JEE Main 2015 Solutions by Aakash.pdf
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Test Booklet Code
A Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472
Time : 3 hrs.
Answers & Solutions for
M.M. : 360
JEE (MAIN)-2015 (Physics, Chemistry and Mathematics) Important Instructions : 1.
The test is of 3 hours duration.
2.
The Test Booklet consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response.
4.
Candidates will be awarded marks as stated above in Instructions No. 3 for correct response of each question. ¼ (one-fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.
5.
There is only one correct response for each question. Filling up more than one response in each question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 4 above.
6.
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.
7.
No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc. except the Admit Card inside the examination room/hall.
8.
The CODE for this Booklet is
A. Make sure that the CODE printed on Side-2 of the Answer Sheet
and also tally the serial number of the Test Booklet and Answer Sheet are 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.
JEE (MAIN)-2015
PART–A : PHYSICS 1.
2.
Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first?
L g . Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. The accuracy in the determination of g is T 2
(Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/s2) (The figures are schematic and not drawn to scale)
240
The period of oscillation of a simple pendulum is
(y2 – y1) m
(1) 2%
(2) 3%
(3) 1%
(4) 5%
Answer (2) 2 Sol. g 4 .
(1)
t
8
12
t (s)
(y2 – y1) m 240 (2)
240
3.
(y2 – y1) m
(3)
8
240
g l T 100 100 2 100 g l T =
l t 100 2. 100 l t
=
0.1 1 100 2 100 20.0 90
=
100 200 1 20 3% 200 90 2 9
t (s)
12
12
F
l T2
A B
Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is
t (s)
(y2 – y1) m
(4)
(1) 100 N
(2) 80 N
(3) 120 N
(4) 150 N
Answer (3)
8
12
t (s) fs
Answer (3) Sol.
Sol. Till both are in air (From t = 0 to t = 8 sec) x = x2 – x1 = 30t
F
A 20 N
x t When second stone hits ground and first stone is in air x decreases.
B
N
100 N
Clearly fs = 120 N (for vertical equilibrium of the system) 2
JEE (MAIN)-2015
4.
A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction wth speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to (1) 44%
(2) 50%
(3) 56%
(4) 62%
6.
From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is (1)
MR 2 32 2
(2)
MR 2 16 2
(3)
4 MR 2 9 3
(4)
4 MR 2 3 3
Answer (3) Sol. m
2v v
v'
Answer (3)
2mv 2 v' 3m
=
Sol. d 2 R a 3
2m KE loss =
1 1 2 m 2 v 2m v 2 2 2 2
⎛ 2mv 2 ⎞ 1 5 2 3m ⎜⎜ ⎟⎟ mv 2 3 m 3 ⎝ ⎠ 5 mv 2 3 Required % = 100 56% 2mv 2 mv 2 5.
h2 4R
(2)
3h 4
(3)
5h 8
(4)
3h 2 8R
2 R 3
4 3 R M 3 3 M ⎛ 2 ⎞3 2 R⎟ ⎜ ⎝ 3 ⎠
Distance of the centre of mass of a solid uniform cone from its vertex is z0. If the radius of its base is R and its height is h then z0 is equal to (1)
a
M'
I
Answer (2) Sol. dm r 2 .dy.
2M 3
M ' a2 2 M 4 2 1 R 6 6 3 3
4 MR 2 9 3 From a solid sphere of mass M and radius R, a R is removed, as spherical portion of radius 2 shown in the figure. Taking gravitational potential V = 0 at r = , the potential at the centre of the cavity thus formed is (G = gravitational constant) I
y
7.
r
h
R yCM
∫ ydm ∫ ∫ dm
h
0
r 2 dy y 1 2 R h 3
3h 4 3
(1)
GM 2R
(2)
GM R
(3)
2GM 3R
(4)
2GM R
JEE (MAIN)-2015
Answer (2)
Answer (3)
Sol. V = V1 – V2
1⎛U ⎞ 1 4 Sol. P ⎜ ⎟ kT 3⎝V ⎠ 3
2⎤
GM ⎡ V1 3 ⎢ 3R 2 ⎛⎜ R ⎞⎟ ⎥ 2R ⎣ ⎝2⎠ ⎦
PV = RT
M 3G ⎛⎜ ⎞⎟ ⎝ 8 ⎠ V2 R ⎛ 2 ⎜ ⎞⎟ ⎝2⎠
V T–3
GM R A pendulum made of a uniform wire of crosssectional area A has time period T. When an additional mass M is added to its bob, the time period changes to TM. If the Young's modulus of the material of the wire is Y then
R
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat.
1 is equal to Y
⎡⎛ TM ⎞ 2 ⎤ A (1) ⎢⎜ T ⎟ 1⎥ Mg ⎠ ⎢⎣⎝ ⎥⎦
⎡⎛ TM ⎞ 2 ⎤ Mg (2) ⎢⎜ T ⎟ 1⎥ A ⎠ ⎢⎣⎝ ⎥⎦
⎡ ⎛ TM ⎞ 2 ⎤ A (3) ⎢1 ⎜ T ⎟ ⎥ Mg ⎠ ⎥⎦ ⎢⎣ ⎝
⎡ ⎛ T ⎞2 ⎤ A (4) ⎢1 ⎜ T ⎟ ⎥ Mg ⎢⎣ ⎝ M ⎠ ⎥⎦
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat. In both the cases body is brought from initial temperature 100°C to final temperature 200°C. Entropy change of the body in the two cases respectively is
Answer (1)
⇒ 9.
(1) ln 2, 4ln 2
(2) ln 2, ln 2
(3) ln 2, 2ln 2
(4) 2ln 2, 8ln 2
Answer (None)
l Sol. T 2 g
Y
1 T
10. A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways :
(g = gravitational acceleration)
Sol. ds
...(1) l l g
...(2)
Mgl Fl ⇒ l Al AY
...(3)
TM 2
...(ii)
RT 1 4 kT V 3
V 8.
...(i)
dQ dT ms T T
s ∫ ds ms ∫
T 473 dT 1log e 2 log e 373 T T1
11. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as Vq, where V is the volume of the gas. The value of q is
2 ⎤ 1 A ⎡⎛ TM ⎞ 1⎥ ⎢⎜ ⎟ Y Mg ⎢⎣⎝ T ⎠ ⎥⎦
⎛ CP ⎞ ⎜ ⎟ Cv ⎠ ⎝
Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume u
(1)
U T 4 and V
1 2 Answer (3)
(3)
1 U pressure P ⎛⎜ ⎞⎟ . If the shell now undergoes an 3⎝V ⎠
Sol.
adiabatic expansion the relation between T and R is (1) T e–R
(2) T e–3R
(3) T 1 R
(4) T
3 5 6
1 R3
vrms
4
V T
(2)
3 5 6
(4)
1 2
1 3 RT N 2 d 2 ⎛⎜ ⎞⎟ ⎝V ⎠ M
...(i)
...(ii)
JEE (MAIN)-2015
TV – 1 = k
Answer (2) ⎡ v ⎤ 320 ⎡ 320 ⎤ Sol. f 1 f ⎢ v v ⎥ f ⎢ 320 20 ⎥ f 300 Hz ⎣ ⎦ s ⎦ ⎣
1 2
12. For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly?
⎡ v ⎤ 320 f2 f ⎢ Hz ⎥ f v v 340 s ⎦ ⎣ ⎛f ⎞ ⎛ f f1 ⎞ 100 ⎜ 2 1 ⎟ ⎜ 2 ⎟ 100 ⎝ f1 ⎠ ⎝ f1 ⎠
(Graphs are schematic and not drawn to scale)
⎡ 300 ⎤ 100 ⎢ 1⎥ 12% ⎣ 340 ⎦ 14. A long cylindrical shell carries positive surface charge in the upper half and negative surface charge – in the lower half. The electric field lines around the cylinder will look like figure given in
E KE PE d E
(2)
KE E
(figures are schematic and not drawn to scale)
PE
d
++++ + + –– ––
(1)
KE
d (3)
(2)
+ ++ –– ––
–– –– ++ +
(1)
–– ––
V
...(iii)
PE
E (3)
PE (4)
KE
(4)
Answer (2) Sol. KE
++ ++ + + –– –– –– ––
1 m2 ( A2 d 2 ) 2
Answer (1)
1 m2 d 2 2 At d = ± A,
Sol. The field line should resemble that of a dipole.
PE
15. A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ) on its surface. For this sphere the equipotential surfaces 3V0 5V0 3V0 V , , and 0 have radius with potentials 2 4 4 4 R1, R2, R3 and R4 respectively. Then
PE = maximum while KE = 0. 13. A train is moving on a straight track with speed 20 ms–1. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 ms–1) close to (1) 6%
(2) 12%
(3) 18%
(4) 24%
(1) R1 = 0 and R2 > (R4 – R3) (2) R1 0 and (R2 – R1) > (R4 – R3) (3) R1 = 0 and R2 < (R4 – R3) (4) 2R < R4 5
JEE (MAIN)-2015
Answer (3, 4)
Charge
Q Sol. V0 k R
...(i)
VI
kQ 3R 2 r 2 2 R3
Q2 (4)
1 F
3 V V0 R1 = 0 2
R2
C
Answer (2)
2 2 5 kQ kQ 3R 3 r 4 R 2R
3 F
Sol. C aq =
3C 3C
...(i)
⎛ 3C ⎞ ⎟E Total charges q ⎜ ⎝ 3C ⎠
R 2
...(ii)
3 kQ kQ 3 4 R R
Charge upon capacitor 2 F,
4R 3 1 kQ kQ 4 R R4
q'
R3
R4 = 4R R4 > 2R
2 3CE 2CE 2E 3 (3 C ) 3 C 1 3 C
Now,
16. In the given circuit, charge Q2 on the 2 F capacitor changes as C is varied from 1 F to 3 F. Q2 as a function of C is given properly by : (Figures are drawn schematically and are not to scale)
dQ dQ 2 0, 2 0 dC dC
17. When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10–4 ms–1. If the electron density in the wire is 8 × 1028 m–3, the resistivity of the material is close to
1 F C
(1) 1.6 × 10–8 m
(2) 1.6 × 10–7 m
(3) 1.6 × 10–6 m
(4) 1.6 × 10–5 m
Answer (4)
2 F
Sol. V IR I
Charge
⇒
E
VA VA V Il ln eA d l n e d
5 0.1 2.5 10 19 1.6 10 19 8 10 28 = 1.6 × 10–5 m ⇒
Q2 (1)
1 F
l A
3 F
18. In the circuit shown, the current in the 1 resistor is
C
6V
P
Charge Q2
2
1
(2)
1 F
3 F
C
3
3 Q
Charge
(1) 1.3 A, from P to Q
Q2
(2) 0 A
(3)
(3) 0.13 A, from Q to P
1 F
3 F
C
(4) 0.13 A, from P to Q 6
9V
JEE (MAIN)-2015
Answer (3)
Answer (2)
Sol. From KVL,
9 = 6I1 – I2 …(1)
T
6 = 4I2 – I1 …(2) Solving, I1 – I2 = –0.13A
6V
I2
Sol.
2
P
I1
(l)g
I1–I2
3
F
Tcos = gl
9V
1
Tsin =
Q
3 19. Two coaxial solenoids of different radii carry current I in the same direction. Let F1 be the magnetic force on the inner solenoid due to the outer one and F2 be the magnetic force on the outer solenoid due to the inner one. Then (1) F1 = F2 = 0
…(1)
0 I Il . 2 2 L sin
…
gL 0 cos
⇒ I 2 sin
21. A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orientations as shown in the figures below:
z
(2) F1 is radially inwards and F2 is radially outwards (3) F1 is radially inwards and F2 = 0
I
B
I
(a)
y
I
x
(4) F1 is radially outwards and F2 = 0
I
z
Answer (1)
B
Sol. Net force on each of them would be zero.
(b)
20. Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle with the vertical. If wires have mass per unit length then the value of I is
I
I z
x
I
(g = gravitational acceleration)
I
I
I
I
(c)
x
y
B y
I
z
L
B
gL (1) sin cos 0
(d)
I
I
x
gL (2) 2 sin cos 0
I
I I
I
y
If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the loop
(3) 2
gL tan 0
(4)
gL tan 0
would be in (i) stable equilibrium and (ii) unstable equilibrium? 7
JEE (MAIN)-2015
23. A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is
(1) (a) and (b), respectively (2) (a) and (c), respectively (3) (b) and (d), respectively
(1) 1.73 V/m
(2) 2.45 V/m
(3) 5.48 V/m
(4) 7.75 V/m
Answer (2)
(4) (b) and (c), respectively Answer (3)
Sol. I
Stable equilibrium M||B
U av
z
⇒
B I
I
y
I
I
x
B
I
x
...(2)
P 1 0 E02 c 4 r 2 2
2P 2.45 V/m 4 r 2 0 c
24. Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is , a ray, incident at an angle , on the face AB would get transmitted through the face AC of the prism provided.
z
I
A
y
I
22. An inductor (L = 0.03 H) and a resistor (R = 0.15 k) are connected in series to a battery of 15 V EMF in a circuit shown below. The key K1 has been kept closed for a long time. Then at t = 0, K1 is opened and key K 2 is closed simultaneously. At t = 1 ms, the current in the circuit will be ( e 5 150)
0.03 H
B
⎡ ⎤ ⎛ 1 1 ⎛ 1 ⎞ ⎞ (2) sin ⎢ sin ⎜ A sin ⎜ ⎟ ⎟ ⎥ ⎝ ⎠ ⎠ ⎦⎥ ⎝ ⎣⎢
0.15 k
⎡ ⎤ ⎛ 1 1 ⎛ 1 ⎞ ⎞ (3) cos ⎢ sin ⎜ A sin ⎜ ⎟ ⎟ ⎥ ⎝ ⎠ ⎠ ⎦⎥ ⎝ ⎣⎢
⎡ ⎤ ⎛ 1 1 ⎛ 1 ⎞ ⎞ (4) cos ⎢ sin ⎜ A sin ⎜ ⎟ ⎟ ⎥ ⎢⎣ ⎝ ⎠ ⎠ ⎥⎦ ⎝
K1
15 V (1) 100 mA
(2) 67 mA
(3) 6.7 mA
(4) 0.67 mA
Answer (1)
Answer (4)
t
C
⎡ ⎤ ⎛ 1 1 ⎛ 1 ⎞ ⎞ (1) sin ⎢ sin ⎜ A sin ⎜ ⎟ ⎟ ⎥ ⎢⎣ ⎝ ⎠ ⎠ ⎥⎦ ⎝
K2
Sol. I I 0 e ,
...(1)
1 0 E02 2
⇒ E0
Unstable equilibrium M|| B
I
P U av c 4 r 2
L R
Sol.
r 1
r2
1103
15 1/5103 e 0.67 mA 150
sin = sin r1 8
JEE (MAIN)-2015
26. Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is
sin sin r1 = 1 ⎛ sin ⎞ r1 = sin ⎜ ⎟ ⎝ ⎠
⎛ sin ⎞ r2 = A – sin 1 ⎜ ⎟ ⎝ ⎠
(2) 30 m
(3) 100 m
(4) 300 m
Answer (2)
Sol. RP
1 ⎛ 1 ⎞ r2 < sin ⎜ ⎟ ⎝⎠
1.22 (500 10 9 m) 1.22 2 sin ⎛ 1 ⎞ 2 1 ⎜ ⎟ ⎝ 100 ⎠
0.25 cm
⎛ sin ⎞ 1 ⎛ 1 ⎞ A sin 1 ⎜ ⎟ sin ⎜ ⎟ ⎝ ⎠ ⎝⎠
(1) 1 m
25 cm
= 3.05 × 10–5 m
⎛1⎞ ⎛ sin ⎞ A sin 1 ⎜ ⎟ sin 1 ⎜ ⎟ ⎝⎠ ⎝ ⎠
= 30 m 27. As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ion
⎛ sin 1 ⎛ 1 ⎞ ⎞ sin ⎜ A sin ⎜ ⎟ ⎟ ⎝ ⎠⎠ ⎝
(1) Its kinetic energy increases but potential energy and total energy decrease
⎛ ⎞ ⎛ 1 ⎛ 1 ⎞ ⎞ ⎜⎜ sin ⎜ A sin ⎜ ⎟ ⎟ ⎟⎟ sin ⎝ ⎠⎠⎠ ⎝ ⎝
(2) Kinetic energy, potential energy and total energy decrease (3) Kinetic energy decreases, potential energy increases but total energy remains same
⎛ ⎞ ⎛ 1 1 ⎛ 1 ⎞ ⎞ sin ⎜⎜ sin ⎜ A sin ⎜ ⎟ ⎟ ⎟⎟ ⎝ ⎠⎠⎠ ⎝ ⎝
(4) Kinetic energy and total energy decrease but potential energy increases
25. On a hot summer night, the refractive index of air is smallest near the ground and increases with height form the ground. When a light beam is directed horizontally, the Huygen's principle leads us to conclude that as it travels, the light beam
Answer (1) Sol. PE 27.2
(1) Becomes narrower
TE
(2) Goes horizontally without any deflection (3) Bends downwards (4) Bends upwards Answer (4) Sol. Consider a plane wavefront travelling horizontally. As it moves, its different parts move with different speeds. So, its shape will change as shown
13.6 z 2 eV n2
KE
13.6 z 2 eV n2
KE
13.6 eV , As n decreases, KE n2
Light bends upward
9
z2 eV n2
PE
27.2 eV , as n decreases, PE n2
TE
13.6 eV , as n decreases, TE n2
JEE (MAIN)-2015
R
28. Match List-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the choices given below the list:
List -I (A) Franck-Hertz experiment (B) Photo-electric experiment (C) Davison-Germer experiment
List-II Particle nature of light (ii) Discrete energy levels of atom (iii) Wave nature of electron (iv) Structure of atom
L
(i)
(1) (A) - (i)
(B) - (iv)
(C) - (iii)
(2) (A) - (ii)
(B) - (iv)
(C) - (iii)
(3) (A) - (ii)
(B) - (i)
(C) - (iii)
(4) (A) - (iv)
(B) - (iii)
(C) - (ii)
C If a student plots graphs of the square of maximum 2 charge QMax on the capacitor with time (t) for two
different values L1 and L2 (L1 > L2) of L then which of the following represents this graph correctly? (Plots are schematic and not drawn to scale) 2
QMax
L1
(1)
L2
t
Answer (3) Sol. Franck-Hertz exp.– Discrete energy level. 2
QMax
Photo-electric effect– Particle nature of light
L2
(2)
Davison-Germer exp.– Diffraction of electron beam.
L1
t
29. A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are
2
QMax
L1
(1) 2 MHz only (3)
(2) 2005 kHz and 1995 kHz
L2
t
(3) 2005 kHz, 2000 kHz and 1995 kHz (4) 2000 kHz and 1995 kHz 2
QMax
Answer (3)
Q0 (For both L1 and L2)
(4)
Sol. Frequencies of resultant signal are
t
fe + fm, fe and fe – fm (2000 + 5) kHz, 2000 kHz, (2000 – 5) kHz,
Answer (1)
2005 kHz, 2000 kHz, 1995 kHz
Sol. For a damped pendulum, A = A0e–bt/2m
30. An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is
charged to Q0 and then connected to the L and R as shown below :
A A0 e
R⎞ ⎛⎜ ⎟t ⎝ 2L ⎠
(Since L plays the same role as m) 10
JEE (MAIN)-2015
PART–B : CHEMISTRY 31. The molecular formula of a commercial resin used for exchanging ions in water softening is C8H 7SO 3Na (mol. wt. 206). What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin? (1)
1 103
(2)
2 (3) 309
4 =2
34. The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is
1 206
1 (4) 412
(1) Ion-ion interaction
(2) Ion-dipole interaction
(3) London force
(4) Hydrogen bond
Answer (4) Sol. H-bond is one of the dipole-dipole interaction and dependent on inverse cube of distance between the molecules.
Answer (4) Sol. Ca+2 + 2C8H7SO3–Na+ Ca(C8H7SO3–)2 + 2Na+ 1 mol
–13.6 = –3.4
=
2 mol
35. The following reaction is performed at 298 K.
1 1 mol/g The maximum uptake = 206 2 412
2NO (g) 2NO(g) + O2(g) 2
The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO2(g) at 298 K?
32. Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately
(Kp = 1.6 × 1012)
(1) 1.86 Å
(2) 3.22 Å
(1) R(298) ln(1.6 × 1012) – 86600
(3) 5.72 Å
(4) 0.93 Å
(2) 86600 + R(298) ln(1.6 × 1012)
Answer (1)
(3) 86600
Sol. Edge length of BCC is 4.29 Å. In BCC, edge length =
4.29
4 3
4 3
r
Answer (4) 2NO (g) Sol. 2NO(g) + O2(g) 2
r
G
4.29 3 1.86 Å 4
reaction
⎡⎢ G ⎣
⎡⎢ G ⎣
33. Which of the following is the energy of a possible excited state of hydrogen?
⎤
formation ⎥ ⎦ product
⎤
⎥ reactant formation ⎦
⇒ RT ln K P 2 G
(1) +13.6 eV (2) –6.8 eV
⇒ G
(3) –3.4 eV (4) +6.8 eV
⇒ G
Answer (3) Sol. Energy of excited state is negative and correspond to n > 1. n=
R 298
12 (4) 0.5 ⎡⎣ 2 86, 600 R 298 ln1.6 10 ⎤⎦
r
ln 1.6 10 12
–13.6
NO 2
NO 2
2 G
NO
NO 2
2 G
NO
RT ln K P
2 86600 R 298 ln K P 2
2 86600 R 298 ln1.6 10 12 2
0.5 ⎡⎣ 2 86, 600 R 298 ln 1.6 1012 ⎤⎦
E excited state
11
JEE (MAIN)-2015
36. The vapour pressure of acetone at 20°C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20°C, its vapour pressure was 183 torr. The molar mass (g mol–1) of the substance is
Now Q C
(2) 64
38. Two faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is (at. mass of Cu = 63.5 amu)
(3) 128 (4) 488 Answer (2)
(1) 0 g
Sol. Vapour pressure of pure acetone P°A 185 torr
(2) 63.5 g (3) 2 g
Vapour pressure of solution, PS = 183 torr
(4) 127 g
Molar mass of solvent, MA = 58 g/mole
Answer (2)
P°A PS n B as we know PS nA
Sol. Cu+2 + 2e Cu So, 2 F charge deposite 1 mol of Cu. Mass deposited = 63.5 g.
185 183 WB M A ⇒ 183 M B WA
39. Higher order (>3) reactions are rare due to
2 1.2 58 183 M B 100
⇒ MB
A 2
⎛1⎞ ⎜ ⎟ 2 2 ⎝ ⎠ 2 4 ⎛1⎞ ⎜ ⎟ ⎝2⎠
as QC > KC, hence reaction will shift in backward direction.
(1) 32
⇒
C B
(1) Low probability of simultaneous collision of all the reacting species (2) Increase in entropy and activation energy as more molecules are involved
1.2 58 183 2 100
(3) Shifting of equilibrium towards reactants due to elastic collisions
63.68 g/mole
(4) Loss of active species on collision
37. The standard Gibbs energy change at 300 K for the
Answer (1)
reaction 2A B + C is 2494.2 J. At a given time,
the composition of the reaction mixture
Sol. Higher order greater than 3 for reaction is rare because there is low probability of simultaneous collision of all the reacting species.
is
1 1 , B 2 and C . The reaction proceeds 2 2 in the : [R = 8.314 J/K/mol, e = 2.718]
A
40. 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is
(1) Forward direction because Q > KC (2) Reverse direction because Q > KC
(1) 18 mg
(3) Forward direction because Q < KC
(2) 36 mg
(4) Reverse direction because Q < KC
(3) 42 mg
Answer (2)
(4) 54 mg
Sol. 2A B + C, G° = 2494.2 J
Answer (1) Sol. Number of moles of acetic acid adsorbed
As we know G° = –2.303 RT logKC
50 50 ⎞ ⎛ ⎜ 0.06 0.042 ⎟ 1000 1000 ⎠ ⎝
2494.2 = –2.303 × 8.314 × 300 log KC –0.434 = log KC
0.9 moles 1000 Weight of acetic acid adsorbed
KC = anti log (–0.434) KC = 0.367 Now A
= 0.9 × 60 mg = 54 mg
1 1 , B 2 and C 2 2
Hence, the amount of acetic acid adsorbed per g of 12
JEE (MAIN)-2015
charcoal =
54 mg 3
Answer (2) Sol. BeSO4 has hydration energy greater than its lattice energy.
= 18 mg
45. Which among the following is the most reactive?
Hence, option (1) is correct.
(1) Cl2
41. The ionic radii (in Å) of N 3– , O 2– and F – are respectively
(2) Br2
(1) 1.36, 1.40 and 1.71
(3) I2
(2) 1.36, 1.71 and 1.40
(4) ICl
(3) 1.71, 1.40 and 1.36
Answer (4)
(4) 1.71, 1.36 and 1.40
Sol. Because of polarity and weak bond interhalogen compounds are more reactive.
Answer (3)
46. Match the catalysts to the correct processes :
Sol. Radius of N3–, O2– and F– follow order
Catalyst
N3– > O2– > F– As per inequality only option (3) is correct
a.
TiCl3
(i) Wacker process
b.
PdCl2
(ii) Ziegler-Natta polymerization
c.
CuCl2
(iii) Contact process
that is 1.71 Å, 1.40 Å and 1.36 Å 42. In the context of the Hall-Heroult process for the extraction of Al, which of the following statement is false?
d. V2O5 (2) a(ii), b(i), c(iv), d(iii)
(2) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity (3)
(iv) Deacon's process
(1) a(iii), b(ii), c(iv), d(i)
(1) CO and CO2 are produced in this process
Al 3+
Process
(3) a(ii), b(iii), c(iv), d(i) (4) a(iii), b(i), c(ii), d(iv)
is reduced at the cathode to form Al
Answer (2)
(4) Na3AlF6 serves as the electrolyte
Sol. TiCl3
-
Ziegler-Natta polymerisation
Answer (4)
V2O5
-
Contact process
Sol. In Hall-Heroult process Al2O3 (molten) is electrolyte.
PdCl2
-
Wacker process
43. From the following statement regarding H 2O 2, choose the incorrect statement
CuCl2 -
Deacon's process
47. Which one has the highest boiling point?
(1) It can act only as an oxidizing agent
(1) He
(2) It decomposes on exposure to light
(2) Ne
(3) It has to be stored in plastic or wax lined glass bottles in dark.
(3) Kr (4) Xe
(4) It has to be kept away from dust
Answer (4)
Answer (1) Sol. H2O2 can be reduced or oxidised. Hence, it can act as reducing as well as oxidising agent.
Sol. Down the group strength of van der Waal's force of attraction increases hence Xe have highest boiling point.
44. Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy?
48. The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH 3 )(NH 2 OH)] + is (py = pyridine)
(1) CaSO4
(1) 2
(2) BeSO4
(2) 3
(3) BaSO4
(3) 4
(4) SrSO4
(4) 6 13
JEE (MAIN)-2015
52. Which of the following compounds will exhibit geometrical isomerism?
Answer (2)
a Sol.
d
Pt
b
a
c
d
Pt
c
a
b
c
Pt
b
(1) 1 - Phenyl - 2 - butene (2) 3 - Phenyl - 1 - butene
d
(3) 2 - Phenyl - 1 - butene
as per question a = Cl, b = py, c = NH 3 and d = NH2OH are assumed.
(4) 1, 1 - Diphenyl - 1 propane Answer (1)
49. The color of KMnO4 is due to (1) M L charge transfer transition
Sol. For geometrical isomerism doubly bonded carbon must be bonded to two different groups which is only satisfied by 1 - Phenyl - 2 - butene.
(2) d - d transition (3) L M charge transfer transition
H
(4) - * transition
Ph – CH2
Answer (3)
50. Assertion : Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen.
CH3 (2)
CH3
CH3
(3)
(4)
H3C
CH3 Answer (2)
(4) Both the assertion and reason are incorrect
Sol. 5-keto-2-methylhexanal is
Answer (1)
O
Sol. N2 + O2 2NO
O H
Required temperature for above reaction is around 3000°C which is a quite high temperature. This reaction is observed during thunderstorm.
O3
Zn
O
+
H
51. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is (At. mass Ag = 108; Br = 80)
O H
54. The synthesis of alkyl fluorides is best accomplished by (1) Free radical fluorination
Answer (1)
(2) Sandmeyer's reaction
Sol. Percentage of Br
(3) Finkelstein reaction (4) Swarts reaction
Weight of AgBr Mol. mass of Br = Mol. mass of AgBr Weight of O.C. 0
Answer (4) Sol. Swart's reaction
=
H
CH3
(3) The assertion is incorrect, but the reason is correct
(4) 60
CH3
CH3
(2) Both assertion and reason are correct, but the reason is not the correct explanation for the assertion
(3) 48
Ph – CH2
(1)
(1) Both assertion and reason are correct, and the reason is the correct explanation for the assertion
(2) 36
C=C
CH3
The reaction between nitrogen and oxygen requires high temperature.
(1) 24
CH3
H
cis trans 53. Which compound would give 5-keto-2-methyl hexanal upon ozonolysis?
Sol. Charge transfer spectra from ligand (L) to metal (M) is responsible for color of KMnO4.
Reason :
H C=C
141 80 100 = 24% 188 250
CH 3 Cl AgF CH 3 F AgCl
14
JEE (MAIN)-2015
55. In the following sequence of reactions : KMnO
SOCl
57. Which polymer is used in the manufacture of paints
H /Pd
4 2 2 Toluene A B C, BaSO
and lacquers?
4
the product C is
(1) Bakelite
(1) C6H5COOH
(2) Glyptal
(2) C6H5CH3
(3) Polypropene
(3) C6H5CH2OH
(4) Poly vinyl chloride
(4) C6H5CHO
Answer (2)
Answer (4) CH3
COOH KMnO4
Sol.
COCl
SOCl2
Sol. Glyptal is used in manufacture of paints and
CHO
lacquires.
H2/Pd BaSO4
(A)
(B)
58. Which of the vitamins given below is water soluble?
(C)
56. In the reaction
(1) Vitamin C
NH2
(2) Vitamin D NaNO2/HCl 0-5°C
D
CuCN/KCN
(3) Vitamin E
E + N2 ,
(4) Vitamin K
CH3
Answer (1)
the product E is
Sol. Vitamin C is water soluble vitamin.
COOH
59. Which of the following compounds is not an antacid?
(1)
(1) Aluminium Hydroxide
CH3
(2) Cimetidine (2) H3C
CH3
(3) Phenelzine (4) Ranitidine
CN
Answer (3) (3)
Sol. Phenelzine is not antacid, it is anti-depressant.
CH3
60. Which of the following compounds is not colored yellow?
CH3
(1) Zn2[Fe(CN)6] (4)
(2) K3[Co(NO2)6] (3) (NH4)3[As (Mo3O10)4]
Answer (3) +
NH2 Sol.
CH3
–
N2Cl
CN
NaNO2/HCl
CuCN/KCN
0°C - 5°C
CH3 (D)
(4) BaCrO4 Answer (1)
+ N2
Sol. (NH4)3[As (Mo3O10)4], BaCrO4 and K3[Co(NO2)6] are
CH3
yellow colored compounds but Zn2[Fe(CN)6] is not
(E)
yellow colored compound. 15
JEE (MAIN)-2015
PART–C : MATHEMATICS 61. Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set A × B, each having at least three elements is (1) 219
(2) 256
(3) 275
(4) 510
63. Let and be the roots of equation x2 – 6x – 2 = 0. a10 – 2 a8 If an = n – n, for n 1, then the value of 2 a9 is equal to (1) 6 (2) –6 (3) 3
Answer (1)
Answer (3)
Sol. n(A) = 4, n(B) = 2
Sol. From equation,
(4) –3
+=6
n(A × B) = 8
= –2
Required numbers = 8C3 + 8C4 + ...... + 8C8
10 10 8 8 a10 – 2 a8 ( ) The value of 2( 9 9 ) 2 a9
= 28 – (8C0 + 8C1 + 8C2) = 256 – 37
= 219
62. A complex number z is said to be unimodular if |z| = 1. Suppose z1 and z2 are complex numbers such that
z1 2 z2 is unimodular and z 2 is not 2 z1 z2
6 3 2 2
equation AAT = 9I, where I is 3 × 3 identity matrix, then the ordered pair (a, b) is equal to
(1) Straight line parallel to x-axis (2) Straight line parallel to y-axis (3) Circle of radius 2
2
(1) (2, –1)
(2) (–2, 1)
(3) (2, 1)
(4) (–2, –1)
Answer (4)
Answer (3)
2 ⎤ ⎡ 1 2 a⎤ ⎡9 0 0⎤ ⎡1 2 ⎢ 2 1 2 ⎥ ⎢ 2 1 2 ⎥⎥ ⎢⎢0 9 0 ⎥⎥ ⎥⎢ Sol. ⎢ ⎢⎣ a 2 b ⎥⎦ ⎢⎣ 2 2 b ⎥⎦ ⎢⎣0 0 9 ⎥⎦
⎛ z1 2 z2 ⎞ Sol. ⎜ 2 z z ⎟ 1 1 2 ⎠ ⎝
⎛ z1 2 z2 ⎞ ⎛ z1 2 z2 ⎜ ⎟⎜ ⎝ 2 z1 z2 ⎠ ⎝ 2 z1 z2
2(9 9 )
⎡1 2 2 ⎤ 64. If A = ⎢⎢ 2 1 2 ⎥⎥ is a matrix satisfying the ⎢⎣ a 2 b ⎥⎦
unimodular. Then the point z1 lies on a
(4) Circle of radius
9 ( ) 9 ( )
a 4 2b 0
⎞ ⎟1 ⎠
2 a 2 2b 0
z1 z1 2 z1 z2 2 z2 z1 4 z2 z2
a1b 0
4 2 z1 z2 2 z1 z2 z1 z1 z2 z2
2 a 2 b 2 a 2 b 4
z1 z1 4 z2 z2 4 z1 z1 z2 z2
3 a 6
zz1 1 z2 z2 4 1 z2 z2 0
a 2
z1 z1 4 1 z2 z2 0
2 1 b 0
b=–1
z1 z1 4
a=–2
|z| = 2 i.e. z lies on circle of radius 2.
(–2, –1) 16
JEE (MAIN)-2015
65. The set of all values of for which the system of linear equations
5 digit numbers
2x1 – 2x2 + x3 = x1 2x1 – 3x2 + 2x3 = x2
5
–x1 + 2x2 = x3
5 × 4 × 3 × 2 × 1 = 120
has a non-trivial solution
Total number of integers = 72 + 120 = 192
(1) Is an empty set
67. The sum of coefficients of integral powers of x in
(2) Is a singleton (3) Contains two elements (4) Contains more than two elements Answer (3) Sol. x1 (2 ) 2 x2 x3 0 2 x1 x2 ( 3) 2 x3 0 x1 2 x2 x3 0
2 1 2 2 3 2 0 2 1
3
Sol.
1 50 (3 1) 2
(2)
1 50 (3 ) 2
(3)
1 50 (3 1) 2
(4)
1 50 (2 1) 2
1 2 x
50
50
C0
50
C1 2 x
50 C 2 (2 x )2 .....
50
C0 2 0 50 C 2 2 2 50 C 4 2 4 50 C 50 2 50
We know that 50
(1 + 2)50 =
3 2 2 2 2 3 3 0
C0 50 C1 2 ..... 50 C 50 2 50
Then,
2 ( 1) 2 ( 1) 3( 1) 0
350 1 2 68. If m is the A.M. of two distinct real numbers l and n (l, n > 1) and G1, G2 and G3 are three geometric
( 1)( 2 2 3) 0
50
( 1)( 3)( 1) 0
1, 1, 3
C0 50 C 2 2 2 ..... 50 C 50 2 50
means between l and n, then G14 2G24 G34 equals.
Two elements. 66. The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is
(1) 4 l2mn
(2) 4 lm2n
(3) 4 lmn2
(4) 4 l2m2n2
Answer (2)
(1) 216 (2) 192
Sol.
(3) 120 (4) 72
ln m 2 l + n = 2m
…(i)
1
⎛ n⎞ 4 G1 l ⎜ ⎟ ⎝ l⎠
Answer (2) Sol. 4 digit numbers
6,
7,
2
8
678
⎛ n⎞ 4 G2 l ⎜ ⎟ ⎝ l⎠
3
⎛ n⎞ 4 G3 l ⎜ ⎟ ⎝ l⎠
3
4
1
Sum of coefficient of integral power of x
3 2 5 3 0
5,
is
(1)
2
3 2 5 3 0
3,
50
50 C 50 ( 2 x )50
2 6 8 3 4 4 4 1 0
Answer (1)
(2 )( 2 3 4) 2 ( 2 2) (4 3) 0 2
the binomial expansion of 1 2 x
5
2
= 72 17
JEE (MAIN)-2015
71. If the function.
Now G14 2G24 G33 2
l4
n ⎛ n⎞ ⎛ n⎞ 2 (l 2 ) ⎜ ⎟ l 4 ⎜ ⎟ ⎝ l⎠ ⎝ l⎠ l
⎪⎧ k x 1 g( x ) ⎨ ⎪⎩ mx 2
3
= nl3 + 2n2l2 + n3l
, 0x3 , 3x5
is differentiable, the value of k + m is
= 2n2l2 + nl(n2 + l2) = 2n2l2 + nl((n + l)2 – 2nl)
(1) 2
= nl(n + l)2 = nl (2m)2
(3)
=4
nlm2
3
3
3
3
(1) 71
(2) 96
(3) 142
(4) 192
lim h0
g(3 h ) g(3) h
= lim h0
⎡ n n 1 ⎤ ⎢ ⎥ 2 ⎣ ⎦ 2 n
2
lim h0
m(3 h ) 2 2 k h
(3m 2 k ) mh 2 m h
and 3m – 2k + 2 = 0 L.H.D.
n 1 2 4
lim
k (3 h ) 1 2 k h
h0
1 ⎡⎣n2 2n 1⎤⎦ 4
lim h0
1 ⎡ n n 1 2n 1 2 n n 1 ⎤ 1⎥ ⎢ 4⎣ 6 2 ⎦
h0
1 cos 2 x 3 cos x x tan 4x
x 0
h( 4 h 2)
k 4
k m and 3m – 2k + 2 = 0 4 m
is equal to
2 8 and k 5 5
km
(1) 4
8 2 10 2 5 5 5
Alternative Answer
(2) 3
⎪⎧ k x 1 g( x ) ⎨ ⎪⎩ mx 2
(3) 2 (4)
4h4
From above,
= 96
lim
k[ 4 h 2] h
lim k
1 ⎡ 9 10 19 ⎤ ⎢ 9 10 9 ⎥ 4⎣ 6 ⎦
70.
, 0x3 , 3x5
R.H.D.
Answer (2)
tn
(4) 4
⎧⎪ k x 1 Sol. g( x ) ⎨ ⎪⎩ mx 2
3
1 1 2 1 2 3 ........ is 1 1 3 1 3 5
Sol.
10 3
16 5
Answer (1)
69. The sum of first 9 terms of the series 3
(2)
1 2
, 0x3 , 3x5
g is constant at x = 3
Answer (3)
k 4 3m 2
2 sin 2 x 3 cos x x 2 =2 Sol. lim tan 4 x x0 x x2 4x 4x
2k = 3m + 2 k ⎞ ⎛ m Also ⎜⎝ ⎟ 2 x 1 ⎠ x3
18
…(i)
JEE (MAIN)-2015
Now, f (x) = 4x + 3a3x2 + 4a4x3
k m 4
= x[4 + 3a3x + 4a4x2]
k=4m
Given, f (1) = 0 and f (2) = 0
…(ii)
8m=3m+2
2 8 m ,k 5 5 2 8 m k 2 5 5 72. The normal to the curve, x2 + 2xy – 3y2 = 0 at (1,1)
3a3 + 4a4 + 4 = 0
…(i)
and 6a3 + 16a4 + 4 = 0
1 , a = –2 2 3
Solving, a4
2 3 i.e., f ( x ) 2 x – 2 x
(1) Does not meet the curve again
…(ii)
1 4 x 2
i.e., f (2) 0
(2) Meets the curve again in the second quadrant (3) Meets the curve again in the third quadrant
74. The integral
(4) Meets the curve again in the fourth quadrant Answer (4)
∫x
2
dx equals ( x 4 1)3/4
1
⎛ x4 1⎞ 4 (1) ⎜ 4 ⎟ c ⎝ x ⎠
Sol. Curve is x2 + 2xy – 3y2 = 0 dy ⎡ dy ⎤ y ⎥ 6y 0 Differentiate wr.t. x, 2 x 2 ⎢ x dx ⎣ dx ⎦
1
(2) ( x 4 1) 4 c 1
⎛ x4 1⎞ 4 (4) ⎜ 4 ⎟ c ⎝ x ⎠
1 4
(3) ( x 1) c 4
⎛ dy ⎞ 1 ⎜ ⎟ ⎝ dx ⎠ (1, 1)
Answer (4)
So equation of normal at (1, 1) is Sol. I ∫
y – 1 = – 1 (x – 1) y=2–x
dx x 2 ( x 4 1)3/4 ∫
Solving it with the curve, we get x2 + 2x(2 – x) – 3(2 – x)2 = 0 –4x2 + 16x – 12 = 0 x2 – 4x + 3 = 0 x = 1, 3
4 t ⇒ 5 dx dt 4 x x
So, I
1 dt 1 3/4 t dt 3/4 ∫ 4 t 4 ∫
73. Let f(x) be a polynomial of degree four having extreme
(1) –8
(2) –4
(3) 0
(4) 4
1⎞ ⎛ x5 ⎜ 1 4 ⎟ ⎝ x ⎠
Let 1
So points of intersections are (1, 1) & (3, –1) i.e. normal cuts the curve again in fourth quadrant. f (x) ⎤ ⎡ values at x = 1 and x = 2. If lim ⎢ 1 ⎥ 3 , then x 0 ⎣ x2 ⎦ f(2) is equal to
dx 3/4
1 ⎛ t 1/4 ⎞ c 4 ⎜⎝ 1 / 4 ⎟⎠
1 ⎞ ⎛ = ⎜1 4 ⎟ x ⎠ ⎝
1/4
c
So, option (4). 4
75. The integral
Answer (3) Sol. Let f(x) = a0 + a1x + a2x2 + a3x3 + a4x4
to
f (x) ⎤ ⎡ Using lim ⎢ 1 2 ⎥ 3 x 0 ⎣ x ⎦ f (x) lim 2 2 x 0 x
∫
2 log x
log x 2 2
log(36 – 12 x x 2 )
(1) 2
(2) 4
(3) 1
(4) 6
Answer (3) 4
log x 2 dx
Sol. I ∫
a a x a2 x 2 a3 x 3 a4 x 4 lim 0 1 2 x 0 x2 So, a0 = 0, a1 = 0, a2 = 2 i.e., f(x) = 2x2 + a3x3 + a4x4
2 log x
4
I∫
log(36 – 12 x x 2 )
log(6 – x )2 dx
2 log x
19
2
2
log(6 – x )2
dx is equal
JEE (MAIN)-2015 4
Answer (3)*
2
Sol. It is best option. Theoretically question is wrong, because initial condition is not given.
2 I ∫ 1 dx 2I = 2 I=1 76. The area (in sq. units) of the region described by {(x, y) : y2 2x and y 4x – 1} is 7 32
(1)
dy + y = 2x logx dx
x log x
(2)
15 (3) 64
dy y 2 dx x log x
5 64
1
I.F. e
9 (4) 32
Sol.
1
∫ x log x dx
e log log x log x
Solution is y log x ∫ 2 log x dx c
Answer (4)
y=1
If x = 1 then y = 0
y log x 2( x log x x ) c
y=1
x = 1, y = 0 Then, c = 2, y(e) = 2 78. The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0), is
1 2
After solving y = 4x – 1 and y2 = 2x
y 4
y 1 2
–y–1=0
y
1 18 1 3 4 4 1
(3) 820
(4) 780
y 1, 1
1
(0, 41)
Sol.
A39
1 2
2
y ⎛ y1⎞ ∫ ⎜⎝ 4 ⎟⎠ dy ∫ 2 dy 1/2 1/2
A2
⎤ 1 ⎡ y2 1 ⎡y ⎢ ⎥ ⎢ y⎥ 4 ⎢⎣ 2 ⎥⎦ 1/2 2 ⎢⎣ 3 ⎥⎦ 1/2
1 ⎡ 4 8 1 4 ⎤ 1 ⎡8 1⎤ ⎥ 2 ⎢ 24 ⎥ 4 ⎢⎣ 8 ⎦ ⎣ ⎦
1 ⎡ 15 ⎤ 9 4 ⎢⎣ 8 ⎥⎦ 48
=
15 6 9 32 32 32
B2 B1
A1
3 ⎤1
(41, 0)
(0, 0)
Total number of integral coordinates as required = 39 + 38 + 37 + ....... + 1
39 40 780 2
79. Locus of the image of the point (2, 3) in the line (2x – 3y + 4) + k(x – 2y + 3) = 0, k R, is a
77. Let y(x) be the solution of the differential equation
x log x
(2) 861
Answer (4)
2
2y2
A
(1) 901
(1) Straight line parallel to x-axis
dy y 2 x log x , ( x 1). dx
(2) Straight line parallel to y-axis
Then y(e) is equal to (1) e
(2) 0
(3) 2
(4) 2e 20
(3) Circle of radius
2
(4) Circle of radius
3
JEE (MAIN)-2015
82. Let O be the vertex and Q be any point on the parabola, x2 = 8y. If the point P divides the line segment OQ internally in the ratio 1 : 3, then the locus of P is
Answer (3) Sol. After solving equation (i) & (ii) 2x – 3y + 4 = 0
...(i)
2x – 4y + 6 = 0
...(ii)
x = 1 and y = 2 Slope of AB × Slope of MN = – 1 b3 2 b3 2 1 a2 a2 1 2
A
(2, 3)
(2) y 2 x
(3) y 2 2 x
(4) x 2 2 y
Answer (4)
⎛ a 2 b 3⎞ M⎜ , ⎟ ⎝ 2 2 ⎠
N (1, 2)
(1) x 2 y
Sol. x2 = 8y Let Q be (4t, 2t2)
y
(y – 3)(y – 1) = –(x – 2)x (a, b) B (Image of A)
y2 – 4y + 3 = –x2 + 2x
x2 + y2 – 2x – 4y + 3 = 0 Circle of radius =
2
1
Let P be (h, k)
2
80. The number of common tangents to the circles x2 + y2 – 4x – 6y – 12 = 0 and x2 + y2 + 6x + 18y + 26 = 0, is (1) 1
(2) 2
(3) 3
(4) 4
h = t, k
P
3
Q(4t, 2t )
O
x
t2 2
2k h 2
Locus of (h, k) is x2 = 2y. 83. The distance of the point (1, 0, 2) from the point of
Answer (3) Sol. x2 + y2 – 4x – 6y – 12 = 0 C1(center) = (2, 3), r =
intersection of the line
2 2 32 12 5
C2(center) (– 3, –9), r 9 81 26 64 8
C1C2 = 13, C1C2 = r1 + r2 81. The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to
Sol.
27 4 27 (3) 2 Answer (4)
(2) 8
(3) 3 21
(4) 13
x2 y1 z2 3 4 12 P 3 2, 4 1, 12 2
x2 y2 1 , is 9 5
Lies on plane x – y + z = 16 Then,
(2) 18
(1)
(1) 2 14
Answer (4)
Number of common tangent is 3.
the ellipse
x2 y1 z2 and the 3 4 12
plane x – y + z = 16, is
x2 + y2 + 6x + 18y + 26 = 0
3 2 4 1 12 2 16
(4) 27 2
11 5 16
2
y x 1 9 5 i.e., a2 = 9, b2 = 5
Sol. Ellipse is
So, e
⎛ t2 ⎞ P ⎜ t, 2 ⎟⎠ ⎝
1
P 5, 3, 14
Distance =
16 9 144 169 13
84. The equation of the plane containing the line 2x – 5y + z = 3; x + y + 4z = 5, and parallel to the plane, x + 3y + 6z = 1, is
2 3
2 As, required area 2a 2 9 27 e (2/3)
21
(1) 2 x 6 y 12 z 13
(2) x 3y 6 z 7
(3) x 3y 6 z 7
(4) 2 x 6 y 12 z 13
JEE (MAIN)-2015
Answer (3)
Answer (1)*
Sol. Required plane is
Sol. Question is wrong but the best suitable option is (1).
(2x – 5y + z – 3) + (x + y + 4z – 5) = 0
Required probability =
It is parallel to x + 3y + 6z = 1
C3
29 312
=
55 ⎛ 2 ⎞ ⎜ ⎟ 3 ⎝3⎠
11
87. The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is
2 5 1 4 1 3 6
11 2
Solving =
12
(1) 16.8
Required plane is
(2) 16.0 (3) 15.8
11 (x + y + 4z – 5) = 0 (2x – 5y + z – 3) – 2
(4) 14.0 Answer (4)
x + 3y + 6z – 7 = 0
Sol. Mean = 16
85. Let a , b and c be three non-zero vectors such that no two of them are collinear and
Sum = 16 × 16 = 256 New sum = 256 – 16 + 3 + 4 + 5 = 252
1 ( a b ) c |b ||c | a . If is the angle between 3 vectors b and c , then a value of sin is
(1)
2 2 3
(2)
2 3
(3)
2 3
(4)
2 3 3
252 = 14 18
Mean =
88. If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading to the foot of the tower, are 30º, 45º and 60º respectively, then the ratio, AB : BC, is (1)
3 : 1
(2)
3 :
2
Answer (1) (3) 1 :
1 Sol. ( a c ) b (b c ) a |b ||c | a 3
1 (b c ) |b ||c | 3
1 cos 3
sin
(4) 2 : 3 Answer (1)
55 ⎛ 2 ⎞ ⎜ ⎟ 3 ⎝3⎠
h 3
⎛1⎞ (3) 220 ⎜ ⎟ ⎝3⎠
12
h
BO = h
2 2 3
11
P
Sol. AO = h cot30º
CO
86. If 12 identical balls are to be placed in 3 identical boxes, then the probability that one the boxes contains exactly 3 balls is (1)
3
2 (2) 55 ⎛⎜ ⎞⎟ ⎝3⎠
10
⎛1⎞ (4) 22 ⎜ ⎟ ⎝3⎠
11
h 3
A
AB AO BO BC BO CO
h 3 h h h 3
3
22
30º
60º
45º B
C
O
JEE (MAIN)-2015
⎛ 2x ⎞ 89. Let tan 1 y tan 1 x tan 1 ⎜ ⎝ 1 x 2 ⎟⎠ where |x|
1 3
y
90. The negation of ~ s (~ r s) is equivalent to
. Then a value of y is
3x x 3 (1) 1 3x 2
3x x 3 (2) 1 3x 2
3x x 3 (3) 1 3x 2 Answer (1)
3x x 3 (4) 1 3x 2
(1) s ~ r (2) s (r ~ s) (3) s (r ~ s) (4) s r Answer (4) Sol. ∼ ( ∼ s ( ∼ r s ))
⎛ 2x ⎞ Sol. tan y tan x tan ⎜ ⎝ 1 x 2 ⎟⎠ 1
1
3x x 3 1 3x 2
1
= s (r ∼ s ) = (s r ) (s ∼ s )
3 1 ⎛ 3x x ⎞ 3tan–1 x = tan ⎜ 2 ⎟ ⎝ 1 3x ⎠
= s r
23
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