NARAYANA IIT/PMT ACADEMY - Narayanaicc.com

NARAYANA IIT/PMT ACADEMY 

JEE ADVANCE : 2013 

(PAPER – II) CODE: 0 

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PAPER -1 : PHYSIC, CHEMISTY & MATHEMATICS 

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1




PART I : PHYSICS 

SECTION -1 : (One or more options correct Type) 

The section contains 8 multiple choice questions Each question has four choices (A), (B), (C) and (D) 

out of which ONE and MORE are correct 

1 The figure below shows the variation of specific heat capacity (C) of a solid as a function of 

temperature (T) The temperature is increased continuously from 0 to 500 K at a constant rate 

Ignoring any volume change, the following statement(s) is (are) correct to a reasonable 

approximation 

(A) the rate at which heat is absorbed in the range 0-100 K varies linearly with temperature T 

(B) heat absorbed in increasing the temperature from 0-100 K is less than the heat required for 

increasing the temperature from 400-500 K 

(C) there is no change in the rate of heat absorption in the range 400-500 K 

(D) the rate of heat absorption increases in the range 200-300 K 

C 

100 200 300 400 500 

T(K) 

Ans A, B, C, D 

Sol: dQ CdT and C varies linearly with T 

(A) is correct 

Area under the graph in the range O-100 k is less than area under 400-500 K range 

(B) is correct 

In the range 400 – 500 K, C is constant 

(C) is correct 

In the range 200 – 300 K, C is increasing 

(D) is correct 

2 The radius of the orbit of an electron in a hydrogen-like atom is 45 a 0 , where a 0 is the Bohr 

radius Its orbital angular momentum is 3h It is given that h is Planck constant and R is Rydberg 

2 

constant The possible wavelength(s), when the atom de-excites, is (are) 

9 

9 

(A) 

(B) 

32R 

16R 

2 

NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS - INDIA

Ans 

Sol: 

9 

(C) 

5R 

A, C 

2 

an 

o 

Z 

4.5 a o 

nh 3h 

2 2 

n 3, z 2 

Now, 



1 2 1 1 

Rz 

x x 

2 2 

1 2 

Possible wavelengths are 

9 9 1 

, and 


4 

(D) 

3R 

5R 32R 3R 

3 Using the expression 2d sin = , one calculates the values of d by measuring the corresponding 

angles in the range 0 to 90 o The wavelength is exactly known and the error in is constant 

for all values of As increases from 0 o 

Ans 

Sol: 

(A) the absolute error in d remains constant (B) the absolute error in d increases 

(C) the fractional error in d remains constant (D) the fractional error in d decreases 

D 

d 

2sin 

d cosec 

2 

cot 

Now 

d 

d 

cot 

As increases, fractional error decreases 

4 Two non-conducting spheres of radii R 1 and R 2 and carrying uniform volume charge densities + 

and – , respectively, are placed such that they partially overlap, as shown in the figure At all 

points in the overlapping region, 

(A) the electrostatic field is zero 

(B) the electrostatic potential is constant 

(C) the electrostatic field is constant in magnitude (D) the electrostatic field has same direction 

R 1 R 2 

- 


3

Ans 

Sol: 

C, D 

In the overlapping region 

r r 

1 2 




E 3 

o 3 r o 

r 

1r 2 

o 

r 

l 

l 

E 3 

C and D are correct 

5 A steady current I flows along an infinitely long hollow cylindrical conductor of radius R This 

cylinder is placed coaxialy inside an infinite solenoid of radius 2 R The solenoid has n turns per 

unit length and carries a steady current I Consider a point P at a distance r from the common axis 

The correct statement (s) is (are) 

(A) In the region 0 < r < R, the magnetic field is non-zero 

(B) In the region R < r < 2R, the magnetic field is along the common axis 

(C) In the region R < r < 2R, the magnetic field is tangential to the circle of radius r, centered on 

the axis 

(D) In the region r > 2R, the magnetic field is non-zero 

Ans A, C, D 

Sol: B Bhollow B 

solenoid 

conductor 

B due to hollow conductor is zero inside the conductor and non-zero outside (co-axial) while B 

due to solenoid is zero outside the solenoid and non-zero inside the solenoid (axial) 

R 

2R 

A, C & D 

6 Two vehicles, each moving with speed u on the same horizontal straight road, are approaching 

each other Wind blows along the road with velocity w One of these vehicles blows a whistle of 

frequency f 1 An observer in the other vehicle hears the frequency of the whistle to be f 2 The 

speed of sound in still air is V The correct statement (s) is (are) 

(A) If the wind blows from the observer to the source, f 2 > f 1 

(B) If the wind blows from the source to the observer, f 2 > f 1 

(C) If the wind blows from observer to the source, f 2 < f 1 

(D) If the wind blows from the source to the observer, f 2




7 Two bodies, each of mass M, are kept fixed with a separation 2L A particle of mass m is 

projected from the midpoint of the line joining their centres, perpendicular to the line The 

gravitational constant is G The correct statement(s) is (are) 

(A) The minimum initial velocity of the mass m to escape the gravitational field of the two 

Ans 

Sol: 

GM 

bodies is 4 

L 

(B) The minimum initial velocity of the mass m to escape the gravitational field of the two 

GM 

bodies is 2 

L 

(C) The minimum initial velocity of the mass m to escape the gravitational field of the two 

2GM 

bodies is 

L 

(D) The energy of the mass m remains constant 

B,D 

1 2 2GMm 

mve 

0 

2 L 

GM 

ve 

2 

L 

M 

v e 

M 

8 A particle of mass m is attached to one end of a mass-less spring of force constant k, lying on a 

frictionless horizontal plane The other end of the spring is fixed The particle starts moving 

horizontally from its equilibrium position at time t = 0 with an initial velocity u 0 When the speed 

of the particle is 05 u 0 , it collides elastically with a rigid wall After this collision, 

(A) the speed of the particle when it returns to its equilibrium position is u 0 

(B) the time at which the particle passes through the equilibrium position for the first time is 

Ans 

Sol: 

m 

t . 

k 

4 m 

(C) the time at which the maximum compression of the spring occurs is t . 

3 k 

(D) the time at which the particle passes through the equilibrium position for the second time is 

m 

t . 

3 k 

A, D 

2 2 

3u 

using v A x , we get x 

2 

(x : distance between mean position & the wall) 

Now using x = A sin t 


5

t 

2 m 

3 k 



time for option (B) = 2 m 2 m 4 m 

3 k 3 k 3 k 

time for option (C) 

time for option (D) = 

4 m m 7 m 

3 k 2 k 6 k 

7 m m 5 m 

6 k 2 k 3 k 


SECTION – 2 : (Paragraph Type) 

This section contains 4 paragraphs each describing theory, experiment, data etc Eight questions relate to 

four paragraphs with two questions on each paragraph Each question of a paragraph has only one correct 

answer among the four choices (A), (B), (C) and (D) 

A small block of mass 1 kg is released from rest at the top of a rough track the track is a circular arc of 

radius 40 m the block slides along the track without toppling and a frictional force acts on it in the 

direction opposite to the instantaneous velocity The work done in overcoming the friction up to the point 

Q, as shown in the figure below, is 150 J (Take the acceleration due to gravity, g = 10 ms –2 ) 

y 

30 o 

R 

P 

Q 

R 

O 

x 

9 The speed of the block when it reaches the point Q is 

(A) 5 ms –1 (B) 10 ms –1 

1 

(C) 10 3ms 

1 

(D) 20ms 

Ans B 

Sol: From work-kinetic energy theorem 

o 1 2 

mgR sin30 Wf 

mv 

2 

v = 10 m/s 

10 The magnitude of the normal reaction that acts on the block at the point Q is 

(A) 75 N 

(B) 86 N 

(C) 115 N 

(D) 225 N 

Ans A 

Sol: At Q: 


6




2 

o mv 

N mgcos60 

R 

N 7.5N 

Paragraph for questions 11 and 12 

A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a 

place 20 km away from the power plant for consumers’ usage It can be transported either directly with a 

cable of large current carrying capacity or by using a combination of step-up and step-down 

transformers at the two ends The drawback of the direct transmission is the large energy dissipation In 

the method using transformers, the dissipation is much smaller In this method, a step-up transformer is 

used at the plant side so that the current is reduced to a smaller value At the consumers’ end, a stepdown 

transformer is used to supply power to the consumers at the specified lower voltage It is 

reasonable to assume that the power cable is purely resistive and the transformers are ideal with a power 

factor unity All the currents and voltages mentioned are rms values 

11 If the direct transmission method with a cable of resistance 04 km -1 is used, the power 

dissipation (in%) during transmission is 

(A) 20 (B) 30 (C) 40 (D) 50 

Ans (B) 

Sol P 

3 

600 10 watt 

Vrms 

4000 volt 

Total resistance R = 04 20 = 80 

Power lost P 

2 

I R 

L 

P 

V 

rms 

rms 

2 

R 

= 180000 watt 

= 180 kW 

So % loss = 180 600 

100 30 

= 30% 

3 

600 10 

4000 

12 In the method using the transformers, assume that the ratio of the number of turns in the primary 

to that in the secondary in the step-up transformer is 1 : 10 If the power to the consumers has to 

be supplied at 200 V, the ratio of the number of turns in the primary to that in the secondary in 

the step-down transformer is 

(A) 200 : 1 (B) 150 : 1 (C) 100 : 1 (D) 50 : 1 

Ans (A) 

Sol Step – up 

2 

8 


7




For Step – down 

N P 

4000V 

VS 

10 

4000 1 

P 

input 

V 

S 

40,000volt 

N V 

S out 

200 1 

N V 40,000 200 

So N P : N S = 200 : 1 

N S 

V =? S 

Paragraph for Questions 13 and 14 

A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity 

This can be considered as equivalent to a loop carrying a steady current 2 

A uniform magnetic field 

along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one 

second Assume that the radius of the orbit remains constant The application of the magnetic field 

induces an emf in the orbit The induced emf is defined as the work done by an induced electric field in 

moving a unit positive charge around a closed loop It is known that, for an orbiting charge, the magnetic 

dipole moment is proportional to the angular momentum with a proportionality constant 

13 The magnitude of the induced electric field in the orbit at any instant of time during the time 

interval of the magnetic field change is 

(A) BR 4 

(B) BR 2 

(C) BR 

(D) 2BR 

Ans(B) 

Sol Induced emf = d dt 

dB 2 

e R 

dt 

2 

e B R 

E.2 R B. R 

BR 

E 

2 

2 

14 The change in the magnetic dipole moment associated with the orbit, at the end of the time 

interval of the magnetic field change, is 


8

(A) 

2 

BQR 

Ans (C or B) 

Sol e 

B 2 

R 

t 

E.2 R 

B 

t 

R 

E 

2 

B R 

2 R. t 

L t 

m L 

m 

2 

QBR 



2 

(B) 

2 

BQR 

2 


(C) 

2 

BQR 

2 

2 

Sign may be taken as +ve/-ve also depending upon direction of angular velocity 

Paragraph for questions 15 and 16 

(D) 

2 

BQR 

The mass of a nucleus A X 

Z 

is less than the sum of the masses of (A-Z) number of neutrons and Z number 

of protons in the nucleus The energy equivalent to the corresponding mass difference is known as the 

binding energy of the nucleus A heavy nucleus of mass M can break into two light nuclei of masses m 1 

and m 2 only if (m 1 + m 2 ) < M Also two light nuclei of masses m 3 and m 4 can undergo complete fusion 

and form a heavy nucleus of mass M' only if (m 3 + m 4 ) > M' The masses of some neutral atoms are 

given in the table below : 

1 

H 2 

1007825 u H 3 

2014102 u H 4 

3016050 u He 4002603 u 

1 1 1 2 

6 

Li 7 

6015123 u Li 70 

7016004 u Zn 82 

69925325 u Se 81916709 u 

3 3 30 34 

152 

Gd 206 

151919803 u Pb 209 

205974455 u Bi 210 

208980388 u Po 209982876 u 

64 82 83 84 

15 The correct statement is 

(A) The nucleus 6 Li 

3 

can emit an alpha particle 

(B) The nucleus 210 

Po 

84 

can emit a proton 

(C) Deuteron and alpha particle can undergo complete fusion 

(D) The nuclei 70 

82 

Zn 

30 

and Se 

34 

can undergo complete fusion 

Ans (C) 

Sol from data given only option C is correct 

m m M ' 

3 4 


9




16 The kinetic energy (in keV) of the alpha particle, when the nucleus 210 

Po 

84 

at rest undergoes alpha 

decay, is 

(A) 5319 (B) 5422 (C) 5707 (D) 5818 

Ans (A) 

Sol From data given 

Po Pb He 

210 206 4 

84 82 2 

m=00058184 u 

E = 0005818 932 

=5422376 MeV 

KE of 

particle 

E 206 

210 

= 5319 MeV = 5319 KeV 

SECTION – 3 : (Matching List Type) 

This section contains 4 multiple choice questions Each question has matching lists 

The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct 

17 One mole of a monatomic ideal gas is taken along two cyclic processes E F G E and 

E F H E as shown in the PV diagram The processes involved are purely isochoric, 

isobaric, isothermal or adiabatic 

Match the paths in List I with the magnitudes of the work done in List II and select the correct 

answer using the codes given below the lists 

List I 

List II 

P G E 1 160 P 0 V 0 ln2 

Q G H 2 36 P 0 V 0 

R F H 3 24 P 0 V 0 


10



S F G 4 31 P 0 V 0 

Codes : 

P Q R S 

(A) 4 3 2 1 

(B) 4 3 1 2 

(C) 3 1 2 4 

(D) 1 3 2 4 

Ans (A) 

Sol = 5/3 

FH is steeper therefore it is adiabatic process FG is isothermal 

P V P V 

FH 

F F H H 

(32P )V P V V 8V 

0 0 H H 

H 0 

FG 


P F V F = P G V G 

(32P 0 )V 0 = P 0 V G V G = 32 V 0 

GE 

Isobaric Process 

Work = P V = P 0 (32V 0 – V 0 ) = 31 P 0 V 0 

GH Work = P 0 (32V 0 – 8V 0 ) = 24 P 0 V 0 

PV P V (32P )V P 8V 

1 1 2 2 

0 0 0 0 

FH Work 36P V 

0 0 

1 5/ 3 1 

VG 

FG 

Work P V ln 32P V ln 32 160ln 2 

F F 0 0 

V 

F 

18 Match List I of the nuclear processes with List II containing parent nucleus and one of the end 

products of each process and then select the correct answer using the codes given below the lists 

List I 

List II 

15 15 

P Alpha decay 1 O N 

8 7 

.... 

238 234 

Q decay 2 U Th 

98 90 

.... 

185 184 

R Fission 3 Bi Pb 

83 82 

.... 

239 140 

S Proton emission 4 Pu La 

94 57 

.... 

Codes : 

P Q R S 

(A) 4 2 1 3 

(B) 1 3 2 4 

(C) 2 1 4 3 

(D) 4 3 2 1 


11

Ans 

Sol 

(C) 

1 

2 

3 

4 



15 15 0 

O N X x is particle 

8 7 1 

U Th x is particle 

238 234 4 

92 90 2 

Bi Pb X Xis Proton 

185 184 1 

83 82 1 

Pu La X Fission 

239 140 99 

94 57 47 


19 A right angled prism of refractive index 

1 

is placed in a rectangular block of refractive index 

2 

, which is surrounded by a medium of refractive index 

3 

, as shown in the figure A ray of 

light ‘e’ enters the rectangular block at normal incidence Depending upon the relationships 

between and , it takes one of the four possible paths ‘ef’, ‘eg’, ‘eh’ or ‘ei’ 

1, 2 3 

Match the paths in List I with conditions of refractive indices in List II and select the correct 

answer using the codes given below the lists : 

List I 

List II 

P e f 1 

Codes : 

P Q R S 

(A) 2 3 1 4 

(B) 1 2 4 3 

(C) 4 1 2 3 

(D) 2 3 4 1 

2 

1 2 

Q e g 2 and 

2 1 2 3 

R e h 3 

1 2 

S e i 4 

2 and 

2 1 2 2 3 

Ans (D) 

Sol 

S - 1 

e – i case of TIR 

which is possible only when 

o 

i 45 is i 

c 


12




sini if 2 , i is 45 

2 

C 1 2 C 

1 

Q – 3 e g is [ 

1 2]rays undergo undeviated for any angle of incidence 

R – 4 e h is 

1 2 

and not undergoing (TIR) light ray bends towards base of 

prism, (prism block) 

P – 2 e f 

1 2 

and 

2 3(Snell’s law) 

o 

20 Match List I with List II and select the correct answer using the codes given below the lists : 

Codes : 

P Q R S 

(A) 3 1 2 4 

(B) 3 2 1 4 

(C) 4 2 1 3 

(D) 4 1 2 3 

List I 

List II 

P Boltzmann constant 1 [ML 2 T -1 ] 

Q Coefficient of viscosity 2 [ML -1 T -1 ] 

R Planck constant 3 [MLT -3 K -1 ] 

S Thermal conductivity 4 ML 2 T -2 K -1 ] 

Ans (C) 

Sol 

2 2 

energy ML T 

Boltzmann constant = 

Kelvin K 

= ML 2 T -2 K -1 

Plank’s constant = energy time = ML 2 T -2 T = ML 2 T -1 

2 

F MLT 

1 1 

Co-efficient of viscosity = 

ML T 

1 2 1 

L LT L T 

Thermal conductivity = 

Energy Length 

Area Kelvin Time MLT-3 K -1 

P 4 

Q 2 

R 1 

S 3 


13




PART II : CHEMISTRY 

SECTION – 1 : (One or more options correct Type) 

This section contains 8 multiple choice questions Each question has four choices (A), (B), (C) and (D) 

out of which ONE or MORE are correct 

21 The carbon-based reduction method is NOT used for the extraction of 

(A) tin from SnO 2 (B) iron from Fe 2 O 3 

(C) aluminium from Al 2 O 3 (D) magnesium from MgCO 3 CaCO 3 

Ans 

Sol 

(C), (D) 

Al is extracted by electrolytic reduction 

MgCO 3 , CaCO 3 are extracted by electrolytic reduction 

22 The thermal dissociation equilibrium of CaCO 3 (s) is studied under different conditions 

CaCO 3 (s) CaO (s) + CO 2 (g) 

For this equilibrium, the correct statement(s) is(are) 

(A) H is dependent of T 

(B) K is independent of the initial amount of CaCO 3 

(C) K is dependent on the pressure of CO 2 at a given T 

(D) H is independent of the catalyst, if any 

Ans (A), (B), (D) 

Sol Conceptual 

23 The correct statement(s) about O 3 is(are) 

(A) O O bond lengths are equal 

(B) Thermal decomposition of O 3 is endothermic 

(C) O 3 is diamagnetic in nature 

(D) O 3 has a bent structure 

Ans (A), (B), (C), (D) 

Sol (A) – bond length are equal due to resonance 

O 

O + O - 

No unpaired electrons so diamagnetic 

24 In the nuclear transmutation 

Be X Be Y 

9 8 

4 4 

(X, Y) is(are) 

(A) ,n (B) p,D (C) n, D (D) ,p 

Ans (A), (B) 

Sol (A) 9 8 1 

4Be 4Be 0 

n 


14

(B) 9 1 8 2 

4Be 1 

p 

4Be 1D 



25 The major product(s) of the following reaction is(are) 

OH 


aqueous Br2 

3.0 equivalents 

? 

Br 

SO 3 

H 

OH 

Br 

Br 

OH 

Br 

OH 

OH 

Br 

Ans 

Sol. 

Br 

Br Br Br Br 

SO 3 

H 

Br 

Br 

SO 3 

H 

P 

Q 

R 

S 

(A) P (B) Q (C) R (D) S 

(B) 

OH 

OH 

Br Br 

aqueous Br2 

3.0 equivalents 

SO 3 

H 

Br 

26 After completion of the reactions (I and II), the organic compound(s) in the reaction mixtures 

is(are) 

O 

Reaction I: 

H 3 C CH 3 

1.0 mol 

O 

Reaction II: H 3 C CH 3 

O 

1.0 mol 

O 

Br2 

1.0mol 

aqueous NaOH 

Br2 

1.0mol 

CH3COOH 

O 

O 

O 

O 

C H 3 

CH 2 

Br 

H 3 C CBr 3 

Br 3 

C CBr 3 

CH 2 

Br 

CH 2 

Br 

C H 3 

ONa 

C H 3 

ONa 

Ans 

P Q R S T U 

(A) Reaction I: P and Reaction II : P 

(B) Reaction I : U, acetone and Reaction II: Q, acetone 

(C) Reaction I: T, U, acetone and Reaction II: P 

(D) Reaction I: R, acetone and Reaction II: S, acetone 

(C) 


15

Sol 



O 


Reaction I: 1/3 mole of CH3 C CH react with 1 mole of Br 2 to give Bromofarm and sodium 

3 

acetate and 2/3 mole of acetone itself 

Reaction II: Mono bromination at -carbon in acidic medium 

27 The K sp of Ag 2 CrO 4 is 11 10 -12 at 298 K The solubility (in mol/L) of Ag 2 CrO 4 in a 01 M 

AgNO 3 solution is 

(A) 11 10 -11 (B) 11 10 -10 (C) 11 10 -12 (D) 11 10 -9 

Ans (B) 

Sol 

2 

Ag 

2CrO4 2Ag CrO4 

s' 0.1 2s' s' 

K sp = (01 + 2s ) 2 s 

11 10 -12 = 10 -2 s 

s = 11 10 -10 

28 In the following reaction, the product(s) formed is(are) 

OH 

CHCl 3 

OH 

? 

Ans 

Sol 

CHO 

CH3 

CH 3 

OH 

O 

OH 

OH 

CHO CHO 

CH 3 

H 3 C CHCl 2 

H 3 C CHCl 2 

P Q R S 

(A) P (major) (B) Q (minor) (C) R (minor) (D) S (major) 

(B), (D) 

It is Reimer Tiemann Reaction 

O 

OH 

CHO 

So 

C H 3 

is minor and 

CHCl 2 

CH 3 

is major 

Q 

S 


16




SECTION – 2 : (Paragraph Type) 

This section contains 4 paragraphs each describing theory, experiment, data etc Eight questions relate 

to four paragraphs with two questions on each paragraph Each question of a paragraph has only one 

correct answer among the four choices (A), (B), (C) and (D) 

Paragraph for Questions 29 and 30 

An aqueous solution of a mixture of two inorganic salts, when treated with dilute HCl, gave a precipitate 

(P) and a filtrate (Q) The precipitate P was found to dissolve in hot water The filtrate (Q) remained 

unchanged, when treated with H 2 S in a dilute mineral acid medium However, it gave a precipitate (R) 

with H 2 S in an ammoniacal medium The precipitate R gave a coloured solution (S), when treated with 

H 2 O 2 in an aqueous NaOH medium 

29 The precipitate P contains 

(A) Pb 2+ 

(B) 

Ans 

Sol 

(A) 

Salt will be of PbCl 2 as it is soluble in hot water 

2 

Hg 

2 

(C) Ag + (D) Hg 2+ 

30 The coloured solution S contains 

(A) Fe 2 (SO 4 ) 3 (B) CuSO 4 (C) ZnSO 4 (D) Na 2 CrO 4 

Ans (D) 

Sol Cr 3+ + H 2 S/NH 4 OH, NH 4 Cl Cr(OH) 3 

Cr(OH) 3 + H 2 O 2 + NaOH Na 2 CrO 4 

(yellow colour) 

Paragraph for Question 31 and 32 

P and Q are isomers of dicarboxylic acid C 4 H 4 O 4 Both decolorize Br 2 / H 2 O On heating, P forms the 

cyclic anhydride 

Upon treatment with dilute alkaline KMnO 4 , P as well as Q could produce one or more than one from S, 

T and U 

31 Compounds formed from P and Q are, respectively 

(A) Optically active S and optically active pair (T, U) 

(B) Optically inactive S and optically inactive pair (T, U) 

(C) Optically active pair (T, U) and optically active S 

(D) Optically inactive pair (T, U) and optically inactive S 

31 (B) 


17

Sol 

‘P’ and ‘Q’ are 

H H 

HOOC 

C 

C 

(Maleic -acid) 



COOH 

, 

H 

HOOC 

C 


C 

COOH 

H 

(Fumaric -acid) 

Baeyer’s reagent, ie dilute alkaline KMnO 4 will give syn addition product of alkene 

COOH 

H H 

KMnO 4 

/OH H OH 

C C 

dilute 

HOOC 

H OH 

COOH 

(cis) 

COOH 

(meso) 

H 

HOOC 

C 

(trans) 

C 

COOH 

H 

KMnO 4 

/OH 

dilute 

H 

HO 

COOH 

OH 

H 

COOH 

HO 

H 

COOH 

H 

OH 

COOH 

(Racemic mixture) 

32 In the following reaction sequences V and W are, respectively 

H Ni 

Q 

2 / 

V 

+ V 

O 

AlCl 3 

(anhydrous) 

1. Zn-Hg/HCl 

2. H 3 

PO 4 

W 

(A) 

O 

and 

V 

O 

W 

O 

CH 2 OH 

(B) 

CH 2 OH 

and 

V 

W 


18

(C) 

(D) 

O 

O 

O 

V 

HOH 2 C 



and 

W 

and 


Ans 

Sol. 

(A) 

HOOC 

V 

CH 2 OH 

COOH 

Fumaric acid 

(Q) 

O 

H 2 

/Ni 

W 

CH 2 OH 

O 

O 

(V) 

O 

O 

+ O 

O 

AlCl 3 

anhydrous 

HO 

C 

O 

CH 2 

CH 2 

1. Zn-Hg/HCl 

2. H 3 

PO 4 

O 


A fixed mass ‘m’ of a gas is subjected to transformation of states from K to L to M to N and back to K 

as shown in the figure 

K 

L 

Pressure 

N 

M 

Volume 


19




33 The succeeding operations that enable this transformation of states are 

(A) Heating, cooling, heating, cooling (B) Cooling, heating, cooling, heating 

(C) Heating, cooling, cooling, heating (D) Cooling, heating, heating, cooling 

Ans 

Sol 

(C) 

K 

L 

Pressure 

N 

M 

Volume 

34 The pair of isochoric processes among the transformation of states is 

(A) K to L and L to M 

(B) L to M and N to K 

(C) L to M and M to N 

(D) M to N and N to K 

Ans (B) 

Sol According to following graph the proven L M isochoric and N K also isochoric 


The reactions of Cl 2 gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two 

(different) oxoacids of chlorine, P and Q, respectively The Cl 2 gas reacts with SO 2 gas, in presence of 

charcoal, to give a product R R reacts with white phosphorous to give a compound S On hydrolysis, S 

gives an oxoacid of phosphorus, T 

35 P and Q, respectively, are the sodium salts of 

(A) hypochlorus and chloric acids (B) hypochlorous and chlorous acids 

(C) chloric and perchloric acids 

(D) chloric and hypochlorous acids 

Ans (A) 

Sol Cl 2 + cold NaOH(aq) NaCl + NaOCl, (oxiacid HOCl) hypochlorous acid 

Cl 2 + (hot) NaOH NaCl + NaClO 3 , (oxiacid HClO 3 ) chloric acid 

P – HOCl 

Q – HClO 3 

P4 

HO 2 

Cl 2 + SO 2 + Coke SO 2 Cl 2 PCl 5 H 3 PO 4 


20




36 R, S and T, respectively, are 

(A) SO 2 Cl 2 , PCl 5 and H 3 PO 4 (B) SO 2 Cl 2 , PCl 3 and H 3 PO 3 

(C) SOCl 2 , PCl 3 and H 3 PO 2 (D) SOCl 2 , PCl 5 and H 3 PO 4 

Ans 

(A) 

SECTION – 3 (Matching List Type) 

This section contains 4 multiple choice questions Each question has matching lists The codes for the 

lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct 

37 An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List I The 

variation in conductivity of these reaction is given in List II Match List I with List II and select 

the correct answer using the code given below the lists: 

List I 

P (C 2 H 5 ) 3 N + CH 3 COOH 

X 

Y 

Q KI(01M) + AgNO 3 (001M) 

X 

Y 

R CH 3 COOH + KOH 

X Y 

S NaOH + HI 

X Y 

Code : 

P Q R S 

(A) 3 4 2 1 

(B) 4 3 2 1 

(C) 2 3 4 1 

(D) 1 4 3 2 

List II 

1 Conductivity decreases and then increases 

2 Conductivity decreases and then doest not 

change much 

3 Conductivity increases and then does not 

change much 

4 Conductivity does not change much and then 

increases 

Ans 

(A) Factual 

38 The standard reduction potential data at 25 0 C is given below 

E 0 (Fe 3+ , Fe 2+ ) = +077 V 

E 0 (Fe 2+ , Fe) = 044 V 

E 0 (Cu 2+ , Cu) = +034 V; 

E 0 (Cu 2+ , Cu) = +052 V 

E 0 [O 2 (g) + 4H + + 4e 2H 2 O] = +123 V 

E 0 [O 2 (g) + 2H 2 O + 4e 4OH ] = +040V 

E 0 (Cr 3+ , Cr) = 074 V; 

E 0 (Cr 2+ , Cr) = 091 V; 

Match E 0 of the redox pair in List I with the values given in List II and select the correct answer 


21

Ans 

Sol 



using the code given below the lists: 

List I 

P E 0 ( Fe 3+ , Fe) 1 018 V 

Q E 0 (4H 2 O 4H + + 4OH ) 2 04 V 

R E 0 (Cu 2+ + Cu 2Cu + ) 3 004 V 

S E 0 (Cr 3+ , Cr 2+ ) 4 083 V 

Code : 

P Q R S 

(A) 4 1 2 3 

(B) 2 3 4 1 

(C) 1 2 3 4 

(D) 3 4 1 2 

(D) 

(P) E 0 (Fe 3+ , Fe) 

Fe 3+ + e Fe 2+ E 0 1 

0.77V 

Fe 2+ 0 

+ 2e Fe E2 

0.44V 


List II 

Fe 3+ + 3e Fe 

G3 G1 G 

2 

3E 1 0.77 (2 0.44) 

0 

3 

0 

E 

3 

0 0.11 

E3 

3 

0.0366 0.04 V 

(Q) E 0 (4H 2 O 2H + + 4OH ) 

2H 2 O O 2 + 4H + + 4e E 0 = 123 V 

O 2 + 2H 2 O + 4e 4OH E 0 = +04 V 

4H 2 O 4H + + 4OH E 0 = 083 V 

(R) E 0 (Cu 2+ + Cu 2Cu + ) 

Cu 2+ + 2e Cu E 0 = +034 V 

2(Cu Cu + + e) E 0 = 052 V 

Cu 2+ + Cu 2Cu + E 0 = 034 – 052 = 018 V 

(S) E 0 (Cr 3+ , Cr 2+ ) 

Cr 3+ + 3e Cr 

E = 074 V 

Cr 2+ + 2e Cr 

0 

1 

E = 091V 

0 

2 

Cr 3+ + e Cr 2+ E 

0 3 

G3 G1 G 

2 


22

1 

0 

3 

0 

3 



0 

E 

3 

= 3 (074) ( 2 091) 

E = 222 – 182 

E = 04V 


39 The unbalanced chemical reactions given in List I show missing reagent or condition (?) which 

are provided in List II Match List I with List II and select the correct answer using the code 

given below the lists 

List I 

List II 

P PbO 2 + H 2 SO 4 

Q Na 2 S 2 O 3 + H 2 O 

R N 2 H 4 

S XeF 2 

Code : 

P Q R S 

(A) 4 2 3 1 

(B) 3 2 1 4 

(C) 1 4 2 3 

(D) 3 4 2 1 

? 

? 

? 

? 

N 2 + other product 

PbSO 4 + O 2 + other product 

1 NO 

NaHSO 4 + other product 2 I 2 

3 warm 

Xe + other product 4 Cl 2 

Ans (D) 

warm 

1 

Sol (i) PbO 2 + H 2 SO 4 PbSO 4 + H 2 O 2 H 2 O + O 

2 

2 

Cl2 

(ii) Na 2 S 2 O 3 + H 2 O 2NaHSO 4 + NaCl 

(iii) N 2 H I 2 

4 N 2 + 2HI 

(iv) XeF 2 + NO Xe + NOF 

40 The unbalanced chemical reactions given in List I show missing reagent or condition (?) which 

are provided in List II Match List I with List II and select the correct answer using the code 

given below the lists 

P. Cl 

List I 

List II 

1 (i) Hg(OAc) 2 ; (ii) NaBH 4 

Q. ONa OEt 

2 NaOEt 

R. 

OH 

3 Et – Br 


23

S. 




4 (i) BH 3 ; (ii) H 2 O 2 /NaOH 

Code : 

P Q R S 

(A) 2 3 1 4 

(B) 3 2 1 4 

(C) 2 3 4 1 

(D) 3 2 4 1 

OH 

Ans 

Sol. 

(A) 

P. Cl 

NaOEt 

Elimination 

Q. ONa 

Et-Br 

Substitution 

OEt 

R. 

(i) Hg(OAc) 2 

(ii) NaBH 4 

Follows 

markownikov's 

product 

OH 

S. 

(i) BH 3 

(ii) H 2 

O 2 

/NaOH 

Follows 

antimarkownikov's 

product 

OH 


24




PART – III : MATHEMATICS 

SECTION – 1 (Only or more options correct Type) 

This section contains 8 multiple choice questions Each question has four choices (A) (B), (C) and (D) 

out of the which ONE or MORE are correct 

41 Let be a complex cube root of unity with 1 and P p 

ij 

be a n n matrix with 

Ans 

Sol 

2 

Then P 0, when n = 

(A) 57 (B) 55 (C) 58 (D) 56 

(B,C,D) 

1 1 ... ... 1 1 ... ... 

2 2 2 2 

2 2 

1 ... 1 ... 

1 ... 1 ... 

2 2 

... ... ... ... ... ... 

2 2 

1 ... 1 ... 

..... ..... 

... ... 

p 

ij 

i 

j 

Clearly P 2 will be O if n is a multiple of 3, then only, the elements of P 2 will be reduced to 

2 

1 0 

2 

If P 0 n is not multiple of 3 

42 The function y 2 | x | | x 2 | | x 2 | 2 | x | has a local minimum or a local maximum at x = 

(A) 2 

(B) 

Ans (A, B) 

Sol y 2 | x | | x 2 | | x 2 | 2 | x | 

2 

3 

Case (i) x 2 

y 2x x 2 | x 2 2x | 

y 3x 2 | x 2| 

y 2x 4 … (1) 

Case (ii) 2 x 0 

y x 2 | 3x 2| 

So for x 

2 

3 

y 2x 4 … (2) 

(C) 2 (D) 2 3 


25



And for x 

2 

,0 

3 

y 4x … (3) 

Case (iii) 0 x 

y 2x x 2 | x 2 2x | 

y 3x 2 | x 2| 

For 0 x 2 

y 4x … (4) 

And for, x 2 

y 2x 4 … (5) 

So graph is 


(0,8) 

y = 2x+4 

y= -2x-4 

y= 2x+4 

y= -4x 

(0,8/3) y=4x 

(-2, 0) (-2/3,0) (0, 0) (2,0) 

x 

From graph 

x 2,0 , are point of minimum, 

& x 

2 

is point of maxima 

3 

43 Let w 

3 i 

n 

1 

and P : n 1,2,3,... Further H1 

z C : Rez 

2 

2 and 

1 

H2 

z C : Rez 

2 , where C is the set of all complex numbers If z1 P H 

1, z2 P H 

2 

and O represents the origin, then z1O z 

2 

(A) 2 

(B) 6 

(C) 2 3 

(D) 5 6 

Ans (C, D) 

Sol 


26




2 5 

Clearly from the figure, z1Oz 2 

or or 

3 6 

x x 1 

44 If 3 4 , then x = 

2log3 

2 

2 

(A) (B) 

2log3 

2 1 

2 log2 

3 

Ans (A, B, C) 

x x 1 

Sol 3 4 

Taking log both sides with base 3 

x 1 

log 3 log 4 

3 3 

x log 3 x 1 log 4 

3 3 

x x 1 log3 

4 

x x log34 log34 

log34 x log34 1 

log3 

4 

x 

log3 

4 1 

2log3 

2 

x 

2log 2 1 

3 

x x 1 

3 4 

Taking log both sides with base 4 

log 3 log 4 

x x 1 

4 4 

x log 3 x 1 

4 

1 x x log 

43 x 1 log 

43 1 

1 2 

x or x 

1 log 3 2 log 3 

4 2 

(C) 

1 

1 log 3 

4 

2log2 

3 

(D) 

2log 3 1 

2 


27



y z 

45 Two lines L 

1 

: x 5, and 

2 

3 2 


y z 

L : x , 

are coplanar Then 

1 2 

value(s) 

(A) 1 (B) 2 (C) 3 (D) 4 

Ans (A, D) 

Sol 

x 5 y z 

L: 

1 

0 3 2 

x y z 

L 

2 

: 

0 1 2 

5 0 0 

0 3 2 0 

0 1 2 

5, 3 2 2 0 

2 

2 

5;6 5 2 0 

5 4 0 

1 4 0 

can take 

Cos P 

1,4,5 

46 In a triangle PQR, P is the largest angle and cos P 

1 

Further the incircle of the triangle touches 

3 

the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM 

are consecutive even integers Then possible length(s) of the side(s) of the triangle is (are) 

(A) 16 (B) 18 (C) 24 (D) 22 

Ans (B, D) 

Sol 

1 

3 

2 2 2 

1 4n 2 4n 4 4n 6 

3 2 4n 2 4n 4 

n 4 or n 1 

Hence, the possible length of side of PQR will be 18, 20 or 22 

47 For a R (the set of all real numbers), a 1, 

n 

Lt 

Ans (B, D) 

a 1 

a a a 

1 2 .... n 1 

n 1 na 1 na 2 ... na n 

60 

then a = 

(A) 5 (B) 7 (C) 

15 

2 

(D) 

17 

2 


28

Sol 

n 

Lt 

1 

0 

n 

n 

n 

Lt 

Lt 

Lt 

1 

n 1 

n 

a 1 

a 

n 1 

a 1 



n 

r 1 

n 

r 1 

n 

r 

na 

n 

a 

r 1 

n 

r 1 

r 

r 

n 

a 

n 

1 r 

a 1 

n n r 1 n 

a 1 n 

n 1 

1 

1 

a 

x dx 

0 

a 

x dx 

r 1 

a 

r 

n 

a 

1 r 

a 

n n 

n 

1 r 

1 n r 1 n 

a 1 n 

1 1 r 

a 

n n r 1 n 

1 

60 

a 

2 

2a 3a 119 0 


17 

a 7,a 

2 

48 Circles(s) touching x – axis at a distance 3 from the origin and having an intercept of length 

2 7 on y-axis is (are) 

2 2 

2 2 

(A) x y 6x 8y 9 0 (B) x y 6x 7y 9 0 

2 

(C) x 

2 

y 6x 8y 9 0 (D) 

Ans (A, C) 

Sol Let circle be 

2 

x 

2 

y 2gx 2fy c 0 

2 2 

g c,2 f c 2 7 

g 3,c 9,f 4 

So circle will be 

2 2 

x y 6x 8y 9 0 

2 2 

x y 6x 7y 9 0 


29




SECTION – 2 (Paragraph Type) 

This section contains 4 paragraphs each describing theory, experiment, data etc Eight questions relate to 

four paragraphs with two questions on each paragraph Each question of a paragraph has only one correct 

answer among the four choices (A), (B), (C) and (D) 

Let f : 0,1 

Paragraph for Questions 49 to 50 

R (the set of all real numbers) be a function Suppose the function f is twice 

differentiable, f 0 f 1 0 and satisfies f x 2f ' x f x f x x 0,1 

49 Which of the following is true for 0 < x < 1 ? 

(A) 0 f x (B) 

1 1 

f x 

2 2 

Ans (D) 

Sol 

x 

e f " x 2f ' x e 

x x 

e f x 1 

x x x x 

e f " x f ' x e e f ' x e f x 1 

d 

dx 

Let 

d 

dx 

x 

x 

e f ' x e f x 1 

d d e 

x f x 1 

dx dx 

x 

e f x g x 

g" x 1 

Since g 0 g 1 0 and g" x positive 

Possible graph for g(x) is 

(C) 

1 

4 

f x 1 

(D) f x 0 

O 

1 

Since g x 0for x 0,1 

Therefore f x is negativefor x 0,1 

Which satisfies f x 0 

50 If the function 

following is true? 

1 3 

(A) f ' x f x , x 

4 4 

e 

x 

f x assumes its minimum in the interval {0, 1] at 

1 

, which of the 

4 


x 

(B) f ' x f x ,0 x 

1 

4 

30

Ans 

Sol 

(C) f ' x f x ,0 x 

(C) 

For 

f ' x 



1 

4 

1 

x 0, g ' x 0 

4 

x 

x 

e f ' x e f x 0 

f x for 


3 

(D) f ' x f x , x 1 

4 

1 

x 0, 4 


Let PQ be a focal chord of the parallel 

lying on the line y 2 x a, a 0. 

y 

2 

4 ax . The tangents of the parabola at P and Q meet at a point 

51 Length of chord PQ is 

(A) 7a (B) 5a (C) 2a (D) 3a 

Ans (B) 

Sol 

, 2at) 

P (at2 y=2x+a 

(0, 0) 

(-a, -a) 

S 

Q 

a 2a 

Q , t 

2 t 

2 

1 

PQ a t 

………(i) 

t 

Eq of tangent at P 

2 

ty x at 

As it passes through (–a, –a) 

– at = – a + at 2 

t 2 + t – 1 = 0 

5 1 

t (for p, t > 0) 

2 

2 

5 1 2 5 1 

PQ a 

2 5 1 5 1 


31

a 



5 1 5 1 

2 2 

= 5a 

52 If chord PQ subtends an angle at the vertex of 


2 

y 4 ax , then tan 

Ans 

Sol 

(A) 2 7 

3 

(D) 

m 1 = slope of OP 

2 

t 

(B) 

2 7 

3 

5 1 

m 2 = slope of OQ 2t 

5 1 

(C) 2 5 

3 

(D) 

2 5 

3 

tan 

m1 m2 

1 mm 

1 2 

2 5 

3 

Where is Acute angle between line OP and OQ but as POQ obtuse 

2 

tan 5 

3 

Let S S S S , where 

1 2 3 


z 1 3i 

S1 z : z 4 , S2 

z : Im 0 

1 3i 

S3 z : Re z 0 . 

53 Area of S = 

(A) 10 3 

(B) 20 3 

(C) 16 3 

(D) 32 3 

Ans 

Sol 

(B) 

|z|=4 

60 0 

B 

A (1, -3) 

3x 

y 0 


32



S 1 represent the inside part of circle |z| = 4 

By putting z = x + iy, we get S2 3x y 0 

S 3 represent the right part of y-axis 

S is the shaded region in above diagram 

Now A 

2 2 2 

r 2 r r 20 

4 3 4 3 4 3 

54 min 1 3i z 

z 

S 


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

2 

2 

2 

2 

Ans (C) 

Sol Min |1 – 3i – z| = minimum distance of point A 1, 3 z S from region S 

= AB distance in diagram 

3 3 3 3 

2 2 


A box B 1 contains 1 white ball, 3 red balls and 2 block balls Another box B 2 contains 2 white balls, 3 red 

balls and 4 black balls A third box B 3 contains 3 white balls, 4 red balls and 5 black balls 

55 If 1 balls is drawn from each of the boxed B 1 , B 2 the probability that all 3 drawn balls are of the 

same colour is 

(A) 82 

(B) 90 

(C) 558 

(D) 566 

648 

648 

648 

648 

Ans 

Sol 

(A) 

Required probability 

1 2 3 3 3 4 2 4 5 

6 9 12 6 9 12 6 9 12 

82 

648 

56 If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is 

white and the other balls is red, the probability that these 2 balls are drawn from box B 2 is 

(A) 116 

(B) 126 

(C) 65 

(D) 55 

181 

181 

181 

181 

Ans (D) 

Sol Required probability 


33

1 

3 

1 

3 



C. 

C 

2 3 

1 1 

9 

C2 

C . C C . C C . C 

1 3 2 3 3 4 

1 1 1 1 1 1 

6 9 12 

C2 C2 C2 


55 

181 

SECTION – 3: (Matching list Type) 

This section contains 4 multiple choice question Each question has matching lists The codes for the lists 

have choice (A), (B), (C) and (D) out of which ONLY ONE is correct 

57 

P 

List-I 

Volume of parallelepiped determined by vector 

ab , and c is 2 Then the volume of the 

parallelepiped determined by vectors 

2 a b , 3 b c and c a is 

Q Volume of parallelepiped by vectors ab , and c is 

5 Then the volume of the parallelepiped determined 

R 

S 

by vectors 3 a b , 3 b c and 2 c a is 

Area of a triangle with adjacent sides determined 

by vectors a and b is 20 Then the area of the 

triangle with adjacent sides determined by vectors 

2a 3b and a b is 

Area of a parallelogram with adjacent sides 

determined by vectors a and b is 30 Then the area 

of the parallelogram with adjacent sides determined 

by vectors a 

b and a is 

Codes 

P Q R S 

(A) 4 2 3 1 

(B) 2 3 1 4 

(C) 3 4 1 2 

(D) 1 4 3 2 

Ans (C) 

Sol (P) volume of parallelepiped = 2 

Volume 2 a b . 3 b c c a 

List-II 

1 100 

2 30 

3 24 

4 60 

6 a b . b c . a c b c . c a 


34



6 a b . a b c c 


= 24 

6 a b c 

(Q) a b c 5 

2 

Volume 3 a b . b c 2 c a 

= 60 

6 a b . b c b a c a 

(R) Area of triangle 

40 a b 

1 

2 a b 

1 

Required area 2 a 

2 3 b a b 

1 2 

2 a a 2 b 3 a 3 b 

= 100 

(S) Area of pallelogram a b 

30 a b 

Required area a b a 

a a b a 

= 30 

0 a b 

58 

x 1 y z 3 x 4 y 3 z 3 

Consider the lines L1: , L 

2: 

and the planes 

2 1 1 1 1 2 

P1: 7x y 2z 3, P2: 3x 5y 6z 4. Let ax by cz d be the equation of the plane passing 

through the point of intersection of lines L 1 and L 2 , and perpendicular to planes P 1 and P 2 

Match List-I with List-II and selected the correct answer using the code given below the lists: 

List-I 

List-II 

P a = 1 13 

Q b = 2 3 

R c = 3 1 

S d = 4 2 

Codes 


35



P Q R S 

(A) 3 2 4 1 

(B) 1 3 4 2 

(C) 3 2 1 4 

(D) 2 4 1 3 

Ans (A) 

Sol Point of intersection of the lines L 1 and L 2 

x 1 y z 3 

x 4 y 3 z 3 


r 

1 

r 

2 

2 1 1 

1 1 2 

x = 2r 1 + 1 x = r 2 + 4 

y = - r 1 y = r 2 – 3 

z = r 1 – 3 z = 2r 2 – 3 

2r 1 + 1 = r 2 + 4 

2r 1 – r 2 = 3 ………(i) 

- r 1 = r 2 - 3 ……………(ii) 

Solving (i) and (2) r 1 = 2, r 2 = 1 

So point is (5, - 2, -1) 

Now the equation of the plane is 

a(x – 5) + b(y + 2) + c(z + 1) = 0 

Also this plane is perpendicular to P 1 & P 2 

So 7a + b + 2c = 0 

3a + 5b – 6c = 0 

a b c 

1 3 2 

- 1(x – 5) + 3(y + 2) + 2(z – 1) = 0 

- x + 3y + 2z + 13 = 0 

x – 3y – 2z = 13 

by comparing 

we get, a = 1, b = - 3, c = - 2, d = 13 

59 Match List I with List II and select the correct answer using the code given below the lists 

List-I 

List-II 

P 

1/2 

1 1 

2 

1 1 5 

1 cos tan y y sin tan y 

4 

y 

2 3 13 

2 1 1 

y cot sin y tan sin y 

Q If cos x cos y cos z 0 sin x sin y sin z then 2 2 

x y 

possible value of cos is 

2 

R If 

3 1 

cos x cos 2x sin xsin 2xsexx cos xsin 2xsec 

x 

2 

4 


36




Codes 

Ans 

Sol 

S 

cos x cos 2x then possible value of sec x is 

4 

If 

1 2 1 

cot sin 1 x sin tan x 6 , x 0, then 

possible value of x is 

P Q R S 

(A) 4 3 1 2 

(B) 4 3 2 1 

(C) 3 4 2 1 

(D) 3 4 1 2 

(B) 

(P) 

1 

cot tan 

y y sin tan y 

1 1 

2 1 1 

y cot sin y tan sin y 

2 

y 

4 

1/2 

4 1 

1 

1 

1 y 1 

2 

2 2 

2 

y 

2 

1 y y 

1 

2 

y 

1 

y 

1 

y 

1 

y 

1 

y 

2 

2 

y 

y 

2 2 

y 

2 

1 

2 

y 

y 

y 

y 

2 

4 

1/2 

y 

4 

=1 

(Q) cos x + cosy = - cos z, sin x + sin y = - sin z 

2 2 2 

cos x cos y 2cos x cos y cos z 

1/2 

2 2 2 

sin sin y 2sin xsin y sin z 

1 1 2cos x y 1 

1 

cos x y 

2 

2 x y 1 

2cos 1 

2 2 

2 x y 1 

cos 

2 4 

x y 1 

cos 

2 2 


37




cos x sin x cos x sin x 

(R) cos 2x sin 2xsec x cos x sin x 

2 2 2 2 

2sin xcos 

x 

cos 2x 2 sin x cos x sin x 

cos x 

(S) 

cos x sin x 2 sin 2x 1 x 

sec 2 

4 

1 2 1 

cot sin 1 x sin tan x 6 . 

cot cos 

x 

sin sin 

1 1 

1 6 

2 2 

1 x 6x 

1 

12x 2 = 5 

2 2 

1 x 

6x 

1 

x 

6 

4 

1 5 

x 

2 3 

60 A line L: y mx 3 meets y-axis at E(0, 2) and the arc of the parabola y 2 = 16x, 0 y 6 at the 

point F(x 0 , y 0 ) The tangent to the parabola at F(x 0 , y 0 ) intersects the y-axis at G(0, y 1 ) The slop m 

of the line L is chosen such that the area of the triangle EFG has a local maximum 

Match List I with List II and select the correct answer using the code given below the lists: 

List-I 

List-II 

P m = 1 1 

2 

Q Maximum area of EFG is 2 4 

R y 0 = 3 2 

S y 1 = 4 1 

Ans 

Sol 

Codes 

P Q R S 

(A) 4 1 2 3 

(B) 3 4 1 2 

(C) 1 3 2 4 

(D) 1 3 4 2 

(A) 


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(0, 3) 

(0, 3) 

E 

G 

0, 0 

F(x , y ) 

0 0 

(0, y ) 1 

2 

y =16x 

Let Co-ordinate of F= (x 0 , y 0 ) = (4t 2 8t) 

Co-ordinate of E = (0, 3) ……… given 

Equation of tangent at F ty = x + 4t 2 

Co-ordinate of G (0, 4t) 

0 3 1 

Area of EFG 

1 0 4 t 1 

2 4 

2 

t 8 t 1 

1 12 2 3 

t t 

2 

Let 

t 

1 12 

2 16 

3 

t t 

2 

' t 

1 

2 

24t 48t 

2 

' t 0 t 

1 

2 

" t 

1 

24 

2 

96t 

" 

1 1 1 

24 96 

2 2 2 

12 

Since (4t 2 , 8t) lies on the lie y = mx + 3 

8t 

3 

m 

2 

4t 

1 

at t , t will be maximum 

2 

2 

8t 4mt 

3 

At t = 1 2 

We get 1 m 

Maximum area will be at t = 1/2 

maximum area of triangle = 1/2 

y 0 = 8t = 4 

y 1 = 4t = 2 


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Answers Key 

PHYSICS CHEMISTRY MATHEMATICS 

Q. No. Answer Q. No. Answer Q. No. Answer 

1 A, B,C,D 21 C,D 41 B,C,D 

2 A,C 22 A,B,D 42 A,B 

3 D 23 A,B,C,D 43 C,D 

4 C,D 24 A,B 44 A,B,C 

5 A,C,D 25 B 45 A,D 

6 A,B 26 C 46 B,D 

7 BD 27 B 47 B,D 

8 A,D 28 B,D 48 A,C 

9 B 29 A 49 D 

10 A 30 B 50 C 

11 B 31 B 51 B 

12 A 32 A 52 D 

13 B 33 C 53 B 

14 C or B 34 B 54 C 

15 C 35 A 55 A 

16 A 36 A 56 D 

17 A 37 A 57 C 

18 C 38 D 58 A 

19 D 39 D 59 B 

20 C 40 A 60 A 


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