎪⎭ ⎦ 

⎡ 

n–1 

⎢using a + ar + ....... + ar 

⎢⎣ 

⎡ ⎤ 

2 

⎢ e −1 

= lim e –1 ⎢ 

h→0 

⎢ 

⎛ 

h 

⎞ 

⎜ 

e −1 

⎟ 

⎢ 

⎢⎝ 

h ⎠ ⎥ ⎥⎥⎥⎥ ⎣ ⎦ 

⎛ 

n 

⎞ ⎤ 

⎜ 

r −1 

= a ⎟ 

⎥ 

⎝ r −1 

⎠ ⎥⎦ 

[Q h = 2/n ⇒ nh = 2] 

⎛ 

2 

= e –1 ⎟ ⎞ 

⎜ 

e −1 

⎡ 

= e – e –1 e h −1 

⎝ 1 ⎠ 

⎥ ⎤ 

⎢Q 

lim = 1 

h→0 ⎣ h ⎦ 


29. The given information can be exhibited 

diagrammatically as follows 

Factory 

P 

8 units 

Rs.16 

x units 

Rs.10 

y units 

Rs.15 

8–(x + y) units 

Depot 

A 

5 units 

Depot 

B 

5 units 

Depot 

C 

4 units 

Rs.10 

(5–x) units 

Rs.12 

(5–y) units 

Rs.10 

Factory 

Q 

6 units 

6 – (5 – x + 5 – y) 

= x + y – 4 units 

Let the factory at P transports x units of commodity to 

depot at A and y units to depot at B. Since the factory 

at P has the capacity of 8 units of the commodity. 

Therefore, the left out (8 – x – y) units will be 

transported to depot at C. 

Since the requirements are always non-negative 

quantities. 

Therefore, 

x ≥ 0, y ≥ 0 and 8 – x – y ≥ 0 

⇒ x ≥ 0, y ≥ 0 and x + y ≤ 8. 

Since the weekly requirement of the depot at A is 5 

units of the commodity and x units are transported 

from the factory at P. Therefore the remaining (5 – x) 

units are to be transported from the factory at Q. 

Similarly, 5 – y units of the commodity will be 

transported from the factory at Q to the depot at B. 

But the factory at Q has the capacity of 6 units only, 

therefore the remaining 6 – (5 – x + 5 – y) = x + y – 4 

units will be transported to the depot at C. As the 

requirements at the depots at A, B and C are always 

non-negative. 

∴ 5 – x ≥ 0, 5 – y ≥ 0 and x + y – 4 ≥ 0 

⇒ x ≤ 5, y ≤ 5 and x + y ≥ 4 

The transportation cost from the factory at P to the 

depots at A, B and C are respectively Rs.16x, 10y and 

15(8 – x –y). Similarly, the transportation cost from 

the factory at Q to the depots at A, B and C are 

respectively Rs.10 (5 – x), 12(5 – y) and 10(x + y –4). 

Therefore, the total transportation cost Z is given by 

Z = 16x + 10y + 15(8 – x – y) + 10 (5 – x) + 12 (5 – y) 

+ 10 (x + y – 4) = x –7y + 190 

Z = 16x + 10y + 15(8 – x – y) + 10(5 – x) + 12(5 – y) 

+ 10(x + y – 4) = x –7y + 190) 

Hence, the above LPP can be stated mathematically 

as follows, find x & y which Minimize 

Z = x – 7y + 190 

x + y ≤ 8 

x + y ≥ 4 

x ≤ 5 

y ≤ 5 

and x ≥ 0, y ≥ 0 

OR 

Let the depot A transport x thousand bricks to builder 

P and y thousand bricks to builder Q. Then the above 

LPP can be stated mathematically as follows. 

Minimize Z = 30x –30y + 1800 

Subject to x + y ≤ 30 

x ≤ 15 

y ≤ 20 

x + y ≥ 15 

and, x ≥ 0, y ≥ 0 

To solve this LPP graphically, we first convert 

inequations into equations and then draw the 

corresponding lines. The feasible region of the LPP is 

shaded in fig. The coordinates of the corner points of 

the feasible region A 2 PQB 3 B 2 are A 2 (15, 0), 

P(15, 15), Q (10, 20), B 3 (0, 20) and B 2 (0, 15). These 

points have been obtained by solving the 

corresponding intersecting lines simultaneously. 

y 

B (0, 30) 

B 3 (0, 20) 

O 

• • 

B 2 (0, 15) • 

A 2 (15, 0) 

x = 15 

Q (10, 20) 

• 

P (15, 15) 

• 

A (30, 0) 

x + y = 15 

y = 20 

x 

x + y = 30 

The values of the objective function at the corner 

points of the feasible region are given in the 

following table. 

Point (x, y) Value of the objective function 

Z = 30x –30y + 1800 

A 2 (15, 0) Z = 30 × 15 –30 × 0 + 1800 = 2250 

P(15,15) Z = 30 × 15 –30 × 15 + 1800 = 1800 

Q(10,20) Z = 30 × 10 –30 × 20 + 1800 = 1500 

B 3 (0, 20) Z = 30 × 0 –30 × 20 + 1800 = 1200 

B 2 (0, 15) Z = 30 × 0 –30 × 15 + 1800 = 1350 

Clearly, Z is minimum at x = 0, y = 20 and the 

minimum value of Z is 1200. Thus, the manufacturer 

should supply 0, 20 and 10 thousand bricks to 

builders P, Q and R from depot A and 15, 0, and 5 

thousand bricks to builders P, Q and R from depot B 

respectively. In this case the minimum transportation 

cost will be Rs.1200. 


XtraEdge Test Series 

ANSWER KEY 

IIT- JEE 2011 (February issue) 

PHYSICS 

Ques 1 2 3 4 5 6 7 8 9 10 

Ans D A C A A C B C B B,C 

Ques 11 12 13 14 15 16 17 18 19 20 

Ans A,B,C A,C,D A,B,C,D A,B B A B B D A 

Column 21 A → P,Q,R,S B → P,Q,R,S C → P,Q,R D → P,Q,R,S 

Matching 22 A → P B → R C → Q D → S 

CHEMISTRY 

Ques 1 2 3 4 5 6 7 8 9 10 

Ans C C A D A D D C A A,C 

Ques 11 12 13 14 15 16 17 18 19 20 

Ans B,C,D A,D B,C,D B,C,D C B D B B C 

Column 21 A → Q B → R C → S D → P 

Matching 22 A → Q,S B → R,S C → P,Q D → R,S 

MATHEMATICS 

Ques 1 2 3 4 5 6 7 8 9 10 

Ans C D D C B C A C C B,C,D 

Ques 11 12 13 14 15 16 17 18 19 20 

Ans A,D C,D A,B,D A,B,C B C D B A C 

Column 21 A → R B → P C → Q D → S 

Matching 22 A → S B → Q C → P D → R 

IIT- JEE 2012 (February issue) 

PHYSICS 

Ques 1 2 3 4 5 6 7 8 9 10 

Ans C C C D C C B C B A,C 

Ques 11 12 13 14 15 16 17 18 19 20 

Ans A,B,D A,B B, D A,D A B C C A D 

Column 21 A →R B → P C → S D → Q 

Matching 22 A → R B → S C → Q D → P 

CHEMISTRY 

Ques 1 2 3 4 5 6 7 8 9 10 

Ans D D B A C A C C B D 

Ques 11 12 13 14 15 16 17 18 19 20 

Ans A,B, D A,C A,C B C A A B B C 

Column 21 A → P,R,S B → Q C → P,R,S D → S 

Matching 22 A → S B → R C → P D → Q 

MATHEMATICS 

Ques 1 2 3 4 5 6 7 8 9 10 

Ans C B B D A B B B A A,B,D 

Ques 11 12 13 14 15 16 17 18 19 20 

Ans A,B,C B,D D A,B,C,D A C B B C C 

Column 21 A → S B → R C → Q D → P 

Matching 22 A → P B → R C → Q D → S 


Subscription Offer for Students 

'XtraEdge for IIT-JEE 

IIT JEE becoming more competitive examination day by day. 

Regular change in pattern making it more challenging. 

C 

"XtraEdge for IIT JEE" magazine makes sure you're updated & at the forefront. 

Every month get the XtraEdge Advantage at your door step. 

✓ Magazine content is prepared by highly experienced faculty members on the latest trend of IIT JEE. 

✓ Predict future paper trends with XtraEdge Test Series every month to give students practice, practice & more practice. 

✓ Take advantage of experts' articles on concepts development and problem solving skills 

✓ Stay informed about latest exam dates, syllabus, new study techniques, time management skills and much more XtraFunda. 

✓ Confidence building exercises with Self Tests and success stories of IITians 

✓ Elevate you to the international arena with international Olympiad/Contests problems and Challenging Questions. 

SUBSCRIPTION FORM FOR “EXTRAEDGE FOR IIT-JEE 

The Manager-Subscription, 

“XtraEdge for IIT-JEE” 

Career Point Infosystems Ltd, 

4 th Floor, CP-Tower, 

IPIA, Kota (Raj)-324005 

I wish to subscribe for the monthly Magazine “XtraEdge for IIT-JEE” 

Half Yearly Subscription (Rs. 100/-) One Year subscription (Rs. 200/-) Two year Subscription (Rs. 400/-) 

I am paying R. …………………….through 

Money Order (M.O) 

Bank Demand Draft of No………………………..Bank………………………………………………………..Dated 

(Note: Demand Draft should be in favour of "Career Point Infosystems Ltd" payable at Kota.) 

Name: 

Special 

Offer 

_______________________________________________________________________________________ 

Father's Name: _______________________________________________________________________________________ 

Address: _______________________________________________________________________________________ 

________________________City_____________________________State__________________________ 

PIN_________________________________________Ph with STD Code __________________________ 

Class Studying in ________________E-Mail: ________________________________________________ 

From months: ____________________to ________________________________________________ 

C 


Subscription Offer for Schools 

XtraEdge for IIT-JEE 

IIT JEE becoming more competitive examination day by day. 

Regular change in pattern making it more challenging. 

C 

"XtraEdge for IIT JEE" magazine makes sure you're updated & at the forefront. 

Every month get the XtraEdge Advantage at your door step. 

✓ Magazine content is prepared by highly experienced faculty members on the latest trend of the IIT JEE. 

✓ Predict future paper trends with XtraEdge Test Series every month to give students practice, practice & more practice. 

✓ Take advantage of experts' articles on concepts development and problem solving skills 

✓ Stay informed about latest exam dates, syllabus, new study techniques, time management skills and much more XtraFunda. 

✓ Confidence building exercises with Self Tests and success stories of IITians 

✓ Elevate you to the international arena with international Olympiad/ Contests problems and Challenging Questions. 

FREE SUBSCRIPTION FORM FOR “EXTRAEDGE FOR IIT-JEE 

The Manager-Subscription, 

“XtraEdge for IIT-JEE” 

Career Point Infosystems Ltd, 

4 th Floor, CP-Tower, 

IPIA, Kota (Raj)-324005 

We wish to subscribe for the monthly Magazine “XtraEdge for IIT-JEE” 

Half Yearly Subscription One Year subscription Two year Subscription 

Institution Detail: 

Graduate Collage Senior Secondary School Higher Secondary School 

Name of the Institute: 

_____________________________________________________________________________ 

Name of the Principal: _____________________________________________________________________________ 

Mailing Address: _____________________________________________________________________________ 

__________________City_________________________State__________________________ 

PIN_____________________Ph with STD Code_____________________________________ 

Fax_______________________________ E-Mail_____________________________________ 

Board/ University: _____________________________________________________________________________________ 

✂ 

C 

School Seal with Signature

February 2011 - Career Point

Create successful ePaper yourself

Delete template?

Save as template?