Answers & Solutions

Test Booklet Code

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Answers & Solutions for

M.M. : 360

JEE (MAIN)-2018 (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 of the total marks allotted to the question (i.e. 1 mark) 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 any question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction No. 4 above.

6.

For writing particulars/marking responses on Side-1 and Side-2 of the Answer Sheet use only Black Ball Point Pen provided in the examination hall.

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 hall/room.

1

JEE (MAIN)-2018 (Code-A)

PART–A : PHYSICS 1.

The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is (1) 2.5%

(2) 3.5%

(3) 4.5%

(4) 6%

3.

Two masses m1 = 5 kg and m2 = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is

m m2

Answer (3) Sol.  

T

m

T

l3

d  dm dl  3  m l

m1 m1g

= (1.5 + 3 × 1) (1) 18.3 kg

= 4.5% 2.

(2) 27.3 kg

All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

(3) 43.3 kg (4) 10.3 kg

Velocity

Answer (2) Sol. To stop the moving block m2, acceleration of m2 should be opposite to velocity of m2

Position

(1)

m1g < (m + m2)g  5 < 0.15(10 + m2)

Distance

 m2 > 23.33 kg

Time

(2)

 Minimum mass = 27.3 kg (according to given options) 4.

Position

under the action of an attractive potential U  –

Time

(3)

A particle is moving in a circular path of radius a

Its total energy is Velocity

(4)

(1) –

k 4a 2

Time

(2)

k 2a2

Answer (2) (3) Zero Sol. Options (1), (3) and (4) correspond to uniformly accelerated motion in a straight line with positive initial velocity and constant negative acceleration, whereas option (2) does not correspond to this motion.

(4) –

3 k 2 a2

Answer (3) 2

k 2r 2

.


Sol. F 

–dU dr

6.

k ⎤ ⎡ ⎢U  – 2 ⎥ 2r ⎦ ⎣

mv 2 k  3 r r

[This force provides necessary

Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is

centripetal force] 2

 mv   K .E 

P

k r2

O

k 2r 2

 P .E  –

k 2r 2

(1)

Total energy = Zero 5.

73 MR 2 2 Answer (4)

(3)

In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is (1)

(3) (4)

Sol. I0 

v0 4

=

v0 2

7.

v0

Sol. It is a case of superelastic collision mv0 = mv1 + mv2

...(i)

181 MR 2 2

⎛ MR 2 ⎞ MR 2  6⎜  M (2R )2 ⎟ ⎜ ⎟ 2 ⎝ 2 ⎠

181 MR 2 2

From a uniform circular disc of radius R and mass





1 3⎛1 ⎞ m v12  v 22  ⎜ mv 02 ⎟ 2 2⎝2 ⎠  v 22



3  v 02 2

(2)

...(ii) (4)

v 02 2

2R 3 R

37 MR 2 9

Answer (1)

3v 02  2v1v 2 2

 2v1v 2  –

40 MR 2 9

(3) 10MR2

 (v1  v 2 )2  v12  v 22  2v1v 2  v 02 

R is removed as shown 3

(1) 4MR2

 v1 + v2 = v0



(4)

in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is

Answer (2)



55 MR 2 2

9M, a small disc of radius

2

v12

(2)

IP = I0 + 7M(3R)2

2v 0

(2)

19 MR 2 2

m

Sol.

9M

...(iii)

2 2  (v1 – v2)2 = (v1 + v2)2 – 4v1v2 = v 0  v 0

 v1 – v 2  2 v 0

I1  3

(9M )  R 2 2

m

(9M ) M 9


10. Two moles of an ideal monoatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

2

⎛R⎞ M ⎜ ⎟ 2 2 ⎝ 3 ⎠  M  ⎛ 2R ⎞  MR I2  ⎜ ⎟ 2 2 ⎝ 3 ⎠

 Ireq = I1 – I2

9 MR 2  MR 2 – 2 2 2 = 4MR 8.

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then (1) T  R 3/2 for any n (3) T  R

( n 1)/2

n

(2) T  R 2 (4) T  R

T2

(3) (a) 189 K

(b) –2.7 kJ

(4) (a) 195 K

(b) 2.7 kJ

5

1

⎛ V ⎞3 Tf  300 ⎜ ⎟ ⎝ 2V ⎠

n /2

–1

 189 K



k

11. The mass of a hydrogen molecule is 3.32 × 10–27 kg. If 1023 hydrogen molecules strike, per second, a fixed wall of area 2 cm2 at an angle of 45° to the normal, and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly

1 R n 1

A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area of a floats on the surface of the liquid, covering entire cross-section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the

mg 3Ka Answer (3)

(4)

(2) 4.70 × 103 N/m2

(3) 2.35 × 102 N/m2

(4) 4.70 × 102 N/m2

Sol. F = nmvcos × 2 P

Ka (2) 3mg

(3)

(1) 2.35 × 103 N/m2

Answer (1)

⎛ dr ⎞ radius of the sphere, ⎜ ⎟ , is ⎝ r ⎠

Ka (1) mg

3R  [189 – 300] 2

= –2.7 kJ

Rn

⎛ n 1 ⎞ ⎜ ⎟ 2 ⎠

Sol. K  V

(b) –2.7 kJ

Sol. TV  – 1 = Constant

 T  R⎝ 9.

(2) (a) 195 K

U  nCv T  2 

2 –n Sol. m R  k R 

1

(b) 2.7 kJ

Answer (3)

Answer (3)



(1) (a) 189 K



mg Ka

F 2.nmv cos   A A

2  1023  3.32  10 27  103 2  2  10 4

N/m2

= 2.35 × 103 N/m2 12. A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/second. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avagadro number = 6.02 × 1023 gm mole–1)

dP dV

⇒

dV dP mg   V K Ka

⇒

3dr mg  r Ka

(2) 7.1 N/m

⇒

dr mg  r 3Ka

(4) 5.5 N/m

(1) 6.4 N/m

(3) 2.2 N/m

Answer (2) 4


x

C

Sol. B

Sol.

A a

Kx = ma  a = (K/m)x T  2

f 



⎡ 4a 2 4b2 4c 2 ⎤ VB  ⎢   ⎥ ⎣⎢ 40 b 40 b 40 c ⎦⎥

1 K   1024 42 m

VB 

4  10  108  103  1024 6.02  1023

 0

⎡ a2  b2 ⎤  c⎥ ⎢ ⎥⎦ ⎣⎢ b

15. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric

= 7.1 N/m

material of dielectric constant K 

13. A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 × 103 kg/m3 and its Young's modulus is 9.27 × 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations? (1) 5 kHz

(2) 2.5 kHz

(3) 10 kHz

(4) 7.5 kHz

(1) 1.2 nC

(2) 0.3 nC

(3) 2.4 nC

(4) 0.9 nC

Answer (1) Sol. C' = KC0 Q = KC0V

1⎞ ⎛ Qinduced  Q ⎜ 1– ⎟ K⎠ ⎝

V 1 Y  2L 2L 

 1 9.27  1010  4.88 kHz  5 kHz 2  0.6 2.7  103

⎡ a2 – b2 ⎤  c⎥ ⎢ ⎢⎣ a ⎥⎦

 (2)  0

⎡ a2 – b2 ⎤  c⎥ ⎢ ⎢⎣ b ⎥⎦

5 3⎞ ⎛  90  10 –12  20 ⎜ 1– ⎟ 3 5⎠ ⎝

= 1.2 nC 16. In an a.c. circuit, the instantaneous e.m.f. and current are given by

14. Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities +, – and + respectively. The potential of shell B is

 (1)  0

5 is inserted 3

between the plates, the magnitude of the induced charge will be

Answer (1)

e = 100 sin30 t

⎞ ⎛ i  20 sin ⎜ 30t  ⎟ 4⎠ ⎝ In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively (1) 50, 10

 (3)  0

⎡b – c ⎤  a⎥ ⎢ ⎢⎣ b ⎥⎦

(2)

 (4)  0

⎡ b2 – c 2 ⎤  a⎥ ⎢ ⎢⎣ c ⎥⎦

(3)

2

+

c

K  42 m  1024 

=

–

b

m K

1 1 K   1012 T 2 m

Sol. f0 

+

2

1000 2 50 2

,0

(4) 50, 0

Answer (2)

Answer (2) 5

, 10


Sol. Pav = Erms Irms cos



100 2



20 2



1 2

iwattless = irms sin  



1000

20 2

2 

1 2

2m

r  rp

 10

(1) 11.6 V and 11.7 V

(2) 11.5 V and 11.6 V

(3) 11.4 V and 11.5 V

(4) 11.7 V and 11.8 V

q



⎡m  4mp ⎤ ⎢ ⎥ ⎣⎢q  2q p ⎦⎥

qp 2mp

=1 Mass of electron is least and charge qe = e

17. Two batteries with e.m.f 12 V and 13 V are connected in parallel across a load resistor of 10 . The internal resistances of the two batteries are 1  and 2  respectively. The voltage across the load lies between

So, re < rp = r 19. The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B 1. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is B2. The B1 ratio B is 2

Answer (2) y

Sol. y

2mk qB

Sol. r 

(1) 2

13 V, 2 

(2)

x x

12 V, 1  x +y

(3)

10 





 12 = 11x + 10y

...(i)

13 = 10x + 12y

...(ii)

req V 

2

7 23 A, y  A 16 32

B2  

2  , R = 10  3

Eeq R  req

(3) re < rp < r

(4) re < r < rp

0I

2



2R



B1  2 B2

1 LC

the current exibits

resonance. The quality factor, Q is given by

18. An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii re, rp, r respectively in a uniform magnetic field B. The relation between re, rp, r is (2) re < rp = r

0 I 2R

and frequency 0 

R  11.56 V

(1) re > rp = r

2

20. For an RLC circuit driven with voltage of amplitude vm

E1 E2 37  ⇒ Eeq  V r1 r2 3





R   2R

B1 

V = 10(x + y) = 11.56 V

Eeq

1

Sol. m = I(R2), m  2m  I   2R

12 – x – 10(x + y) = 0

Aliter : req 

(4)

Answer (3)

Applying KVL in loops

Solving x 

2

3

(1)

0 L R

(2)

0 R L

(3)

R (0C )

(4)

CR 0

Answer (1) Sol. Quality factor, Q 

Q

Answer (2) 6

0 L R

0 (2)


21. An EM wave from air enters a medium. The electric

Sol. Polaroids A and B are oriented with parallel pass axis

⎡ ⎛z ⎞⎤ fields are E1  E01xˆ cos ⎢ 2 ⎜ – t ⎟ ⎥ in air and ⎝c ⎠⎦ ⎣ E2  E02 xˆ cos[k (2z – ct )] in medium, where the wave number k and frequency  refer to their values

Let polaroid C is at angle  with A then it makes  with B also. ∵

in air. The medium is non-magnetic. If r1 and r2 refer to relative permittivities of air and medium respectively, which of the following options is correct? (1)

(3)

r1  r1



r2

2  cos  

(2)   2 r2

1 4

(4)

r1 r2



23. The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 m. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance?

1 2

Answer (3) ⎡ ⎛z ⎞⎤ Sol. E1  E01xˆ cos ⎢ 2 ⎜ – t ⎟ ⎥ c ⎝ ⎠⎦ ⎣ E2  E02 xˆ cos ⎡⎣ k  2z – ct  ⎤⎦

air

(i.e. distance between the centres of each slit.) (1) 25 m

medium

(2) 50 m

During refraction, frequency remains unchanged, whereas wavelength gets changed.  k' = 2k 

(From equations)

(4) 100 m Sol. dsin = 

c 2

1



Answer (1)

0 2

 ' 



(3) 75 m

⎛ 2 ⎞ 2  2⎜ ⎟ ' ⎝ 0 ⎠

 v

1 2

  = 45°

r1

4

r2

I ⎛I ⎞   cos2  ⎟  cos2  8 ⎜⎝ 2 ⎠

0  2

d 



60°

1 1  2 0 1

1 1  2 4

d



22. Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A. The intensity of light beyond B is



I found to be . Now another identical polarizer C is 2 placed between A and B. The intensity beyond B is

now found to be

d 2

Fringe width, B 

and C is (2) 30°

(3) 45°

(4) 60°

[d = 1 × 10–6 m]

  = 5000 Å

I . The angle between polarizer A 8

(1) 0°

30°

10 –2 

D (d ' is slit separation) d'

5000  10 –10  0.5 d'

 d' = 25 × 10–6 m = 25 m

Answer (3) 7


24. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let n, g be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let n be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)



A

 2n

(3) n2  A + Bn2

⎞ 1 ⎟ ⎟  2n ⎠

B  2n

2mc  2g h

, B

2mc  g4 h

(1) 25 L

(4) n2  

(2) 16 L

Answer (1) h h Sol. Pn  , Pg  n g

h2

, Eg  –

2m n2

L/16

(4)

L/25

⎡ 1 1⎤ Sol. hL  E ⎢ – ⎥  E ⎣12  ⎦ ⎡ 1 1⎤ E hP  E ⎢ 2 – ⎥   ⎦ 25 ⎣5

h2 2m g2

 P 

h ⎛ 1 1 ⎞ hc ⎜ ⎟ En – Eg  – 2m ⎜  2g  2n ⎟  n ⎝ ⎠ 2

2 2 h2 ⎛ n – g ⎜ 2m ⎜  2g  2n ⎝

(3)

Answer (4)

P2 h2 h2  , E  –k  – 2 2m 2m 2m 2

En  –

⎛ 2mc  g4 ⎜ ⎜ h ⎝

25. If the series limit frequency of the Lyman series is L, then the series limit frequency of the Pfund series is

(2) n  A + Bn

k

h

 A

B

(1) n A +

2mc  2g

L 25

26. It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively

⎞ hc ⎟ ⎟ n ⎠

(1) (.89, .28) n 

2mc h

⎛  2g  2n ⎜ ⎜  2n –  2g ⎝

⎞ ⎟ ⎟ ⎠

(2) (.28, .89) (3) (0, 0) (4) (0, 1)

n 

2mc  2g

 2n

h

⎛  2g ⎞  n2 ⎜ 1– 2 ⎟ ⎜  n ⎟⎠ ⎝

2mc  2g ⎡  2g ⎤ ⎢1– 2 ⎥  h ⎢⎣  n ⎥⎦



Answer (1) Sol. mu = mv1 + 2m × v2

...(i)

u = (v2 – v1)  v1  

–1



2mc  2g ⎡  2g ⎤ ⎢1  2 ⎥ h ⎢⎣  n ⎥⎦

8

u 3

1 1 ⎛u ⎞ mu 2  m ⎜ ⎟ E 2 2 ⎝3⎠  pd  1 E mu 2 2 

...(ii)

8  0.89 9

2


And mu = mv1 + (12m) × v2

...(iii)

u = (v2 – v1)  v1  

29. In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell.

...(iv)

11 u 13 2

1 1 ⎛ 11 ⎞ mu 2  m ⎜ u ⎟ E 48 2 2 ⎝ 13 ⎠  pc    0.28  1 E 169 2 mu 2 27. The reading of the ammeter for a silicon diode in the given circuit is

(1) 1  (2) 1.5  (3) 2  (4) 2.5 

200 

Answer (2) Sol. ∵ E  l1 and E – ir  l2 

3V (1) 0



(2) 15 mA (3) 11.5 mA (4) 13.5 mA



Answer (3)

200 

V – Vdiode Sol. I  R

⎡ 3 – 0.7 ⎤ ⎢  1000 ⎥ mA 200 ⎣ ⎦ = 11.5 mA

l E  1 E  ir l 2 52 E  40 ⎛ E ⎞ r E ⎜ ⎟ ⎝r 5⎠ r  5 13  5 10

 r = 1.5  30. On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 k. How much was the resistance on the left slot before interchanging the resistances?

3V

(1) 990 

28. A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz?

(2) 505  (3) 550  (4) 910 

(1) 2 × 103

Answer (3)

(2) 2 × 104

R1 l Sol. R  (100 – l ) 2

(3) 2 ×

105

(4) 2 × 106

R2 (l – 10)  R1 (110 – l )

Answer (3) Sol. Frequency of carrier = 10 × 109 Hz

(100 – l)(110 – l) = l(l – 10)

Available bandwidth = 10% of 10 × 109 Hz =

109

11000 + l2 – 210l = l2 – 10l

Hz



Bandwidth for each telephonic channel = 5 kHz  Number of channels 

10

⎛ 55 ⎞ R1  R2 ⎜ ⎟ ⎝ 45 ⎠

9

5  103

=2×

l = 55 cm

R1 + R2 = 1000  R1 = 550 

105 9


PART–B : CHEMISTRY 33. According to molecular orbital theory, which of the following will not be a viable molecule?

31. The ratio of mass percent of C and H of an organic compound (CXHYOZ) is 6 : 1. If one molecule of the above compound (CXHYOZ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O. The empirical formula of compound CXHYOZ is

(1) He22

(2) He2

(3) H2–

(4) H2– 2

(1) C3H6O3

(2) C2H4O

Answer (4)

(3) C3H4O2

(4) C2H4O3

Sol. Electronic configuration Bond order

Answer (4) Sol. Element Relative Relative Simplest whole mass mole number ratio

C

6

H

1

6 = 0.5 12 1 =1 1

1 2

He2



H2–



H2– 2



* 1s2 1s1 * 1s2 1s1

* 1s2 1s2

He22

So, X = 1, Y = 2 Equation for combustion of CXHY



1s2

2–1  0.5 2 2–1  0.5 2 2–2 0 2 2–0 1 2

Molecule having zero bond order will not be a viable molecule.

Y⎞ Y ⎛ C XHY  ⎜ X  ⎟ O2  XCO2  H2 O ⎝ 4⎠ 2

34. Which of the following lines correctly show the temperature dependence of equilibrium constant K, for an exothermic reaction?

Y⎞ ⎛ Oxygen atoms required = 2 ⎜ X  ⎟ ⎝ 4⎠

ln K

As per information,

A B

Y⎞ ⎛ 2 ⎜ X  ⎟  2Z ⎝ 4⎠

(0, 0)

1 T(K)

C

2⎞ ⎛  ⎜1  ⎟  Z ⎝ 4⎠

D

 Z = 1.5

(1) A and B

(2) B and C

Molecule can be written

(3) C and D

(4) A and D

CXHYOZ

Answer (1)

C1H2O3/2

H

⎛A ⎞  Sol. Equilibrium constant K  ⎜ f ⎟ e RT ⎝ Ab ⎠

 C2H4O3 32. Which type of ‘defect’ has the presence of cations in the interstitial sites?

⎞ H ⎛ 1 ⎞ ⎟  R ⎜⎝ T ⎟⎠ ⎠

(1) Schottky defect

⎛A ln K  ln ⎜ f ⎝ Ab

(2) Vacancy defect

y =

(3) Frenkel defect

Comparing with equation of straight line,

C

(4) Metal deficiency defect Slope =

Answer (3)

+ m

x

H R

Since, reaction is exothermic, H° = –ve, therefore, slope = +ve.

Sol. In Frenkel defect, cation is dislocated from its normal lattice site to an interstitial site. 10


ln K

37. An aqueous solution contains 0.10 M H 2S and 0.20 M HCl. If the equilibrium constant for the formation of HS– from H2S is 1.0 × 10–7 and that of S 2– from HS – ions is 1.2 × 10 –13 then the concentration of S2– ions in aqueous solution is

A

(0, 0)

1 T(K)

B

(1) 5 × 10–8 (2) 3 × 10–20

Hence, option (1) is correct.

(3) 6 × 10–21

35. The combustion of benzene (l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 25° C; heat of combustion (in kJ mol–1) of benzene at constant pressure will be

(4) 5 × 10–19 Answer (2) Sol. In presence of external H+,

(R = 8.314 JK–1 mol–1) (1) 4152.6

(2) –452.46

(3) 3260

(4) –3267.6

 2 H2S 2H  S , K a 1  K a2  K eq



Answer (4) 15 O2 (g)  6CO2 (g)  3H2O(l) 2 15 3 ng  6   2 2

H 2 S2   1 10 7  1.2  1013 H S  2  0.22 S2   1.2  10 20 0.1

Sol. C6H6 (l) 

[S2–] = 3 × 10–20

H = U + ngRT

38. An aqueous solution contains an unknown concentration of Ba 2+ . When 50 mL of a 1 M solution of Na2SO4 is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1 × 10–10. What is original concentration of Ba2+?

⎛ 3⎞ 3 = 3263.9  ⎜  ⎟  8.314  298  10 ⎝ 2⎠

= –3263.9 + (–3.71) = –3267.6 kJ mol–1 36. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

(1) 5 × 10–9 M (2) 2 × 10–9 M

(1) [Co(H2O)6]Cl3

(3) 1.1 × 10–9 M

(2) [Co(H2O)5Cl]Cl2  H2O

(4) 1.0 × 10–10 M

(3) [Co(H2O)4Cl2]Cl  2H2O

Answer (3)

(4) [Co(H2O)3Cl3]  3H2O

Sol. Final concentration of [SO4– –] =

Answer (4) Sol. The solution which shows maximum freezing point must have minimum number of solute particles.

Ksp of BaSO4, [Ba2+][SO42–] = 1 × 10–10

(1) [Co(H2O)6]Cl3  [Co(H2O)6]3+ + 3Cl–, i = 4 (2) [Co(H2O)5Cl]Cl2  H2O  [Co(H2O)5Cl]2+ + 2Cl–,

[Ba2+][0.1] =

i=3 (3) [Co(H2O)4Cl2]Cl  2H2O 

[Co(H2O)4Cl2]+

[50  1] = 0.1 M [500]

10 10 = 10–9 M 0.1

Concentration of Ba2+ in final solution = 10–9 M

+ Cl–,

Concentration of Ba2+ in the original solution.

i=2

M1V1 = M2V2

(4) [Co(H2O)3Cl3]  3H2O  [Co(H2O)3Cl3], i = 1

M1 (500 – 50) = 10–9 (500)

So, solution of 1 molal [Co(H2O)3Cl3]  3H2O will have minimum number of particles in aqueous state.

M1 = 1.11 × 10–9 M


So, option (3) is correct. 11


41. The recommended concentration of fluoride ion in drinking water is up to 1 ppm as fluoride ion is required to make teeth enamel harder by converting [3Ca3(PO4)2.Ca(OH)2] to

39. At 518°C, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 torr, was 1.00 torr s–1 when 5% had reacted and 0.5 torr s–1 when 33% had reacted. The order of the reaction is

(1) [CaF2]

(1) 2

(2) [3(CaF2).Ca(OH)2]

(3) [3Ca3(PO4)2.CaF2] (4) [3{Ca(OH)2}.CaF2]

(2) 3

Answer (3)

(3) 1

Sol. F– ions make the teeth enamel harder by converting

(4) 0

[3Ca3 (PO4 )2 .Ca(OH)2 ] to [3Ca3 (PO4 )2 .CaF2 ] Hydroxyapatite

Answer (1)

42. Which of the following compounds contain(s) no covalent bond(s)?

Sol. Assume the order of reaction with respect to acetaldehyde is x.

KCl, PH3, O2, B2H6, H2SO4

Condition-1 : Rate =

k[CH3CHO]x

1 = k[363 ×

0.95]x

(1) KCl, B2H6, PH3

(2) KCl, H2SO4

(3) KCl

(4) KCl, B2H6

Answer (3)

1 = k[344.85]x

Sol. KCl – Ionic bond between K+ and Cl–

...(i)

PH3 – Covalent bond between P and H

Condition-2 :

O2 – Covalent bond between O atoms

0.5 = k[363 × 0.67]x 0.5 = k[243.21]x

B2H6–Covalent bond between B and H atoms

...(ii)

H2SO4 – Covalent bond between S and O and also between O and H.

Divide equation (i) by (ii), 1 ⎛ 344.85 ⎞ ⎜ ⎟ 0.5 ⎝ 243.21 ⎠

Fluorapatite

x

Compound having no covalent bonds is KCl only.

⇒ 2  (1.414)x

43. Which of the following are Lewis acids?

 x=2 40. How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane? (Atomic weight of B = 10.8 u) (1) 6.4 hours

(2) 0.8 hours

(3) 3.2 hours

(4) 1.6 hours

(1) PH3 and BCl3

(2) AlCl3 and SiCl4

(3) PH3 and SiCl4

(4) BCl3 and AlCl3

Answer (4)* Sol. BCl3 – electron deficient, incomplete octet AlCl3 – electron deficient, incomplete octet Ans-(4) BCl3 and AlCl3 SiCl4 can accept lone pair of electron in d-orbital of silicon hence it can act as Lewis acid.

Answer (3) Sol. B2H6 + 3O2  B2O3 + 3H2O

* Although the most suitable answer is (4). However, both option (4) & (2) can be considered as correct answers.

27.66 of B2H6 = 1 mole of B2H6 which requires three moles of oxygen (O2) for complete burning

e.g. hydrolysis of SiCl4

6H2O  6H2+ 3O2 (On electrolysis)

Cl

Number of faradays = 12 = Amount of charge

Cl

12 × 96500 = i × t 12 × 96500 = 100 × t t

12  96500 second 100

t

12  96500 hour 100  3600

Si Cl

Cl Cl

+

H2O

Cl

Si Cl

Cl

O

H H

Cl Cl Si

OH + HCl

Cl Hence option (2), AlCl3 and SiCl4 is also correct.

t = 3.2 hours 12


44. Total number of lone pair of electrons in I3– ion is

47. The oxidation states of

(2) 6

Cr in ⎡⎣Cr H2O 6 ⎤⎦ Cl3 , ⎣⎡Cr  C6H6 2 ⎦⎤ , and K 2 ⎡⎣Cr  CN2  O 2  O2  NH3  ⎤⎦ respectively are

(3) 9

(1) +3, +4 and +6

(4) 12

(2) +3, +2 and +4

(1) 3

Answer (3)

(3) +3, 0 and +6

Sol. Structure of

I3–

(4) +3, 0 and +4

I

–

Answer (3) Sol. ⎣⎡Cr H2 O 6 ⎦⎤ Cl3 ⇒ x  0  6 – 1 3  0

I

 x  3

⎡⎣Cr  C6H6 2 ⎤⎦ ⇒ x  2  0  0

I

x0

Number of lone pairs in I3 is 9.

K 2 ⎣⎡Cr  CN2  O2  O2  NH3 ⎦⎤

45. Which of the following salts is the most basic in aqueous solution?

⇒ 1  2  x – 1 2 – 2  2 – 2  1  0 ⇒x – 6  0

(1) Al(CN)3

x  6

(2) CH3COOK (3) FeCl3

48. The compound that does not produce nitrogen gas by the thermal decomposition is

(4) Pb(CH3COO)2 Answer (2) Sol. CH3COOK + H2O  CH3COOH + KOH Basic

(1) Ba(N3)2

(2) (NH4)2Cr2O7

(3) NH4NO2

(4) (NH4)2SO4

Answer (4)

FeCl3 – Acidic solution

Δ Sol. NH4 2 Cr2O7  N2 + 4H2O + Cr2O3

Al(CN)3 – Salt of weak acid and weak base

Δ NH4NO2   N2 + 2H2O

Pb(CH3COO)2 – Salt of weak acid and weak base CH3COOK is salt of weak acid and strong base.

Δ  2NH3 + H2SO4 NH4 2SO4 

Hence solution of CH3COOK is basic.

Δ Ba N3 2   Ba  3N2

46. Hydrogen peroxide oxidises [Fe(CN) 6 ] 4– to [Fe(CN) 6 ] 3– in acidic medium but reduces [Fe(CN)6]3– to [Fe(CN)6]4– in alkaline medium. The other products formed are, respectively.

Among all the given compounds, only (NH4)2SO4 do not form dinitrogen on heating, it produces ammonia gas.

(1) (H2O + O2) and H2O

49. When metal ‘M’ is treated with NaOH, a white gelatinous precipitate ‘X’ is obtained, which is soluble in excess of NaOH. Compound ‘X’ when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal ‘M’ is

(2) (H2O + O2) and (H2O + OH–) (3) H2O and (H2O + O2) (4) H2O and (H2O + OH–) Answer (3)

(1) Zn

1 3– [Fe(CN)6] + H2O H O + H+ 2 2 2 1 3– [Fe(CN)6] + H2O2 + OH 2 1 4– [Fe(CN)6] + H2O + O 2 2 4–

Sol. [Fe(CN) 6] +

(2) Ca (3) Al (4) Fe 13


Answer (3)

51. Glucose on prolonged heating with HI gives

NaOH Sol. Al3   

Al  OH3 

White gelatinous ppt.

Excess

(1) n-Hexane

NaOH   NaAlO2 Sodium meta alu minate (so lub le)

(2) 1-Hexene

Strong heating 2Al  OH3   Al2O3  3H2O

(3) Hexanoic acid

Al2O3 is used in column chromatography.

(4) 6-iodohexanal

50. Consider the following reaction and statements

Answer (1)

[Co(NH3)4Br2]+ + Br–  [Co(NH3)3Br3] + NH3

CHO

(I) Two isomers are produced if the reactant complex ion is a cis-isomer

Sol. (CH–OH)4

(1) H2 - Pd/C, BaSO4

(IV) Only one isomer is produced if the reactant complex ion is a cis-isomer.

(2) NaBH4

The correct statements are:

(3) (III) and (IV)

(4) (II) and (IV)

(3) Na/liq. NH3 (4) Sn - HCl Answer (3)

Answer (2) Sol.

n-Hexane

52. The trans-alkenes are formed by the reduction of alkynes with

(III) Only one isomer is produced if the reactant complex ion is a trans-isomer.

(2) (I) and (III)

CH3–CH2–CH2–CH2–CH2–CH3

CH2–OH

(II) Two isomers are produced if the reactant complex ion is a trans-isomer.

(1) (I) and (II)

HI, 

Br

NH3

Sol. CH3 – C  C – CH3

Br

Na/liq. NH3

CH3 H

C=C

H CH3

Trans alkene

NH3

NH3

So, option (3) is correct.

NH3 cis-isomer

53. Which of the following compounds will be suitable for Kjeldahl's method for nitrogen estimation?

–

+Br

NH3

Br

NH3

Br

Br Br

(1)

+ NH3

NH3

Br

NH3

NH3

NH3

NH 3

NH2

Br

fac-

NH3

NH3

NH3

(2)

mer-

(2 isomer)

Br

N

Br

NO2

NH3

(3) +

Br

Br

trans

Mer (1 isomer)

–

N2 Cl

Br

NH3

(4) Answer (2)

So option (2) is correct. 14


Answer (3)

Sol. Kjeldahl method is not applicable for compounds containing nitrogen in nitro, and azo groups and nitrogen in ring, as N of these compounds does not change to ammonium sulphate under these conditions. Hence only aniline can be used for estimation of nitrogen by Kjeldahl’s method.

Sol. The pH range of methyl orange is

O

(1)

CH3

H N

O

O

C

N 

NH3 

H N

(4)

(3)

CO2H CH3

O

N H



(4)

OH

OH COOH

CO2, NaOH

Answer (4)

H N

Acidification

(Major)

OH

NH2

N H

O

Answer (1)

Sol.

(2)

CO2H

O OH

(3)

H N

CO2H

CH3

NH2

(1)

CH3

(2)

CO2H

Yellow

56. The predominant form of histamine present in human blood is (pKa, Histidine = 6.0)

O O

4.5

Weak base is having pH greater than 7. When methyl orange is added to weak base solution, the solution becomes yellow. This solution is titrated by strong acid and at the end point pH will be less than 3.1. Therefore solution becomes pinkish red.

54. Phenol on treatment with CO2 in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with (CH3CO)2O in the presence of catalytic amount of H2SO4 produces

O

3.9

Pinkish red

COOH

(CH3CO)2O

Sol. Histamine

O O–C–CH3

H

NH2 At pH (7.4) major form of histamine is protonated at primary amine.

COOH

H2SO4

H N

Acetyl salicylic acid (Aspirin)

55. An alkali is titrated against an acid with methyl orange as indicator, which of the following is a correct combination? Base (1) Weak

Acid

End point

Strong

Colourless to pink

(2) Strong

Strong

Pinkish red to yellow

(3) Weak

Strong

Yellow to pinkish red

(4) Strong

Strong

N

+

NH3 57. Phenol reacts with methyl chloroformate in the presence of NaOH to form product A. A reacts with Br2 to form product B. A and B are respectively

Br

OH (1)

OCH3 O O

(2)

Pink to colourless

OH

and

OCH3 O O

O O

and

O Br

15

O


O

O

O

(3)

and

O

59. The major product formed in the following reaction is

O

Br

OH (4)

O

O

OH and

OCH3

Br

O

OH

(1)

Answer (3) O– OH

Sol.

O – C – O – CH3

O

–

I

(2)

OH

O OH3

Heat

O

OCH3

O

HI

Cl – C – O – CH3

I

OH

(3)

I

(4)

OH

I

Answer (4)

Br2

O O – C – O – CH3

O

Sol.

O Br

(b)

NH

(d)

NHCH3

Br

NH2 (c)

NH

+

I OH

60. The major product of the following reaction is

58. The increasing order of basicity of the following compound is

NH2

OH

+



(a)

I

HI Heat

NaOMe MeOH

OMe

(1)

(2)

(3)

(4)

(1) (a) < (b) < (c) < (d) (2) (b) < (a) < (c) < (d) (3) (b) < (a) < (d) < (c) (4) (d) < (b) < (a) < (c)

OMe

Answer (3) NH2

Sol. (a)

Answer (2)

3

1° & sp NH

(b)

Protonation

Sol. CH3O– is a strong base and strong nucleophile, so favourable condition is SN2/E2.

NH2 2

sp NH2

(c)

NH3

Protonation

Protonation

+ NH2

NH

Given alkyl halide is 2° and C's are 4° and 2°, so sufficiently hindered, therefore, E2 dominates over SN2.

+ NH2

NH2

NH2

Also, polarity of CH3OH (solvent) is not as high as H2O, so E1 is also dominated by E2.

[Equivalent resonance]

4°

Protonation

(d)

NHCH3



NH2–CH3 2° & sp3

Br

 

 Correct order of basicity : b < a < d < c.

(2°)

16

H

CH3O– E2 (Major product)


PART–C : MATHEMATICS 61. Two sets A and B are as under :

(| x – 3 | 3)(| x – 3 | –1)  0

A = {(a, b) R × R : |a – 5| < 1 and |b – 5| < 1}

 | x – 3|  1, | x – 3|  3  0

B = {(a, b)  R × R : 4(a – 6)2 + 9(b – 5)2  36}, then (1) B A (2) A B (3) A B =  (an empty set)

Sol. As, |a – 5| < 1 and |b – 5| < 1

(a  6)2 (b  5)2  1 9 4

⇑

P Q (6, 6)

(2) 0

(3) 1

(4) 2

Roots are –, –2 Let  = –,  = –2 101 + 107 = (–)101 + (–2)107 = –(101 + 214)



(6, 5)

(3, 5)

(1) –1

Sol. x2 – x + 1 = 0

(a  6)2 (b  5)2  1 9 4

(6, 7)

x  4, 2

Answer (3)

Taking axes as a-axis and b-axis

b



63. If ,   C are the distinct roots, of the equation x2 – x + 1 = 0, then 101 + 107 is equal to

Answer (2)

a=6

x – 3  1

x = 16, 4

(4) Neither A B nor B A

 4 < a, b < 6 and



(9, 5)

= –(2 + )

b=5

(4, 5) (6, 4) S R

=1 2x x  4 2x x  4 2 x  ( A  Bx )( x  A)2 , then the 64. If 2 x x4 2x 2x

(6, 3) (0, 0)

a

ordered pair (A, B) is equal to The set A represents square PQRS inside set B representing ellipse and hence A  B. 62. Let S = {x  R : x  0 and

2 x –3  x



(1) (–4, –5)

(2) (–4, 3)

(3) (–4, 5)

(4) (4, 5)

Answer (3)



x – 6  6  0 }. Then S :

x  4 2x 2x x  4 2x Sol.   2 x

(1) Is an empty set

2x

(2) Contains exactly one element

x4

2x

x = –4 makes all three row identical

(3) Contains exactly two elements

hence (x + 4)2 will be factor

(4) Contains exactly four elements

Also, C1  C1  C2  C2

Answer (3)

5 x  4 2x   5x  4 x  4

Sol. 2| x – 3 |  x ( x – 6)  6  0

5x  4

2| x – 3|  ( x – 3  3)( x – 3 – 3)  6  0

2x

 5x – 4 is a factor

2

2| x – 3|  ( x – 3) – 3  0

   (5 x  4)( x  4)2

( x – 3)2  2| x – 3| – 3  0

 B = 5, A = –4 17

2x 2x x4


65. If the system of linear equations

67. The sum of the co-efficients of all odd degree terms



x + ky + 3z = 0 3x + ky – 2z = 0 xz has a non-zero solution (x, y, z), then 2 is equal y to

(1) –10

(2) 10

(3) –30

(4) 30

(1) –1

(2) 0

(3) 1

(4) 2



 



5

3 3 Sol. x  x  1  x  x  1

Sol. ∵ System of equation has non-zero solution.

5

 2 ⎡⎣ 5C0 x 5  5C2 x 3 ( x 3  1)  5C4 x ( x 3  1)2 ⎤⎦

3 –2  0

 2 ⎡⎣x 5  10( x 6  x 3 )  5 x ( x 6  2 x 3  1)⎤⎦

2 4 –3

 2 ⎡⎣x 5  10 x 6  10 x 3  5 x 7  10 x 4  5 x⎤⎦

 44 – 4k = 0  k = 11 Let z = 

 2 ⎡⎣5 x 7  10 x 6  x 5  10 x 4  10 x 3  5 x⎤⎦

 x + 11y = –3

Sum of odd degree terms coefficients

and 3x + 11y = 2 

x

= 2(5 + 1 – 10 + 5) =2

5  ,y – ,z 2 2

68. Let a1, a2, a3, ...., a49 be in A.P. such that 12



,

 x  x3  1

Answer (4)

Answer (2)

1 k 3 k

5

(x > 1) is

2x + 4y – 3z = 0



 

in the expansion of x  x 3  1

∑ a4k 1  416

5 · xz  2  10 y 2 ⎛  ⎞2 ⎜– 2⎟ ⎝ ⎠

k 0

and a9  a43  66 .

2 If a12  a22  ....  a17  140m , then m is equal to

66. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is

(1) 66

(2) 68

(3) 34

(4) 33

Answer (3) Sol. Let a1 = a and common difference = d Given, a1 + a5 + a9 + ..... + a49 = 416

(1) At least 1000 (2) Less then 500

 a + 24d = 32

...(i)

Also, a9 + a43 = 66  a + 25d = 33

...(ii)

Solving (i) & (ii),

(3) At least 500 but less than 750

We get d = 1, a = 8

(4) At least 750 but less than 1000 Answer (1)

2 Now, a12  a22  .....  a17  140m

Sol. Number of ways of selecting 4 novels from 6 novels = 6C4

 82  92  .....  242  140m

Number of ways of selecting 1 dictionary from 3 dictionaries = 3C1 Required arrangements = 6C4 × 3C1 × 4! = 1080  Atleast 1000 18



24  25  49 7  8  15   140m 6 6



m  34

5


69. Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series

71. Let S  {t  R : f ( x )  x   ·(e|x|  1)sin | x | is not differentiable at t}. Then the set S is equal to

12 + 2.22 + 32 + 2.42 + 52 + 2.62 + ..... If B – 2A = 100, then is equal to (1) 232

(2) 248

(3) 464

(4) 496

2

2

(4) {0, }

x = , 0 are repeated roots and also continuous. Hence, 'f' is differentiable at all x. 2

2

2

2

 (1  2  3  ....  20 )  4(1  2  3  ....  10 ) 

(3) {  }

Sol. f ( x )  | x   | (e|x|  1)sin| x |

Sol. A  12  2.22  32  ....  2.202 2

(2) { 0 }

Answer (1)

Answer (2)

2

(1)  (an empty set)

72. If the curves y2 = 6x, 9x2 + by2 = 16 intersect each other at right angles, then the value of b is

20  21 41 4  10  11 21  6 6

(1) 6

(2)

7 2

(3) 4

(4)

9 2

= 2870 + 1540 = 4410 B  12  2.22  32  ....  2.402

 (12  22  32  ....  402 )  4(12  22  32  ....  202 ) 

Answer (4)

40  41 81 4  20  21 41  6 6

Sol. y2 = 6x ; slope of tangent at (x1, y1) is m1 

= 22140 + 11480 = 33620

also 9 x 2  by 2  16; slope of tangent at (x1, y1) is

 B – 2A = 33620 – 8820 = 24800  100 = 24800

m2 

 = 248



⎛ ⎡ 1⎤ ⎡2⎤ ⎡ 15 ⎤ ⎞ lim x ⎜ ⎢ ⎥  ⎢ ⎥  ......  ⎢ ⎥ ⎟  ⎣ x ⎦⎠ ⎝⎣x⎦ ⎣x⎦ x 0

(2) Is equal to 15

(3) Is equal to 120

(4) Does not exist (in R)

1 ⎡ 1⎤ 1 1  ⎢ ⎥  x ⎣x⎦ x

⎛r

⎞ 15 ⎛ r ⎞ r 1

f x

1 x

2

1 and g  x   x  , x  R  {1, 0,1} . x

, then the local minimum value of h(x)

g x

is: (1) 3

(2) –3

(3) 2 2

(4) 2 2

Answer (4)

r

∑ ⎜⎝ x  1⎟⎠  ∑ ⎜⎝ x ⎟⎠  ∑ x

r 1

9  as y12  6 x1  2

If h  x  

15

 1

by12

73. Let f  x   x 2 

2 ⎡2⎤ 2 1  ⎢ ⎥  x ⎣x⎦ x 15

27 x1

 b

Answer (3) Sol. As

9x1 by1

As m1m2  1

70. For each t  R, let [t] be the greatest integer less than or equal to t. Then

(1) Is equal to 0

3 y1

r 1

Sol. h  x  

⎛ 15 ⎡ r ⎤ ⎞ 120  lim x ⎜ ∑ ⎢ ⎥ ⎟  120 ⎜ ⎟ x 0 ⎝ r 1 ⎣ x ⎦ ⎠

x2 

1

x2 x1 x



 x1

⎛ ⎡ 1⎤ ⎡2⎤ ⎡ 15 ⎤ ⎞ ⇒ lim x ⎜ ⎢ ⎥  ⎢ ⎥  ......  ⎢ ⎥ ⎟  120  ⎣ x ⎦⎠ ⎝⎣x⎦ ⎣x⎦ x 0

19

x





2 x1

x




x

x

1  0, x 1  0, x

 x  1x    x  1x  

 

2 x1 2 x1

x



x

 2

 (2 2 , ]

sin2 xdx

∫

Sol. I   ( , 2 2]

Local minimum is 2 2

Also, I 

74. The integral

 2

∫

1 C 1  cot 3 x

(4)

... (ii)

1  2x

 2

2

∫ sin

2I 

1 C (2) 3(1  tan3 x )

1 C (1) 3(1  tan3 x )

2 x sin2 xdx

Adding (i) and (ii)

is equal to

(3)

 2 

sin2 x cos2 x ∫ (sin5 x  cos3 x sin2 x  sin3 x cos2 x  cos5 x )2 dx

... (i)

1  2x

  2

xdx

  2

1 C 1  cot 3 x

 2

 2

0

0

2I  2 ∫ sin2 xdx ⇒ I  ∫ sin2 xdx ... (iii)

(where C is a constant of integration) Answer (2) Sol. I  ∫

(sin

sin2 x.cos2 x dx 2

 2

I  ∫ cos2 xdx



x  cos2 x ) (sin3 x  cos3 x )

2

Dividing the numerator and denominator by cos6x  I∫

Adding (iii) & (iv)

tan2 x sec 2 x dx (1  tan3 x )2

 2

2I  ∫ dx  0

Let, tan3x = z  3tan2x.sec2xdx = dz I

=

the roots of the quadratic equation 18x2 – 9x + 2 = 0. Then the area (in sq. units) bounded by the curve y = (gof)(x) and the lines x = , x =  and y = 0, is

1 C 3(1  tan3 x )

75. Then value of

∫

sin2 x

 1 2  2

x

(1)

1 ( 3  1) 2

(2)

1 ( 3  1) 2

(3)

1 ( 3  2) 2

(4)

1 ( 2  1) 2

dx is :

 8

(2)

 2

(3) 4

(4)

 4

(1)

  ⇒I  2 4

76. Let g ( x )  cos x 2 , f ( x )  x , and ,  ( < ) be

1 dz 1  C 3 ∫ z 2 3z

 2

... (iv)

0

Answer (1) Sol. 18 x 2  9x  2  0 (6 x  )(3 x  )  0    x , 6 3

Answer (4) 20


   ,  6 3

78. A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is

y  (gof )( x )  cos x

Area =

 3  cos x dx 6

∫

=

 sin x

 3  6



(3) 3x + 2y = xy

(4) 3x + 2y = 6xy

Sol. Let the equation of line be

x y  1 a b

...(i)

(i) passes through the fixed point (2, 3)



3  1 sq. units 

77. let y = y(x) be the solution of the differential equation sin x

(2) 2x + 3y = xy

Answer (3)

3 1  2 2

1 = 2

(1) 3x + 2y = 6

⎛⎞ dy  y cos x  4 x , x  (0, ). If y  ⎜ ⎟  0 , dx ⎝2⎠

2 3  1 a b

...(ii)

P(a, 0), Q(0, b), O(0, 0), Let R(h, k),

⎛⎞ then y ⎜ ⎟ is equal to : ⎝6⎠ (1)

4 9 3

2

8 2 (3) –  9

(2)

–8 9 3

2 ⎛h k ⎞ Midpoint of OR is ⎜ , ⎟ ⎝2 2⎠

4 2 (4) –  9

⎛a b⎞ Midpoint of PQ is ⎜ , ⎟ ⇒ h  a, k  b ... (iii) ⎝2 2⎠

Answer (3) Sol. sin x

dy  y cos x  4 x , x  (0, ) dx

From (ii) & (iii),

dy 4x  y cot x  dx sin x



cot x dx I.F.  e ∫  sin x

 Solution is given by y sin x 

4x

y·sinx = 2x2 + c

 ,y=0  2

2 3  1 x y

 3x + 2y = xy

(1) 10 c–

2 2

 Equation is : y sin x  2 x 2 –

(3) 3

2 2

(2) 2 10

5 2

(4)

Answer (3) Sol. A (–3, 5) B (3, 3)

 1 2  2 – when x  then y ·  2· 6 2 36 2



 locus of R(h, k)

79. Let the orthocentre and centroid of a triangle be A(–3, 5) and B(3, 3) respectively. If C is the circumcentre of this triangle, then the radius of the circle having line segment AC as diameter, is

∫ sin x ·sin x dx

when x 

2 3  1 h k

A

8 2 y– 9

21

B

C

3 5 2


mPB = –2

So, AB  2 10 Now, as, AC 

So, radius =

4 2 i.e., tan   3 2 8 1 3

3 AB 2

3 3 5 AB  10  3 4 2 2

P(16, 16)

80. If the tangent at (1, 7) to the curve x2 = y – 6 touches the circle x2 + y2 + 16x + 12y + c = 0 then the value of c is (1) 195

(2) 185

(3) 85

(4) 95

 A

C(4, 0)

B(24, 0)

Answer (4) Sol. Equation of tangent at (1, 7) to curve x2 = y – 6 is

x –1

82. Tangents are drawn to the hyperbola 4x2 – y2 = 36 at the points P and Q. If these tangents intersect at the point T(0, 3) then the area (in sq. units) of PTQ is

1 ( y  7) – 6 2

2x – y + 5 = 0

…(i)

Centre of circle = (–8, –6)

(1) 45 5

(2) 54 3

(3) 60 3

(4) 36 5

Radius of circle  64  36 – c  100 – c

Answer (1)

∵ Line (i) touches the circle

Sol. Clearly PQ is a chord of contact,



2(–8) – (–6)  5 4 1

i.e., equation of PQ is T  0  100 – c

 y = –12 Solving with the curve, 4x2 – y2 = 36

5  100 – c



 c = 95

i.e., P (3 5,  12); Q( 3 5,  12); T (0,3)

81. Tangent and normal are drawn at P(16, 16) on the parabola y2 = 16x, which intersect the axis of the parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and CPB = , then a value of tan  is (1)

1 2

Area of PQT is

(4)

= 45 5

Answer (2) Sol. y2 = 16x ... (1)

Normal at P(16, 16) is y = –2x + 48

... (2)

(1)

i.e., A is (–16, 0); B is (24, 0) Now, Centre of circle is (4, 0) (3) Now, mPC

T (0, 3) x Q

P

83. If L 1 is the line of intersection of the planes 2x – 2y + 3z – 2 = 0, x – y + z + 1 = 0 and L2 is the line of intersection of the planes x + 2y – z – 3 = 0, 3x – y + 2z – 1 = 0, then the distance of the origin from the plane containing the lines L1 and L2, is

4 3

Tangent at P(16, 16) is 2y = x + 16

y

1    6 5  15 2

(2) 2

(3) 3

x  3 5, y  12

4  3

1 4 2 1 2 2

Answer (2) 22

(2)

(4)

1 3 2 1 2


Sol. L1 is parallel to

Length of projection of the line segment on the plane is AC

iˆ jˆ kˆ 2 –2 3  iˆ  jˆ 1 –1 1

AC 2  AB 2  BC 2  2 

AC 2 

iˆ jˆ kˆ L2 is parallel to 1 2 –1  3iˆ – 5 jˆ – 7kˆ 3 –1 2

So, required plane is

1

5 7

y– 1

3

8 7

–5

z 0

Now, perpendicular distance 

162

(3)

2

(2)

3 1 3

 4(iˆ  2 jˆ  4kˆ )  ( ˆj  kˆ )  24   = –1 So, u  4(iˆ  2 ˆj  4kˆ )

2 3

 | u |2  336 86. A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is:

B (4, –1, 3) n=i+j+k

3 10

(2)

2 5

(3)

1 5

(4)

3 4

Answer (2)

Normal to the plane x + y + z = 7 is n  iˆ  ˆj  kˆ

Sol. E1 : Event that first ball drawn is red.

AB  iˆ  kˆ ⇒ | AB |  AB  2

E2 : Event that first ball drawn is black. E : Event that second ball drawn is red.

BC = Length of projection of AB on n  | AB  nˆ |



(1)

C

A (5, –1, 4)



 iˆ  kˆ 

 iˆ  ˆj  kˆ   3



 u  2 (2iˆ  4 ˆj  8kˆ ) as, u  b  24

Answer (4) Sol.

(4) 84



3 2

2 3

(4)

(3) 256

 u  (2a  14b )  2 (2iˆ  3 ˆj  kˆ )  7( ˆj  kˆ )

1

84. The length of the projection of the line segment joining the points (5, –1, 4) and (4, –1, 3) on the plane, x + y + z = 7 is: (1)

(2) 315

Sol. Clearly, u  (a  (a  b ))  u  ((a . b )a  | a |2 b )

–7



(1) 336

Answer (1)

0

 7x – 7y + 8z + 3 = 0

3

2 3

85. Let u be a vector coplanar with the vectors a  2iˆ  3 ˆj  kˆ and b  ˆj  kˆ . If u is perpendicular 2 to a and u  b  24 , then u is equal to

⎛5 8 ⎞ Also, L2 passes through ⎜ , , 0 ⎟ ⎝7 7 ⎠ x–

4 2  3 3

⎛E ⎞ ⎛ E ⎞ P (E )  P (E1 ).P ⎜ ⎟  P (E2 ).P ⎜ ⎟ ⎝ E1 ⎠ ⎝ E2 ⎠

2



3

23

4 6 6 4 2     10 12 10 12 5

JEE (MAIN)-2018 (Code-A) 9

87. If

9

∑ ( xi  5)  9 and ∑ ( xi  5)2  45 ,

i 1



then the

i 1

standard deviation of the 9 items x1, x2, ...., x9 is (1) 9

(2) 4

(3) 2

(4) 3

 k

∑ ( xi  5)2

i 1

9

⎛ 9 ⎞ ⎜ ∑ ( xi  5) ⎟ ⎜ i 1 ⎟ ⎜ ⎟ 9 ⎜ ⎟ ⎜ ⎟ ⎝ ⎠

13 9

13 9

89. PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P, Q and R are respectively 45º, 30º and 30º, then the height of the tower (in m) is

Sol. Standard deviation of xi – 5 is 9

 5 7 , , 9 9 9

 Sum 

Answer (3)



x

2

(1) 100

(2) 50

(3) 100 3

(4) 50 2

Answer (1)

P

   5 1  2

Sol.

As, standard deviation remains constant if observations are added/subtracted by a fixed quantity.

45º

So,  of xi is 2

T

88. If sum of all the solutions of the equation ⎛ ⎛ ⎞ ⎛ ⎞ 1⎞ 8 cos x  ⎜ cos ⎜  x ⎟  cos ⎜  x ⎟  ⎟  1 in [0, ] 6 6 ⎝ ⎠ ⎝ ⎠ 2⎠ ⎝ is k, then k is equal to :

(1)

(3)

2 3

(2)

8 9

(4)

Q

M Let height of tower TM be h

In TQM,

20 9

tan30º 

h QM

QM  3 h In PMQ,

 1⎞ ⎛ 8cos x  ⎜ cos2  sin2 x  ⎟  1 6 2 ⎝ ⎠

PM 2  QM 2  PQ 2

h2  ( 3h )2  2002

⎛3 1 2 ⎞  8cos x ⎜   1  cos x ⎟  1 4 2 ⎝ ⎠



4h 2  2002



h = 100 m

90. The Boolean expression ~ ( p  q )  (~ p  q ) is equivalent to

⎛ 3  4cos2 x ⎞ ⎟ 1  8cos x ⎜⎜ ⎟ 4 ⎝ ⎠  cos 3 x  1

(1) ~p

(2) p

(3) q

(4) ~q

Answer (1)

1  cos3 x  2

 3x 

30º

 PM = h

13 9

Answer (2) Sol.

30º

Sol. ∼ ( p  q )  ( ∼ p  q ) By property, ( ∼ p  ∼ q )  ( ∼ p  q )

 5 7 , , 3 3 3

= ~p

24

R