NARAYANA IIT ACADEMY JEE MAIN – 2013 - Narayana Group

NARAYANA IIT ACADEMY
INDIA
JEE MAIN – 2013
QUESTION PAPER, KEY & SOLUTIONS

Important Instructions :
Time : 3 hr. CODE : P Maximum Marks – 360.
1. Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball
point Pen. Use of pencil is strictly prohibited.
2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test
Booklet, take out the Answer Sheet and fill in the particulars carefully.
3. The test is of 3 hours duration.
4. The test Booklet consists of 90 questions. The maximum marks are 360.
5. There are three parts in the questions paper A, B, C consisting of Physics, Chemistry and
mathematics having 30 questions in each part of equation weightage. Each question is
allotted 4 (four) marks for correct response.
6. Candidates will be awarded marks as stated above in instruction No. 5 correct response of
each question 1/4 (one fourth) marks will be deducted for indicating incorrect response of
each questions. No deduction from the total score will be made if no response is indicated
for an item in the answer sheet.
7. There is only one correct response for each question. Filling up more than one response in
any question will be treated as wrong response and marks for wrong response will be
deducted accordingly as per instruction 6 above.
8. Use Blue/ Black Ball Point Pen only for writing particulars/marking responses on Side -1
and Side – 2 of the Answer Sheet. Use of Pencil is strictly prohibited.
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pager, mobile phone, any electronic device, etc. except the Admit Card inside the
examination hall/room.
10. Rough work is to be done on the space provided for this purpose in the Test Booklet Only.
This space is given at the bottom of each page and in 3 page (Pages 21-23) at the end of the
booklet.
11. On completion of the test, the candidate must hand over the Answer Sheet to the
Invigilator on duty in the Room/Hall. However, the candidates are allowed to take away
this Test Booklet with them.
12. The CODE for this Booklet is P. Make sure that the CODE printed on Side – 2 of the
Answer Sheet is the same as that on this booklet. In case of discrepancy, the candidate
should immediately report the matter to the invigilator for replacement of both the Test
Booklet and the Answer Sheet.
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NARAYANA IIT ACADEMY JEE MAIN – 2013

PHYSICS
1. A uniform cylinder of length L and mass M having cross – sectional area A is suspended,
with its length vertical, from a fixed point by a massless spring, such that it is half submerged
in a liquid of density at equilibrium position. The extension x 0 of the spring when it is in
equilibrium is :
(1) Mg
Mg LA
Mg LA
Mg LA
(2) 1
(3) 1
(4) 1
k
k M k 2M k M
(Here k is spring constant)
Key. 3
L
Sol. kx Ag Mg
0
2
L
kx0
Mg Ag
2
LA
Mg
1
2M
kx 0
F
B
F
B
= L Ag
2
Mg
2. A metallic rod of length ‘1’ is tied to a string of length 2l and made to rotate with angular
speed on a horizontal table with one end of the string fixed. If there is a vertical magnetic
field ‘B’ in the region, the e.m.f. induced across the ends of the rod is :
2l
l
(1)
Key. 4
2Bl
2
2
(2)
3Bl
2
2
(3)
4Bl
2
2
(4)
5Bl
2
2
NARAYANA IIT ACADEMY JEE MAIN – 2013

Sol.
1 2
AC B (3 )
2
1 2
AB
B (2 )
2
5B
BC AC AB
2
2
A B C
2l
l
3. This question has Statement I and Statement II. Of the four choices given after the
Statements, choose the one that best describes the two Statements.
Statement – I : A point particle of mass m moving with speed v collides with stationary point
particle of mass M. If the maximum energy loss possible is given as
1 2
m
f m
then f
2 M m .
Statement – II : Maximum energy loss occurs when the particles get stuck together as a
result of the collision.
(1) Statement – I is true, Statement – II is true, Statement – II is a correct explanation of
Statement – I.
(2) Statement – I is true, Statement – II is true, Statement – II is Not a correct explanation of
Statement – I.
(3) Statement – I is true, Statement-II is false.
(4) Statement – I is false, Statement – II is true.
Key. 4
Sol. mV + M(0) = mV 1 + MV 2
V V
2 1
e
O
2 1
mV
V V
2 1
m M
1 2 1 2 1 2
Loss mV mV MV
1 2
2
2 2
2 2
1 2 (m M) m V
mV
2
2 2
(m M)
2
MmV
; So statement I is wrong.
2
2(m M)
4. Let 0 denote the dimensional formula of the permittivity of vacuum. If M = mass, L =
length, T = time and A = electric current, then :
1 3 2
1 3 4 2
(1) [ ] [M L T A]
(2) [ ] [M L T A ]
0
1 2 1 2
1 2 1
(3) [ ] [M L T A ]
(4) [ ] [M L T A]
0
NARAYANA IIT ACADEMY JEE MAIN – 2013
0
0

Key. 2
Sol. [ ] [M] [L] [T] [A]
0
M L A T
a b c d
[M] [L] [T] [A]
1 3 2 4 a b c d
5. A projectile is given an initial velocity of ˆ i 2j ˆ
m/s, where
î is along the ground and ĵ is
along the vertical. If g = 10 m/s 2 , the equation of its trajectory is :
(1) y = x – 5x 2 (2) y = 2x – 5x 2 (3) 4y = 2x – 5x 2 (4) 4y = 2x – 25x 2
Key. 2
Sol. x = (1) t
1 2
y 2t gt
2
2
gx
y 2x
2
6. The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In
another 10s it will decrease to times its original magnitude, where equals :
(1) 0.7 (2) 0.81 (3) 0.729 (4) 0.6
Key. 3
Sol.
kt
A A e
0
0
A 1 A0
ln kt t ln
A k A
1 10
5 ln
k 9
…(1)
1 1
15 ln
k
….(2)
3
1 10
9
3
(0.9) 0.729
7. Two capacitors C 1 and C 2 are charged to 120 V and 200 V respectively. It is found that by
connecting them together the potential on each one can be made zero. Then
(1) 5C1 3C2
(2) 3C1 5C2
(3) 3C1 5C2
0 (4) 9C1 4C2
Ans. (2)
Sol.
C1 120 200C2
3C 5C
1 2
NARAYANA IIT ACADEMY JEE MAIN – 2013

8. A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain
of 1%. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 ×
10 3 kg/m 3 and 2.2 × 10 11 N/m 2 respectively?
(1) 188.5 Hz (2) 178.2 Hz (3) 200.5 Hz (4) 770 Hz
Ans. (2)
Sol. Stress = Y × strain
11 1
2.210
100
F
9
2.2 10
A
f
f
0
v 1 F
2l
2l
A
1 2.210
5
1 20 10
2 1.5 7.7 10 3 7
0 3
9
1 2 3
f0
10
3 7
f0
178.17
Hz
9. A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm.
The centre of the small loop is on the axis of the bigger loop. The distance between their
centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked
with bigger loop is
11
(1) 9.1 10 weber
11
(2) 6 10 weber
11
(3) 3.310 weber g
9
(4) 6.6
10 weber
Ans. (1)
Sol.
2
0i r1
3/2
2 2
1 1
2 r r
r
2
2
Mi
2 2
0 r1 r2
3/2
2 2
1
M
2 r x
Mi
2 2
0 r1 r2
3/2
2 2
1
2 r x
i
NARAYANA IIT ACADEMY JEE MAIN – 2013

2 2
2 2
3.14 2010 7
0.310
2 2
2 2
2
2010 1510
410 2
3/2
11
9.1
10
10. Diameter of a plano-convex lens is 6 cm and thickness at the centre is 3 mm. If speed of light
in material of lens is 2 × 10 8 m/s, the focal length of the lens is
(1) 15 cm (2) 20 cm (3) 30 cm (4) 10 cm
Ans. (3)
2 2
2
Sol. R R 0.3 3
R 15.15cm
2 1 2 1
v u R
3 / 2 1 3 / 2 1
v 15
3 1
2v 2 15
V = 45 cm
v 45
f
3 / 2
f = 30 cm
11. What is the minimum energy required to launch a satellite of mass m from the surface of a
planet of mass M and radius R in a circular orbit at an altitude of 2R?
(1) 5GmM
(2) 2GmM (3) GmM (4) GmM
6R
3R
2R
3R
Ans. (1)
GMm GMm 5GMm
Sol. E
6R R R
12. A diode detector is used to detect an amplitude modulated wave of 60% modulation by using
a condenser of capacity 250 pico farad in parallel with a load resistance 100 kilo ohm. Find
the maximum modulated frequency which could be detected by it.
(1) 10.62 MHz (2) 10.62 kHz (3) 5.31 MHz (4) 5.31 kHz
Ans. NO OPTION
SOL. QUESTION AMBIGUITY
NARAYANA IIT ACADEMY JEE MAIN – 2013

13. A beam of unpolarised light of intensity I 0 is passed through a Polaroid A and then through
another Polaroid B which is oriented so that its principal plane makes an angle of 45°
relative to that of A. The intensity of the emergent light is:
(1) I 0 (2) I 0 /2 (3) I 0 /2 (4) I 0 /8
Ans. (3)
Sol.
I0
I(1st polarisation)
2
I0 2 I0
I(2nd polarisation)
cos 45
2 4
14. The supply voltage to a room is 120 V. The resistance of the lead wires is 6 . A 60 W bulb is
already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is
switched on in parallel to the bulb?
(1) Zero Volt (2) 2.9 Volt (3) 13.3 Volt (4) 10.04 Volt
Ans. (4)
Sol.
2
240
R
0
= 6
60W
R 1
=
120
60
=240
R
0
= 6
S
240 W
R 2
2
120
240
=60
60
15.
120 V
When S is open
120
240
Vbulb
117.073
246
When S is closed
2
120
Peq
300W R
eq
300
48 (Heater and Bulb)
120
48
Vbulb
106.66
54
V 10.4 V no option is correct
The closest option is (4) only
120 V
15.
2p 0
p p 0
v 0
v
2v 0
NARAYANA IIT ACADEMY JEE MAIN – 2013

The above p-v diagram represents the thermodynamic cycle of an engine, operating with an
ideal monatomic gas. The amount of heat, extracted from the source in a single cycle is:
13
11
(1) p0v 0
(2) p
0 v
0
(3) p
0 v
0
(4) 4p0v
0
2
2
Ans. (2)
13
Sol. Only positive q is to be considered hence q
povo
2
16. A hoop of radius r and mass m rotating with an angular velocity
0
is placed on a rough
horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the
velocity of the centre of the hoop when it ceases to slip?
r0
r0
r0
(1) (2) (3) (4) r
0
4
3
2
Ans. (3)
Sol. Applying conservation of angular momentum about point of contact
I0 I
MVcR
0
v0=0
v c
2 2 vc
MR 0 MR MvcR
R
2
MR 2MRv
0 c
initial final
R0
vc
2
17. An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of
mass M. The piston and the cylinder have equal cross sectional area A. When the piston is in
equilibrium, the volume of the gas is V 0 and its pressure is P 0 . The piston is slightly displaced
from the equilibrium position and released. Assuming that the system is completely isolated
from its currounding, the piston executes a simple harmonic motion with frequency:
Ans. (3)
1 AP0
(1)
2
V M
0
1 V0MP
0
(2)
2
2
A
(3)
2
1 A P
2
MV
0
0
1 MV0
(4)
2
AP
0
A M
P 0 ,V 0
Sol.
NARAYANA IIT ACADEMY JEE MAIN – 2013

dV
If displaced by x then Frestoring
A.dP AB V
2
Ax
2 BA x
Frestoring
A.B M x
V0 V0
2
P0
A x
V
0
A P 1 A P
2
MV 2 MV
2 2
0 0
0 0
18. If a piece of metal is heated to temperature and then allowed to cool in a room which is at
temperature 0, the graph between the temperature T of the metal and time t will be closed
to :
T
T
T
T
1.
2. 0
3. 0
4. 0
0 t
0 t
0 t
0 t
Ans. (3)
d 4 4
Sol. As per Stefan’s Law
0 which decrease with t
dt
19. This question has Statement – I and Statement – II. Of the four choices given after the
Statements, choose the one that best describes the two Statements.
Statement – I : Higher the range, greater is the resistance of ammeter.
Statement – II : To increase the range of ammeter, additional shunt needs to be used across
it.
(1) Statement – I is true, Statement – II is true, Statement – II is the correct explanation of
Statement – I
(2) Statement – I is true, Statement – II is true, Statement – II is not the correct explanation
of Statement – I
(3) Statement – I is true, Statement – II is false
(4) Statement – I is false, Statement – II is true
Ans. (4)
Range of the ammeter is given by
R
g
I Ig
1
R
and resistance of the ammeter is given by
R R
gR
R R
g
Hence to have higher range, is the resistance of ammeter should be made smaller.
0
NARAYANA IIT ACADEMY JEE MAIN – 2013

20. In an LCR circuit as shown below both switches are open initially. Now switch S 1 is closed, S 2
kept open. (q is charge on the capacitor and RC is Capacitive time constant). Which of
the following statement is correct?
V
R
C
S 1
L
(1) Work done by the battery is half of the energy dissipated in the resistor
CV
2
1
(2) At t , q (3) At t 2 , q CV(1 e ) (4) At t , q CV(1 e )
2
2
Ans. (3)
As during the charging
t/
q CV(1 e )
Hence at t 2
2
q CV(1 e )
21. Two coherent point sources S 1 and S 2 are separated by a small distance ‘d’ as shown. The
fringes obtained on the screen will be
S 2
d
S 1 S 2
d
O
Screeen
(1) points (2) straight lines (3) semi-circles (4) concentric circles
Ans. (4)
The waves reaching from S 1 and S 2 at any point on the screen which is equidistant from the
central points O will have same path difference. So locus of such points will be concentric
circles on the Screen.
22. The magnetic field in a traveling electromagnetic wave has a peak value of 20 nT. The peak
value of electric field strength is
(1) 3 V/m (2) 6 V/m (3) 9 V/m (4) 12 V/m
Ans. (2)
Sol : As, E BC
9 8
Hence, E 2010 310 6 V / m
NARAYANA IIT ACADEMY JEE MAIN – 2013

23. The anode voltage of a photocell is kept fixed. The wavelength of the light falling on the
cathode is gradually changed. The plate current I of the photocell varies as follows :
I
I
(1)
O
(2)
O
I
I
(3)
O
(4)
O
Ans. (4)
When is increased the cutoff value of current will stop.
24. The I – V characteristic of an LED is
(1)
I
I
O
Red
Yellow
Green
Blue
(R) (Y) (G) (B)
V
R
(3)
(4)
Y
G
O
V
B
Ans. (1)
I
I – V characteristic of an LED is similar to forward I – V characteristics of P – N junction
diode and threshold voltage increases on decreasing the wavelengths.
25. Assume that a drop of liquid evaporates by decrease in its surface energy, so that its
temperature remains unchanged. What should be the minimum radius of the drop for this to
be possible? The surface tension is T, density of liquid is and L is its latent heat of
vaporization.
(1) L/T (2) T / L (3) T/L (4) 2T/L
Key: 4
Sol: (4)
Let the radius of drop is x and ‘dx’ thickness of drop evaporates. Change is Area = A = 4x 2
– 4(x – dx) 2
NARAYANA IIT ACADEMY JEE MAIN – 2013
(2)
I
B
G
Y
R
O
V
V
O

2
= 4x 2 dx
1 1 8xdx
x
So energy released = T. 8xdx
Volume of liquid evaporates = 4x 2 dx
So, T . 8xdx = . 4x 2 2T
dx . L x L
26. In a hydrogen like atom electron makes transition from an energy level with quantum
number n to another with quantum number (n – 1). If n > > 1, the frequency of radiation
emitted is proportional to
(1) 1 1
1
1
(2) (3) (4)
2
3/ 2
3
n
n
n
n
Key : 4
Sol: (4)
2 1 1
RCZ
2 2
(n 1) n
2
2
2
= RCZ 1 RCZ 1
1
1 1
2 2
2
n
1 n n
1
n
2 2
RCZ 2 2RCZ 1
= 1 1
2
3 3
n n n n
27. The graph between angle of deviation () and angle of incidence (i) for a triangular prism is
represented by
d
d
O
(1) i
d
O
(2) i
d
O
(3) i
O
(4) i
Key : 3
Sol: (3)
= i + e – A, when i = e, is minimum.
28. Two charges, each equal to q, are kept at x = – a and x = a on the x-axis. A particle of mass
m and charge q
0
q is placed at the origin. If charge q 0 is given a small displacement (y < <
2
a) along the y – axis, the net force acting on the particle is proportional to
NARAYANA IIT ACADEMY JEE MAIN – 2013

F
2F cos
Y
F
F sin
q 0
F sin
Y
q a a q
X
(1) y (2) – y (3) 1 y
(4)
1
y
Key : 1
Sol: (1)
Force due to each charge is
F kq q0
(a 2 y 2
)
kqq0
y
From figure resultant force is 2Fcos 2. (a y ) a y
2 2 2 2
2k q q0
y
Y < < a, Fr F y (towards +ve y-axis)
3
a
29. Two short bar magnets of length 1 cm each have magnetic moments 1.20 Am 2 and 1.00 Am 2
respectively. They are placed on a horizontal table parallel to each other with their N poles
pointing towards the South. They have a common magnetic equator and are separated by a
distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the midpoint
O of the line joining their centres is close to (Horizontal component of earth’s magnetic
induction is 3.6 10 –5 Wb/m 2 ).
(1) 3.6 10 –5 Wb/m 2 (2) 2.56 10 –4 Wb/m 2
(3) 3.50 10 –4 Wb/m 2 (4) 5.80 10 –4 Wb/m 2
Key : 2
Sol: (2)
Field at centre (O) due to both magnet
0
1 7
1
= (M
3 1
M
2) 10 (1.20 1.00)
1 3
4
r (10 )
= 2.2 10 –4 (towards North)
Earth’s horizontal component of magnetic field at that point is 3.6 10 –5 towards north.
So Net field at that point
= 2.2 10 –4 + 3.6 10 –5 = 2.56 10 –4 Wb/m 2
30. A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure.
The electric potential at the point O lying at a distance L from the end A is
O A B
L
L
(1)
Q
8
L
0
(2)
3Q
4
L
0
NARAYANA IIT ACADEMY JEE MAIN – 2013
(3)
Q
4
L ln 2
0
(4)
Qln 2
4
L
0

Key : 4
Sol: (4)
Let’s consider as elementary length ‘dx’ at a distance x from O.
1 dq 1 Q dx
dq at O is dv
40 x 40
L x
So, net potential at O
2L
1 Q dx Q
V dV ln (2)
4
L
x 4
L
0 L
0
Q
dq dx. Potential due to
L
CHEMISTRY
31. Which of the following complex species is not expected to exhibit optical isomerism ?
(1) [Co(en) 3 ] 3+ (2) [Co(en) 2 Cl 2 ] +
(3) [Co(NH 3 ) 3 Cl 3 ] (4) [Co(en)(NH 3 ) 2 Cl 2 ] +
Key. 3
Sol: Meridional and facial distribution do not show optical isomerism
32. Which one of the following molecules is expected to exhibit diamagnetic behaviour ?
(1) C 2 (2) N 2 (3) O 2 (4) S 2
Key. (1,2)
Sol: C 2 1s 2 < *1s 2 < 2s 2 < *2s 2 2 2
< 2 p 2 p : No unpaired electron hence diamagnetic.
x
N 2 1s 2 < *1s 2 < 2s 2 < *2s 2 2 2 2
< 2 p 2 p < p : No unpaired electron hence
diamagnetic
O 2 1s 2 < *1s 2 < 2s 2 < *2s 2 <
x
2
2 p z
y
y
2 z
2 2
1 1
< 2 p 2 p < *2 p *2 p : This is
paramagnetic
S 2 is also paramagnetic since like O 2 it has unpaired electrons in ABMO
33. A solution of (-)-1- chloro-1- phenylethane in toluene racemises slowly in the presence of a
small amount of SbCl 5 due to the formation of :
(1) carbanion (2) carbene (3) carbocation (4) free radical
Key. (3)
Sol:
x
y
x
y
Toluene, SbCl 5
H 3
C C H
H 3 C
Cl
In presence of SbCl 5, carbocation will be formed.
34. Give
0 0
E 0.74 V ; E 1.51V
3 2
Cr / Cr MnO4
/ Mn
C
+
H
NARAYANA IIT ACADEMY JEE MAIN – 2013

E 1.33; E 1.36V
0 0
2 3
Cr2 O7
/ Cr Cl/
Cl
Based on the data given above, strongest oxidizing agent will be:
(1) Cl (2) Cr 3+ (3) Mn 2+ (4) MnO
4
Key. (4)
Sol: Cr 3+ + 3e Cr(s) :
( 6)
2
Cr2 O7
( 7)
+ 6e 2Cr 3+ :
MnO 4
+ 5e Mn 2+ :
Cl + 1 e Cl : E
0
red
E
E
E
0
red
0
red
0
red
1.36V
0.74V
1.33V
1.51V
Thus tendency of reduction will be most in MnO 4 into Mn 2+ . Hence this will be the strongest
oxidizing agent in given species.
35. A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375 mL at a
constant temperature of 37.0 0 C. As it does so, it absorbs 208J of heat. The values of q and w
for the process will be : (R = 8.314 J/mol K) (ln 7.5 = 2.01)
(1) q = +208 J, w = -208 J (2) q = -208J, w = -208 J
(3) q = 208 J, w = +208J (4) q = +208J, w = +208 J
Key. (1)
Sol: In isothermal expansion process U = 0
q = + 208 J
Thus w = 208 J (expansion)
36. The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2 (M)
HCl will be
(1) 0.875 M (2) 1.00 M (3) 1.75 M (4) 0.975 M
Key. (1)
Sol: Molarity of mixture of identical species is given by
M m (V 1 + V 2 ) = M 1 V 1 + M 2 V 2
M m (750 + 250) = 0.5 750 + 2 250
M m = 875 0.875M
1000
37. Arrange the following compounds in order of decreasing acidity
(1) II > IV > I > III (2) I > II > III > IV (3) III > I > II > IV (4) IV > III > I > II
Ans. (3)
Sol. –R, –I effects increase but +I, +R effects decrease acidic strengths.
NARAYANA IIT ACADEMY JEE MAIN – 2013

38. For gaseous state, if most probable speed is denoted by C*, average speed by C and mean
square speed by C, then for a large number of molecules the ratios of these speeds are
(1) C* : C : C 1.225 : 1.128 : 1
(2) C* : C : C 1.128 : 1.225 : 1
(3) C* : C : C 1 : 1.128 : 1.225 : 1
(4) C* : C : C 1 : 1.225 : 1.128
Ans. (3)
Sol. Ratio is 2 : 8 : 3 .
39. The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation
energy of such a reaction will be
(R = 8.314 JK –1 mol –1 and log 2 = 0.301)
(1) 53.6 kJ mol –1 (2) 48.6 kJ mol –1 (3) 58.5 kJ mol –1 (4) 60.5 kJ mol –1
Ans. (1)
Sol.
k1
Ea
1 1
log
k
2 2.303R T
2 T
1
log 2 2.3038.314 300310
1
Ea
kJmol .
101000
40. A compound with molecular mass 180 is acylated with CH 3 COCl to get a compound with
molecular mass 390. The number of amino groups present per molecule of the former
compound is
(1) 2 (2) 5 (3) 4 (4) 6
Ans. (2)
Sol. –OH groups of carbohydrates get acctylated COOCH3
No of –OH groups = 390 180 5
42
41. Which of the following arrangements does not represent the correct order of the property
started against it?
(1)
2 2 2 2
V Cr Mn Fe : paramagnetic behavior
(2)
2 2 2 2
Ni Co Fe Mn : ionic size
(3)
3 3 3 3
Co Fe Cr Sc : stability in aqueous solution
(4) Sc < Ti < Cr < Mn : number of oxidation states
Ans. (1)
Sol. Greater the number of unpaired electrons, higher is the paramagnetic behaviour .
42. The order of stability of the following carbocations
is
(1) III > II > I (2) II > III > I (3) I > II > III (4) III > I > II
Ans. (4)
NARAYANA IIT ACADEMY JEE MAIN – 2013

Sol. Benzyl and allyl carbocations are more stable due to resonance.
43. Consider the following reaction:
2
2
z
xMnO4 yC2O4
zH xMn 2yCO
2
H2O
2
The values of x, y and z in the reaction are, respectively:
(1) 5, 2 and 16 (2) 2, 5 and 8 (3) 2, 5 and 16 (4) 5, 2 and 8
Ans. (3)
Sol.
2
2MnO 5C O 16H
4 2 4
2
2Mn 10CO2 8H2O
x = 2
y = 5
z = 16.
44. Which of the following is the wrong statement?
(1) ONCl and ONO – are not isoelectronic (2) O 3 molecule is bent
(3) Ozone is violet-black in solid state (4) Ozone is diamagnetic gas
Ans. No Answer
Sol. All the four given statements are correct.
45. A gaseous hydrocarbon gives upon combustion 0.72 g. of water and 3.08 g. of CO 2 . The
empirical formula of the hydrocarbon is:
(1) C 2 H 4 (2) C 3 H 4 (3) C 6 H 5 (4) C 7 H 8
Ans. (4)
Sol. No. of mole of H 2 O = 0.72 0.04 mole
18
No. of mole of H = 2 × 0.04 = 0.08 mole
No. of mole of CO 2 = 3.08 0.07 mole
44
No. of mole of ‘C’ = 0.07 mole
No. of mole of C 0.07 7
No. of mole of H 0.08 8
Empirical Formula = C 7 H 8 .
46. In which of the following pairs of molecules / ions, both the species are not likely to exist?
2
2
2
2
(1) H
2
, He2
(2) H
2
, He2
(3) H
2
, He2
(4) H
2
, He2
Ans. (3)
Sol.
Species Bond order
H 1
2 [1 0] 0.5
2
He 2 1 2 [4 2] 1
2
NARAYANA IIT ACADEMY JEE MAIN – 2013

Both
H 2 1 2 [2 1] 0.5
2
H 2 1 2 [0 0] 0
2
He 1 2 [2 2] 0
2
2
H 2
and He 2 have zero bond order so they are not likely to exist.
47. Which of the following exists as covalent crystals in the solid state?
(1) Iodine (2) Silicon (3) Sulphur (4) Phosphorous
Ans. (2)
Sol. Silicon crystalises as diamond type covalent solid.
48. Synthesis of each molecule of glucose in photosynthesis involves:
(1) 18 molecules of ATP (2) 10 molecules of ATP
(3) 8 molecules of ATP (4) 6 molecules of ATP
Ans. (1)
Sol. 18 molecules of A.T.P are required.
49. The coagulating power of electrolytes having ions Na 3
, Al 2
and Ba for arsenic sulphide
sol increases in the order
3 2
(1) Al Ba Na
2 3
(2) Na Ba Al
2 3
(3) Ba Na Al
3 2
(4) Al Na Ba
Ans. (2)
Coagulating power is directly proportional to charge density.
50. Which of the following represents the correct order of increasing first ionization enthalpy for
Ca, Ba, S, Se and Ar?
(1) Ca < S < Ba < Se < Ar (2) S < Se < Ca < Ba < Ar
(3) Ba < Ca < Se < S < Ar (4) Ca < Ba < S < Se < Ar
Ans. (3)
(i) Ar being a noble gas has highest IE
(ii) Down the group ionization enthalpy decreases.
(iii) ‘S’ block elements will have lower ionization enthalpy than ‘P’ block.
51.
2
18
Z
Energy of an electron is given by E = 2.17810 J n
2 . Wavelength of light required to
excite an electron in an hydrogen atom from level n = 1 to n = 2 will be
34
8 1
(h = 6.62 10 Js and c = 3.0 10 ms )
7
(1) 1.214 10 m (2)
Ans. (1)
E E E
2 1
hc 2.17810 2.17810
4 1
18 18
7
2.816 10 m
(3)
7
6.500 10 m
7
(4) 8.500
10 m
NARAYANA IIT ACADEMY JEE MAIN – 2013

34 8
6.6210 310 18
3
7
1.214
10 m
2.17810
4
52. Compound (A), C8H9Br , gives a white precipitate when warmed with alcoholic AgNO
3.
Oxidation of (A) gives an acid (B), C8H6O 4. (B) easily forms anhydride on heating. Identify
the compound (A).
(1)
Ans. (4)
CH Br
2
CH 3
(2)
C H 2 5
Br
(3)
CH Br
2
CH 3
(4)
CH Br
2
CH 3
Aryl bromide cannot give ppt. as aryl cations are unstable.
53. Four successive members of the first row transition elements are listed below with atomic
0
numbers. Which one of them is expected to have the highest E 3 2
(1) Cr(Z = 24) (2) Mn(Z = 25) (3) Fe(Z = 26) (4) Co(Z = 27)
Ans. (4)
As it is expected on the basis of electronic configuration.
3 2
Mn e Mn more stable
4 0 5 0
3d 4s 3d 4s
(as it is completely half filled subshell)
However experimental data suggests that
3 2
o
3
2 Co e Co ,E is greater than other M / M couples.
54. How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of
1 to create an aqueous solution with pH of 2?
(1) 0.1 L (2) 0.9 L (3) 2.0 L (4) 9.0 L
NARAYANA IIT ACADEMY JEE MAIN – 2013
M /M

Ans. (4)
M V
M V
1 1 2 2
1 2
10 1lit 10 V 2
V2
10lit
volume of water to be added is 9 litre.
55. The first ionization potential of Na is 5.1 eV. The value of electron gain enthalpy of Na + will
be
(1) – 2.55 eV (2) –5.1 eV (3) –10.2 eV (4) +2.55 eV
Key : 2
Sol: (2)
ionization potential = – electron gain enthalpy of metal cation
56. An organic compound A upon reacting with NH 3 gives B. On heating B gives C. C in
Key : 4
Sol: (4)
presence of KOH reacts with Br 2 to give CH2CH 2NH
2, A is
CH3
CH COOH
|
(1) CH3COOH (2) CH3CH 2CH 2COOH (3) CH
CH3CH 2COOH
A
NH3
CH C H CONH
H O
2 2 2 2
C
CH CH NH
Br2 / KOH
3 2 2
CH CH C OON H
3 2
B
4
3
(4) CH3CH 2COOH
57. Stability of the species Li2 , Li 2
and Li 2
increases in the order of
(1) Li2 Li
2
Li2
(2) Li
2
Li2 Li2
(3) Li2 Li
2
Li2
(4) Li
2
Li2 Li2
Key : 2
Sol: (2)
Normally Stability Bond order but in case of species with same bond order more the
number of electrons in anti bonding molecular orbitals lesser the stability
58. An unknown alcohol is treated with the “Lucas reagent” to determine whether the alcohol is
primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism?
(1)secondary alcohol by S N
1
(2) tertiary alcohol by S N
1
(3) secondary alcohol by S
N
2
(4) tertiary alcohol by S
N
2
Key : 2
Sol: (2)
Tertiary alcohol reacts fastest with Lucas reagent by S N
1 mechanism
59. The gas leaked from a strong tank of the Union Carbide plant in Bhopal gas tragedy was
NARAYANA IIT ACADEMY JEE MAIN – 2013

(1) Methylisocyanate (2) Methylamine (3) Ammonia (4) Phosgene
Key : 1
Sol: (1)
MIC was responsible for Bhopal gas tragedy.
60. Experimentally it was found that a metal oxide has formula M
0.98
O . Metal M, is present as
2
M and
3
M
3
M
in its oxide. Fraction of the metal which exists as would be
(1) 7.01 % (2) 4.08 % (3) 6.05% (4) 5.08 %
Key : 2
Sol: (2)
3
Let the number of M and
x y 0.98 ….(i)
3x
2y
2 ….(ii)
3
No. of M ion = 0.04
2
And M ion = 0.94
So percentage of
2
M ions be x and y respectively
3
M ion 0.04 100 4.08%
0.98
MATHEMATICS
61. Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is
(1) 3 2
(2) 5 2
(3) 7 2
(4) 9 2
Key. (3)
Sol: Required distance =
5
8
2 7
2 2 2
2 1 2 2
62. At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of
dP
production P w.r.t. additional number of workers x is given by 100 12
dx
x.
If firm
employs 25 more workers, then the new level of production of items is
(1) 2500 (2) 3000 (3) 3500 (4) 4500
Key. (3)
Sol: (100 12 x)
dx
dp
24 3/2
100x x P C
3
when x = 0 P = 2000
P = 2000 + 100 x -
As x is 25
P = 3500
24
3 x
3/2
NARAYANA IIT ACADEMY JEE MAIN – 2013

63. Let A and B be two sets containing 2 elements and 4 elements respectively. The number of
subsets of A B having 3 or more elements is:
(1) 256 (2) 220 (3) 219 (4) 211
Key. (3)
Sol: As A B has 8 elements
Number of subsets of A B having
3 or more elements is
8 C 3 + 8 C 4 + . . . . . + 8 C 8
= 2 8 – 1 – 8 – 28
= 219
x 2 y 3 z 4
64. If the line and
1 1 k
(1) any value (2) exactly one value
(3) exactly two values. (4) exactly three values
Key. (3)
1 1 1
Sol:
1 1 k
0
k
2 1
2
x 1 y 4 z 5
are coplanar, then k can have
k 2 1
k 3k 0
k 0, 3
65. If the vectors AB 3iˆ
4kˆ
and AC 5iˆ
2 ˆj 4kˆ
are the sides of a triangle ABC, then the
length of the median through A is
(1) 18 (2) 72 (3) 33 (4) 45
Key. (3)
Sol: Let A (0,0, 0) then B (3, 0, 4) and C (5, –2, 4) so length of median through A will be
16 116 33
66. The real number k for which the equation, 2x 3 + 3x + k = 0 has two distinct real roots in [0,
1]
(1) lies between 1 and 2 (2) lies between 2 and 3
(3) lies between -1 and 0 (4) does not exist.
Key. (4)
Sol: Let f(x) = x 3 + 3x + k
f ' x = 6x 2 + 3
As
f ' x > 0 f(x) is S.I. function
f(x) = 0 will have at most one real root in [0, 1] so no real number k exist.
67. The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ….. is
7 20
(1) 179 10
7
(2) 99 10
20
7
(3) 179 10
20
7
99 10
81
9
81
9
Ans. (3)
20
(4)
NARAYANA IIT ACADEMY JEE MAIN – 2013

7
Sol. 1 10
1 2 20
1 10
....... 1 10
9
7
20 10
9
10 ...... 10
1
7 10 1 10
20
20
9
1
1
10 s
7
179 10
81
1 2 20
20
68. A ray of light along x 3 y 3 gets reflected upon reaching x-axis, the equation of the
reflected ray is
(1) y x 3 (2) 3 y x 3 (3) y 3x 3 (4) 3 y x 1
Ans. (2)
Sol. Slop of reflect ray is
1
3, 0 is
y 0 1
x 3 3
3 y x 3
3 and it passing through
69. The number of values of k, for which the system of equation
k 1 x 8y 4k
kx k 3 y 3k 1
Has no solution, is
(1) infinite (2) 1 (3) 2 (4) 3
Ans. (2)
k 1 k 3 8k
Sol.
2
k 4k 3 0
k 1, 3
but for k = 1 we have infinite solution
70.
2
2
If the equation x 2x 3 0 and 2x bx c 0, z, b, c R , have a common root, then a :
b : c is
(1) 1 : 2 : 3 (2) 3 : 2 : 1 (3) 1 : 3 : 2 (4) 3 : 1 : 2
Ans. (1)
Sol. Here D < 0
NARAYANA IIT ACADEMY JEE MAIN – 2013

Both roots is common
a b c
1 2 3
a : b : c = 1 : 2 : 3
71. The circle passing through (1, –2) and touching the axis of x at (3, 0) also passes through the
point
(1) (–5, 2) (2) (2, –5) (3) (5, –2) (4) (–2, 5)
Ans. (3)
2 2 2
Sol. 3 k
x y k
It passes (1, –2),
2 2 2
13 2 k k
then
2 2
4 4 k 4k k
k 2
Centre (3, –2)
2 2
x 3 y 2 4
Equation of circle
Clearly circle passes (5, –2)
72.
1 1 1
If x, y, z are in A.P. and tan x, tan y and tan z are also in A.P., then
(1) x = y = z (2) 2x = 3y = 6z (3) 6x = 3y = 2z (4) 6x = 4y = 3z
Ans. (1)
Sol. x, y, z are in AP 2y = x + z ……..(i)
1 1 1
2 tan y tan x tan z
2y x z
2
1 y 1 xz
2
1 y 1
xz
2
y xz
.……(ii)
From (i) and (ii)
x = y = z
Remark : Here we have assumed that x, y and z are of the same sign other wise x = – , y = 0,
z = for any positive will be the solution and then no option will be correct
73. Consider :
Statement – I : (p q) ( p q) is a fallacy.
Statement – II : (p q) (~ q p) is a tautology.
NARAYANA IIT ACADEMY JEE MAIN – 2013

(1) Statement-I is true; Statement – II is true; Statement – II is a correct explanation for
Statement – I.
(2) Statement-I is true; Statement – II is true; Statement – II is not a correct explanation for
Statement – I.
(3) Statement – I is true; Statement – II is false.
(4) Statement – I is false; Statement – II is true.
Ans. (2)
Sol.
p q
74. If
~ ~ (p ~ q) (~ p q) (p ~ q)
(~ p q)
T T F F F F F
T F F T T F F
F T T F F T F
F F T T F F F
(p q) (~ q ~ p)
T T T
F F T
T T T
T T T
(p q)
(~ q ~ p)
f (x)dx (x), then x 5 f (x 3
)dx is equal to :
1 x x 3 x x dx C
3
1 x 3 x 3 x 2 x 3
dx C
3
Ans. (3)
1 x x 3 x x dx C
3
1 x 3 x 3 x 3 x 3 dx C
3
3
(1) 3
2 3
3 3 3 3
(2)
(3)
(4)
Sol. Let I =
5 3
x f (x )dx
1 2 3 3
I = 3x .x f (x )dx
3
put x 3 = t
1
I t.f (t)dt
3
1
I t f (t)dt f (t)dt
dt
3
1 3 3
I x . (x ) (t)dt
3
1 3 3 3 2
I x . (x ) (x ).3x dx
3
NARAYANA IIT ACADEMY JEE MAIN – 2013

1 3 3 2 3
I x (x ) x . (x )dx
3
.
(1 cos 2x)(3 cos x)
75. lim
is equal to
x0
x tan 4x
1
(1) (2) 1 4
2
Ans. (4)
2
2 sin x 1
Sol. lim . . .(3 cos x)
x0
2
4 x tan 4x
4x
= 2 4 2
4 .
76. Statement – I : The value of the integral
Statement – II :
b
a
b
a
/3
/6
f (x)dx f (a b x)dx .
(3) 1 (4) 2
dx
is equal to .
1
tan x
6
(1) Statement-I is true; Statement – II is true; Statement – II is a correct explanation for
Statement – I.
(2) Statement-I is true; Statement-II is true; Statement – II is not a correct explanation for
Statement – I.
(3) Statement – I is true; Statement-II is false.
(4) Statement – I is false; Statement – II is true.
Ans. (4)
/3
dx
Sol. Let I
1
tan x
I
/6
/3
/6
dx
1
cot x
b
b
using f (x)dx f (a b x)dx
a
a
/3
2I dx
/6
2I = I .
6 12
77. The equation of the circle passing through the foci of the ellipse
centre at (0, 3) is :
(1) x 2 + y 2 – 6y – 7 = 0 (2) x 2 + y 2 – 6y + 7 = 0
2 2
x y
16 9
1, and having
NARAYANA IIT ACADEMY JEE MAIN – 2013

(3) x 2 + y 2 – 6y – 5 = 0 (4) x -2 + y 2 – 6y + 5 = 0
Ans. (1)
Sol. Foci of the ellipse are
7,0
radius of the circle = 4
equation of circle is
x 2 + y 2 – 6y – 7 = 0.
78. A multiple choice examination has 5 questions. Each question has three alternative answers
of which exactly one is correct. The probability that a student will get 4 or more correct
answers just by guessing is :
17
13
11
(1) (2) (3)
5
5
5
3
3
3
Ans. (3)
Sol. Required probability = P (4 correct answer by guessing)
+ P (5 correct answer by guessing)
4 5
10
(4)
5
3
1 2 1
= 5
C
. . 5
4
C5
3 3 3
11
=
5
3 .
79. The x-coordinate of the incentre of the triangle that has the coordinates of mid points of its
sides as (0, 1), (1, 1) and (1, 0) is
(1) 2 2
(2) 2 2
(3) 1 2 (4) 1
2
Ans. (2)
x coordinate of incentre
0 4 0 4
=
2
2 2 2 2 4 2 2
2
C(0, 2)
2Ö2
2
(0, 1)
(1, 1)
A(0, 0) (1, 0) B(2, 0)
2
80.
x 1 x 1
The term independent of x in the expansion of
2/3 1/3 1/2
x x 1 x x
is
(1) 4 (2) 120 (3) 210 (4) 310
Ans. (3)
NARAYANA IIT ACADEMY JEE MAIN – 2013
10

1/3
Given expression = x 1
x
x
1/3 1/2
= 10
x 1 x 1
x x 1
10 1/2
General term, T
3
r 1
Cr
x
x
10 r r
For term independent of x
10 r r 0
3 2
20 – 2r – 3r = 0
5r = 20 r = 4
Hence the term independent of x = 10 C
4
= 210
10
10
= x 1/3 1 1 x 1/2
81. The area (in square units) bounded by the curves y x , 2y x 3 0, x – axis, and lying in
the first quadrant is
(1) 9 (2) 36 (3) 18 (4) 27 4
Ans. (1)
y x
(1)
2y – x + 3 = 0 (2)
On solving, (1) and (2) we get
A = (9, 3)
Hence, area of the shaded region
9
x dx Area of ABC
0
2 1
= 9
9
3/2
x 6 3
3
0
2
(9, 3)
A
O
C(3, 0)
B(9, 0)
NARAYANA IIT ACADEMY JEE MAIN – 2013

82. Let T
n
be the number of all possible triangles formed by joining vertices of an n-sided
regular polygon. If Tn 1
Tn
10 , then the value of n is
(1) 7 (2) 5 (3) 10 (4) 8
Ans. (2)
n1
C
n
C 10
3 3
n n 1 n 1 n 2 10
6
n(n – 1) = 20
n = 5
83. If z is a complex number of unit modulus and argument , then
1
z
arg
1 z equals
(1) (2) 2
(3) (4)
Ans. (3)
Put, z = cos isin
2cos i
z 1 cos isin
cos isin
2
1 z 1 cos isin =
2 2 2 e
i
= e
i
2cos cos i sin
2
e
2 2 2
1
z
so, arg
1 z
84. ABCD is a trapezium such that AB and CD are parallel and BC CD . If ADB
, BC = p
and CD = q, then AB is equal to
p 2 q 2
sin
2 2
2 2
2 2
p q cos
p q
p q sin
(1) (2)
(3)
(4)
2 2
2
p cos qsin p cos q sin p cos q sin
pcos qsin
Ans. (1)
AB
sin sin
BD
2 2
p q sin
AB sin .cos cos .sin
AB
sin
2 2
p q .sin
q
p
sin
p q p q
2 2 2 2
p
q
, cos
p q p q
2 2 2 2
cos
NARAYANA IIT ACADEMY JEE MAIN – 2013

p 2 q 2
sin
AB =
p cos qsin
B
A
p
2 2
Öp + q
C
q
D
1 3
85. If P
1 3 3
is the adjoint of a 3 × 3 matrix A and |A| = 4, then
is equal to
2 4 4
(1) 4 (2) 11 (3) 5 (4) 0
Key : 2
Sol: (2)
P = Adj A
|P| = | Adj A|
|P| = |A| 2
2 6 16
11
86. The intercepts on x-axis made by tangents to the curve,
y | t | dt, x R,
which are parallel
to the line y 2 x,
are equal to
(1) 1
(2) 2
(3) 3
(4) 4
Key : 1
Sol: (1)
dy
| x |
dx
| x | 2
x 2
If x = 2
2
2 t
y t dt 2
0
2
0
Point is (2, 2)
Equation of tangents
y 2 2 x 2
y 2x
2
x intercept is 1
2
NARAYANA IIT ACADEMY JEE MAIN – 2013
x
0

If x = –2, y = –2
Equation of tangents is y 2 2 x 2
y 2x
2
x intercept is –1
87. Given : A circle
2 2
2x
2y
5
and a parabola, y 4 5x
Statement – I : An equation of a common tangent to these curves is y x 5.
Statement - II : If the line y mx 5 m
0
is their common tangent, then m satisfies
m
4 2
m 3m
2 0
(1) Statement – I is true; Statement – II is true; statement – II is a correct explanation for
statement - I
(2) Statement – I is true; Statement – II is true; statement – II is not a correct explanation for
statement - I
(3) Statement – I is true; Statement – II is false
(4) Statement – I is false; Statement – II is true
Key : 2
Sol: (2)
2
Eq. of tangent to y 4 5x is
5
y mx
….(1)
m
(1) is also the tangent of 2x 2 + 2y 2 = 5
2
5
0 0
5
then,
m
2
1
2
m
m 4 + m 2 – 2 = 0
m = ± 1, which satisfy m 4 – 3m 2 + 2 = 0, and so an equation of common tangent is
y = x + 5 .
88.
1
If y sectan x,
then dy at x=1 is equal to
dx
(1) 1/ 2 (2) 2 (3) 1 (4) 1 / 2
Key : 1
Sol: (1)
y = sec (tan –1 x)
1 2
= sec sec 1 x
y
2
1
x
NARAYANA IIT ACADEMY JEE MAIN – 2013

dy 1
dx
2
2 1
x
2x
dy 1
dx ( at x 1)
2
89.
tan A cot A
The expression can be written as
1
cot A 1
tan A
(1) sin A cos A + 1 (2) secA cosec A + 1 (3) tan A + cot A (4) sec A + cosec A
Key : 2
Sol: (2)
cos A
sin A
sin A
cos A sin A
1
cos A
1
cos A
sin A
sin A sin A cos A cos A
=
cos A sin A cos A sin A cos A sin A
3 3
sin A cos A
=
sin A cos A (sin A cos A)
1 sin A cos A
=
sin A cos A
= sec A cosec A + 1
90. All the students of a class performed poorly in Mathematics. The teacher decided to give
grace marks of 10 to each of the students. Which of the following statistical measures will not
change even after the grace marks were given ?
(1) mean (2) median (3) mode (4) variance
Key : 4
Sol: (4)
Variance
x
x 2
i
N
After given 10 grade marks to each
x x 10
i
2
i
And x x 10
No. change in variance.
NARAYANA IIT ACADEMY JEE MAIN – 2013

NARAYANA IIT ACADEMY JEE MAIN – 2013