Solution Jee-Main (2018) - Prime Classes

Prime Classes Pvt. Ltd. : Kabir Nagar, Durgakund, Varanasi

Jee-Main (2018)

PHYSICS 1. 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 collision with carbon nucleus at rest, fractional loss of energy is pc. The values of prespectively: (1) (0, 0) (2) (0, 1) (3) (89, 28) (4) (28, 89)

Sol. Ans (3) Let initial speed of neutron is v0 and kinetic energy is K 1st collision

by momentum conservation

021210 vv2vmv2mvmv

by e = 1 012 vvv

3

vv;

3

v2v 0

10

2

Fractional loss = 20

2

020

mv2

1

3

vm

2

1mv

2

1

89.9

8Pd

2nd collision

By momentum conservation

210 mv12mvmv 021 vv12v

by e = 1 012 vvv

13

v11v;

13

v2v 0

10

2

Now fraction loss of energy

Kabir Nagar, Durgakund, Varanasi-05, Ph. (0542) – 2314896

Solution

Main (2018) CODE – B

It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional

; while for its similar collision with carbon nucleus at rest, fractional

. The values of pd and pc are

89)

and kinetic energy is K

169

48

mv2

1

13

v11m

2

1mv

2

1P

20

2

020e

2. The mass of a hydrogen molecule is 3.32 1023 hydrogen molecules strike, perof area 2 cm2 at an angle of 45º to the normal, and rebound elastically with a speed of 10pressure on the wall is nearly:(1) 2.35 102 N/m2 (3) 2.35 103 N/m2

Sol. Ans (3)

º45cosnmv2dt

dpF

Pressure Area

º45cosnmv2

A

F

4

32723

102

10103.3102

23 m/N1035.2 3. 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 a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the

sphree

r

dr, is:

(1) Ka3

mg

(3) mg

Ka

2314896

1

28.0169

48

The mass of a hydrogen molecule is 3.32 10—27 kg. If hydrogen molecules strike, per second, a fixed wall

at an angle of 45º to the normal, and rebound elastically with a speed of 103m/s, then the pressure on the wall is nearly:

(2) 4.70 102 N/m2 (4) 4.70 103 N/m2

º

2

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 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 radius of the

(2) Ka

mg

(4) mg3

Ka

Edited with the trial version of Foxit Advanced PDF Editor

To remove this notice, visit:www.foxitsoftware.com/shopping

http://www.foxitsoftware.com/shopping


Sol. Ans (1)

Bulk Modulus = strainvolumetric

stressvolumetric

V

dVa

mgK

)i(...KA

mg

V

dV

Volume of sphere 3R3

4V

Fractional change in volume ...(r

dr3

V

dV

Using sing equation (i) & (ii)

Ka3

mg

r

dr

Ka

mg

r

dr3

4. Two batteries with e.m.f. 12V and 13V are connected in parallel across a load resistor of 10resistances of the two batteries are 1respectively. The voltage across the load lies between:(1) 11.4V and 11.5V (2) 11.7V and 11.8V(3) 11.6V and 11.7V (4) 11.5V and 11.6V

Sol. Ans (4)

volt3

37

2

1

1

12

13

1

12

3

2

12

12req

Now its equivalent circuit is

32

37

3

210

3/37i

volt56.1132

37010

32

3710iV10

5. A particle is moving in a circular path of radius a under

the action of an attractive potential U

energy is:


)ii...(

batteries with e.m.f. 12V and 13V are connected in . The internal

resistances of the two batteries are 1 and 2 respectively. The voltage across the load lies between:

11.7V and 11.8V 11.5V and 11.6V

volt

A particle is moving in a circular path of radius a under

2r2

k . Its total

(1) Zero

(3) 2a4

k

Sol. Ans (1)

3r

K

r

uF

Since it is performing circular motion

2

2

3

2

r

Kmv

r

K

r

mvF

Total energy = P.E. + K.E.

6. Two masses m1 = 5kg and minextensible string over a frictionless 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:

(1) 43.3 kg (3) 18.3kg

Sol. Ans (4) 50 – T = 5 a

T – 0.15(m + 10)g = (10 + m)a a = 0 for rest 50 = 0.15(m + 10)10)

m3

1000)10m(

20

35

kg3.23m

7. If the series limit frequency of the Lyman series is vthen the series limit frequency of the Pfund series is:

(1) vL/16 (3) 25vL Sol. Ans 2

22

1

1

1

h

K

h

E

h

KL

2p

1

5

1

h

K

2314896

2

(2) 2a

k

2

3

(4) 2a2

k

Since it is performing circular motion

2

2

r2

Kmv

2

1.E.K

0r2

K

r2

K22

and m2 = 10kg, 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

to stop the motion is:

(2) 10.3kg

(4) 27.3kg

0.15(m + 10)g = (10 + m)a

10

If the series limit frequency of the Lyman series is vL, then the series limit frequency of the Pfund series is:

(2) vL/25 (4) 16vL

2

1=

25

1

h

Kp





8. 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 found to be

2

I. Now another identical polarizer C is placed between

A and B. The intensity beyond B is now found to be

The angle between polarizer A and C is: (1) 45º (2) 60º (3) 0º (4) 30º

Sol. Ans 1

20 cosII

201 cos

2

II

212 cosII

400 cos2

I

8

I

2

1cos º45

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

(1) 2n

2n BA (2) 2

n

(3) 2n

n

BA

(4) n A

Sol. Ans (3)

2n

2g

2gx

11

m2

h

hchcEEE

2

2g

2n

2

2n

2g B

A

m2

h

..hc

10. The reading of the ammeter for a silicon diode in the given circuit is:

(1) 11.5 mA (2) 13.5 mA(3) 0 (4) 15 mA

Sol. Ans (1) For silicon 7.0Vb

A200

3.2A

200

7.03i

mA5.11mA200

2300


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 found to be

. Now another identical polarizer C is placed between

intensity beyond B is now found to be 8

I.

º

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let n,

be the 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

state to the ground state. For large n, (A, B are

nBA

The reading of the ammeter for a silicon diode in the

13.5 mA

11. An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbital of radii re, rp, r respectively in a uniform magnetic field B. The relation between re, rp, p

(1) re < rp < r (3) re > rp = r

Sol. Ans. (4)

2

Km2R ,

e

Km2R

e

e

e

km2

e2

m4K2R

pp

RRR pe

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

material of dielectric constant

between the plates, the magnitude of the induced charge will be: (1) 2.4 nC (3) 1.2 nC

Sol. Ans. (3)

K

11QQb

5

311020

3

590 12

5

106

5

2103000

912

13. For an RLC circuit driven with voltage of

and frequency LC

10

resonance. The quality factor, Q is given by:

(1) )C

R

0

(3) R

L0

Sol. Ans. (3)

Quality factor

0)(

Where, L

R

R

LQ 0

14. A telephonic communication carrier frequency of 10GHz. Only 10% of it is utilized for transmission. How many telephonic channels cantransmitted simultaneously if each channel requires a bandwidth of 5 kHz:

(1) 2 105

(3) 2 103

2314896

3

An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbital of

respectively in a uniform magnetic field B.

is: (2) re < r < rp (4) re < rp = r

,

e

Km2R

p

p

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

material of dielectric constant 3

5K is inserted

between the plates, the magnitude of the induced charge

(2) 0.9 nC (4) 0.3 nC

nC2.1C

For an RLC circuit driven with voltage of amplitude vm

LC

1the current exhibits

resonance. The quality factor, Q is given by:

(2) 0

CR

(4) L

R0

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

(2) 2 106

(4) 2 104





Sol. Ans. (1)

Utilized frequency GHz110

10

Band width = 5KH

No. channels = 3

9

105

101

widthBand

Range

15. A granite rod of 60cm length is clamped at its middle point and is set into longitudinal vibrations. The densityof 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) 10 kHz (2) 7.5 kHz(3) 5 kHz (d) 2.5 kHz

Sol. Ans. (3)

Wave velocity (v) =

Y3

10

107.2

1027.9

0

0

V

5K Hz

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

(1) 2MR2

73 (2) MR

2

181

(3) 2MR2

19 (4) MR

2

55

Sol. Ans. (2) M.I. about required axis

2cm mdII

222 )R3(m7])R2(m[6mR2

7

222 mR63mR24mR2

7

2mR2

181I

17. 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)

a

b

cb 22

0

(2)

b2

0

(3)

c

a

ba 22

0

(4)

a 2

0


5102

A granite rod of 60cm length is clamped at its middle point and is set into longitudinal vibrations. The density

and its Young’s modulus is Pa. What will be the fundamental frequency

(2) 7.5 kHz (d) 2.5 kHz

10

Seven identical circular planar disks, each of mass M symmetrically as shown. The

moment of inertial of the arrangement about the axis normal to the plane and passing through the point P is:

2MR

2MR

Three concentric metal shells A, B and C of respective radii, a, b and c (a < b < c) have surface charge densities

respectively. The potential of shell B is:

a

c

c2

c

b

b2

Sol. Ans. (4) Potential of shell of Radius b

c

q

b

q

b

qKV CBA

c

q

b

qqK CBA

4

b

b4a4

4

1 22

0

c

b

ba 22

0

18. In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52cm of the potentiometer wire. If the cell is shunted by a resistaof 5, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell: (1) 2 (3) 1

Sol. Ans. (4) 52 ….(1) : potential gradient

40r

5r… (2)

40

5r

5

52

40

5r

5

200r40260

5.12

3r

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

fields are

2cosxEE 011

)]ctz2(kcos[xEE 02

in medium, where the wave

number k and frequency v refer to their values in air. The medium is non-magnetic. If

relative permittivities of air and medium respectively, which of the following is corr

(1) 4

1

2

1

r

r

(3) 42

1

r

r

Sol. Ans. (1)

t2

c

2.zcosEE i01

CKtzK2cosEE2o2

Frequency of light =

2314896

4

Potential of shell of Radius b

c

c4 2

In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52cm of the potentiometer wire. If the cell is shunted by a resistance

, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of

(2) 2.5 (4) 1.5

: potential gradient

An EM wave from air enters a medium. The electric

t

c

zv2 in air and

in medium, where the wave

number k and frequency v refer to their values in air. magnetic. If

1r and

2r refer to

relative permittivities of air and medium respectively, which of the following is correct:

(2) 2

1

2

1

r

r

(4) 22

1

r

r





Wavelength

C

)( o

KK2

2

20

2

2

1

1

221

1

4

1

2

1

20. The angular width of the centre maximum in a single slit diffraction pattern is 60º. The width of the slit is 1m. The slit is illuminated by monochromatic plane waves. If another alit 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 1cm, what is slit separation distance: (1) 75m (b) 100 m(3) 25 m (d) 50 m

Sol. Ans. (3)

2

sinb

m2

1130sinb

Fringe width : d

D

d

1050105.0m10

262

6105.2d

m25d

21. A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 10What 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) 2.2 N/m (2) 5.5 N/m(3) 6.4 N/m (4) 7.1 N/m

Sol. Ans. (4)

m

Kn2

mn4K 20N

mm

23

324

1002.6

10810108.94

.7

22. From a uniform circular disc of radius R

a small disc of radius 3

R removed as shown 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 the disc is:


The angular width of the centre maximum in a single slit diffraction pattern is 60º. The width of the slit is

m. The slit is illuminated by monochromatic plane waves. If another alit 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

ance: m

oscillates in simple harmonic motion in some direction with a frequency of 1012/sec. What is the force constant of the bonds connecting one atom with the other (Mole wt of silver = 108 and

gm mole—1): ) 5.5 N/m ) 7.1 N/m

m/N1.

circular disc of radius R and mass 9M,

as shown in the

figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and

(1) 10 MR2

(3) 4 MR2

Sol. Ans. (3) Moment of inertial of remaining disc

22 2m

3

R

2

m

2

qmRI

9

mR4

18

mR

2

qmR 222

222

2

mR9

18

qmR

2

mR9

23. 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) 2

v0

(3) 4

v0

Sol. Ans. (4)

After collision

20

22

21

4

1mv

2

1mv

2

1mv

2

1

20

22

21 v

2

3vv

021 vvv 221 )vv(

20

2021 v

2

3vvv2 v2 1

2

vv

2

3)vv(

202

02

21 v(

24. The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B1. When the dipole moment is doubled by keeping the current constant, the magnetic field at t

loop is B2. The ratio 2

1

B

Bis:

(1) 2

(3) 2

2314896

5

(2) 2MR9

37

(3) 2MR9

40

Moment of inertial of remaining disc

2

3

R2

22

mR42

mR

In a collinear collision, a particle with an initial speed 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:

(2) 2

v0

(4) 0v2

20mv

4

1

20v

2

vv

20

2

021 v2)vv

circular loop carrying a current I, is m and the magnetic field at the centre of the loop is

. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the

(2) 2

1

(4) 3





Sol. Ans. (1)

Magnetic Moment (m) = i : 2R

R2

iB 0

1

22 .iRi.2'R.m2

New radius of loop is R2'R

R22

iB 0

2

2B

B

2

1

25. 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 error in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is: (1) 4.5% (2) 6% (3) 2.5% (4) 3.5%

Sol. Ans. (1)

3a

m

% error in = % error in m + 3 % error in length = 1.5 + 3 1 = 4.5% 26. On interchanging the resistance, the balance point of a

metre bridge shifts to the left by 10 cm. The resistance of their series combination is 1 k . Howresistance on the left slot before interchanging the resistances: (1) 550 (2) 910 (3) 990 (4) 505

Sol. Ans. (1) On interchanging the resistance shifting is 10cm.

Therefore, –10 = 100 – 551102

1000RR9

11

45

55

R

R21

2

1

1150R1000R11

9R 111 R

27. In an a.c. circuit, the instantaneous e.m.f. and current are given by

e = 100 sin 30t

30sin20i

In one cycle of a.c., the average power consumed by the circuit and the wattles current are, respectively:

(1) 0,2

50 (2) 50, 0


2'R

The density of a material in the shape of a cube is sides of the cube and its

in measuring the mass and length are respectively 1.5% and 1%, the maximum

% error in length

, the balance point of a metre bridge shifts to the left by 10 cm. The resistance

How much was the resistance on the left slot before interchanging the

On interchanging the resistance shifting is 10cm.

550R1

In an a.c. circuit, the instantaneous e.m.f. and current

4t30

e power consumed by the circuit and the wattles current are, respectively:

(3) 50, 10

Sol. Ans. (4)

cos2

10020cos

2

viP 00

avg

sinii rms

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

Sol. Ans. (1) Position time graph is parabolic opening downward. Velocity time graph will be straight line with (

slope and velocity position graph will be parabola on –5 axis.

Distance time graph is incorrect. 29. Two moles of an ideal monoatomic gas occupies a

volume V at 27ºC. The gas expands adiabatically tovolume 2V. Calculate (a) the final temperature of (b) change in its internal energy:

(1) (a) 189K (b) –27kJ (3) (a) 189K (b) 2.7kJ Sol. Ans. (1) For adiabatic process

.constTV 1

1g

22

1

11 VTVT

3

5

23/2

23/2 T)V2(TV300

30. 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:

(a) T 2/)R 1n(

(c) T 2/3R for any n Sol. Ans. (1)

n

2

n R

K

R

mv

R

KF

1n CRmRK

R2

V

R2T

2314896

6

(4) 10,2

1000

º45cos2

1000 ,

intended to represent the same motion. One of them does it incorrectly. Pick it up:

Position time graph is parabolic opening downward. Velocity time graph will be straight line with (–)ve

slope and velocity position graph will be parabola on

Distance time graph is incorrect. Two moles of an ideal monoatomic gas occupies a volume V at 27ºC. The gas expands adiabatically to a volume 2V. Calculate (a) the final temperature of (b)

(2) (a) 195K (b) 2.7 kJ

(4) (a) 195K (b) –2.7k

3

5

3/22

300 and TnR

2

3U

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely

power of R. If the period of rotation of the particle is T, then:

(b) T 2/nR

(d) T 1

2

n

R

1nmR

KV

2/)1n(R.CR





2/)1n(2

21n

CRTCRT

T 2/)1n(R MATHEMATICS

31. If the tangent at (1, 7) to the curve x2 = y the circle x2 +y2 + 16x + 12y+c = 0 then the value of c is: (1) 85 (2) 95 (3) 195 (4) 185

Sol. Ans. (2)

Equation of tangent 62

)7y(1.x

2x – y + 5 = 0

By using condition of tangency for circle

We get c = 95 32. If L1 is the line of intersection of the planes

2x – 2y + 3z – 2 = 0, x – y + z + 1 = 0 and Lline of intersection of the planes x + 2y 3x – y + 2z – 1 = 0, then the distance of the origin from the plane, containing the lines L1 and L2

(1) 22

1 (2)

2

1

(3) 24

1 (4)

23

1

Sol. Ans. (4) Equation of required plane

)1zyx(2z3y2x2P

z)3(y)2(x)2(P

To find direction cosines of line L2 a + 2b – c = 0 3a – b + 2c = 0

13

21

c

32

11

b

21

12

a

7

c

5

b

3

a

Plane P contains line L2, therefore

0)3(7)2(5)2(3

5

Equation of plane is 7x – 7y + 8z + 3 = 0

Length of perpendicular = 877

000

22

33. If C, are the distinct roots, of the equation

1071012 then,01xx is equal to:

(1) 1 (2) 2 (3) – 1 (4) 0

Sol. Ans. (1)

2,


= y – 6 touches the circle x2 +y2 + 16x + 12y+c = 0 then the value of c

By using condition of tangency for circle p = r

is the line of intersection of the planes y + z + 1 = 0 and L2 is the

line of intersection of the planes x + 2y – z – 3 = 0, 1 = 0, then the distance of the origin from

is:

Equation of required plane

0

02

7y + 8z + 3 = 0

23

1

8

3

2

are the distinct roots, of the equation

is equal to:

1072101 )()( ( 2

34. Tangents are drawn to the hyperbola

the points P and Q. If these tangents intersect at the point T(0, 3) then the area (in sq. units) of

(1) 360

(3) 545

Sol. Ans. (3)

136

y

9

x 22

Equation of chord of contact PQ is y + 12 = 0 To find P and Q solve y + 12 = 0

and 36yx4 22

)12,53(P

)12,53(Q

Area TSPQ2

1a

15562

1 = 545

35. If the curves y2 = 6x, 9x2 + byat right angles, then the value of b is:

(1) 4

(3) 6

Sol. Ans. (2)

Point of intersection ),(

y2 = 6x

y

3

dx

dy

3

dx

dy),(

16y6x9 22

b

9

dx

dy),(

1b

93

put 62 b = 9/2

36. If the system of linear equations x + ky + 3z = 0 3x + ky – 2z = 0 2x + 4y – 3z = 0

has a non-zero solution (x, y, z) then

(1) – 30 (3) – 10

2314896

7

) = 1

Tangents are drawn to the hyperbola 36yx4 22 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:

(2) 536

(4) 354

Equation of chord of contact PQ is y + 12 = 0 To find P and Q solve y + 12 = 0

+ by2 = 16 intersect each other at right angles, then the value of b is:

(2) 2

9

(4) 2

7

(say)

If the system of linear equations

zero solution (x, y, z) then 2y

xz is equal to:

(2) 30 (4) 10





Sol. Ans. (4) For non zero solution 0

0

342

2K3

3K1

k = 11

Now equations x + 11y + 3z = 0 ………..(i) 3x + 11y – 2z = 0 ……….(ii) 2x + 4y – 3z = 0 ……….(iii) (i) and (ii) - 2x + 5z = 0

5z = 2x z2

5x

From (iii) 0z3y4z2

52

2z + 4y = 0

Z + 2y = 0 2

zy

Now 2y

xz104

2

5

4

z

zz2

5

2

37. Let 0x:Rx{S and

}.06)6x(x3x2 Then S :

(1) Contains exactly two elements (2) Contains exactly four elements (3) Is an empty set (4) Contains exactly one element

Sol. Ans. (1) 0x

06)6x(x3x2

Case-I : x [9, ]

06x6x)3x(2

0x4x x = 0 (rejected), x = 16 Case-II x [0, 9]

06x6x)x3(2

0)2x()6x(

x = 36 (rejected), x = 4 exactly two elements S = {4, 16} 38. If sum of all the solutions of the equation

12

1x

6cos.x

6cos.xcos8

,k then k is equal to:

(1) 9

8 (2)

9

20


………..(i) ……….(ii) ……….(iii)

Then S :

If sum of all the solutions of the equation

1 in ],0[ is

(3) 3

2

Sol. Ans. (4)

sinxcos6

cos

xsin6

sinxcos6

cosxcos8

xsin4

1xcos

4

3xcos8 22

1)xsin41(xcos2 2

2

1)xcos43(xcos 2

2

1x3cos

9

7,

9

5,

9x

Sum of solutions =

k9

13

39. 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:

(1) 5

1

(3) 10

3

Sol. Ans. (4)

red4

P (Red in IInd drawn) = P(Red in second drawn if in I

drawn black in observed )

+ P(Red in II

drawn red in observed )

=

1

1

21

B

RP).B(P

R

RP).R(P

=12

4

10

6

12

6

10

4 =

5

2

120

48

2314896

8

(4) 9

13

12

1xsin

6sin

x

12

1

= 9

13k

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

bag. If now a ball is drawn at random

from the bag, then the probability that this drawn ball is

(2) 4

3

(4) 5

2

black6

drawn) = P(Red in second drawn if in Ist

+ P(Red in IInd drawn if in Ist

1

2

B

R

5

2





40. Let ,x

1x)x(gand

x

1x)x(f

2

2

}1,0,1{Rx . If ,)x(g

)x(f)x(h then the local

minimum value of h(x) is:

(1) 22 (2) 22 (3) 3 (4) – 3 Sol. Ans. (2)

x/1x

x/1x)x(h

22

x

1x

2

x

1x)x(h

22

x

1x

21

x

11)x('h

For maxima and minima h’(x) = 0

2x

1x

h’(x) changes its sign from (-)ve to (+)ve at

2x

1x

therefore local minimum value of

222

22)x(h

41. Two sets A and B are as under:

5band15a:RR)b,a{(A

)5b(9)6a(4:RR)b,a{(B 22

(1) BA (an empty set)

(2) neither ABnorBA

(3) AB (4) BA Sol. (4) band)6,4(a:RR)b,a{(A

for B

14

)5b(

9

)6a( 22

Put a – 6 = x and b – 5 = y yand)0,2(x:RR)y,x{(A

B

14

y

9

x:}RR)y,x(

22

BA


then the local

)ve to (+)ve at

therefore local minimum value of

}1

}362 . Then:

)}6,4(

)}1,1( &

42. The Boolean expression

equivalent to: (1) q (3) ~ p

Sol. Ans. (3) )qp(~)qp(~

)qp(~)q~p(~

)qq(~p~ p~

43. 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 value of tan

(1) 3

(3) 2

1

Sol. Ans. (4) Equation of tangent

2

16x1616.y

x – 2y + 16 = 0 Equation of normal y – 16 = - 2x + y – 48 = 0 A : (-16, 0), B : (24, 0)

516AP

58BP

AB = 40 APB is a right angle triangle

3

4

416

016mCP

22416

016mBP

2m.m1

mmtan

BPCP

BPCP

44. If A(

4xx2x2

x24xx2

x2x24x

ordered pair (A, B) is equal to: (1) (-4, 5) (3) (-4, -5)

Sol. Ans. (1) To find A, put x = 0 A = - 4

2314896

9

The Boolean expression )qp(~)qp(~ is

(2) ~q (4) p

Tangent and normal are drawn at P(16, 16) on the = 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

tan is:

(2) 3

4

(4) 2

-2(x - 16)

APB is a right angle triangle

,)Ax()BxA 2 then the

ordered pair (A, B) is equal to: (2) (4, 5) (4) (-4, 3)

4





To find B, first differentiate and then put x = 0 B = 5 45. The sum of the co-efficient of all odd degree terms in

the expansion of

)1x(,1xx1xx5

35

3

(1) 1 (2) 2 (3) – 1 (4) 0

Sol. Ans. (2) Using (x +a)5 + (x -a)5

]a.xCa.xCxC[2 44

5232

550

5

53

53 1xx1xx

x(xC)1x(xCxC[2 34

5332

550

5

5x10x5x10x10x[2 47365

Considering odd degree terms,

]x5x10x5x[2 375

sum of coefficient of odd terms is 2 46. Let a1, a2, a3,…,a49 be in A.P. such that

.66aaand416a 439

12

0k1k4

,m140a...aa 217

22

21 then m is equal to:

(1) 34 (2) 33 (3) 66 (4) 68

Sol. Ans. (1)

416a12

0K1k4

416]d48a2[2

131

32d24a1 …………..(1)

66d50a266aa 1439 …………..(2)

64d48a2 1 by …………..(1)

d = 1 and a1 = 8

17

1r

17

1r

22r ]1).1r(8[am140

17

1r

2)7r(m140

24

1r

2rm140

6

15.8.7

6

49.25.24m140

]3105[6

5.8.7m140 280m140

47. A straight line through a fixed point (2, 3) intersect the coordinates axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is: (1) 3x + 2y = xy (2) 3x + 2y = 6xy(3) 3x + 2y = 6 (4) 2x + 3y = xy

Sol. Ans. (1) Equation of line


To find B, first differentiate and then put x = 0

efficient of all odd degree terms in

) is:

])1 23

]x5

be in A.P. such that

If

then m is equal to:

…………..(1)

…………..(2)

by …………..(1)

7

1r

2r

34m17.280

line through a fixed point (2, 3) intersect the coordinates axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the

(2) 3x + 2y = 6xy (4) 2x + 3y = xy

y – 3 = m (x - 2)

0,

m

3m2:P ,

)m23,0(:Q

m

3m2h

k = 3 – 2m Eliminate ‘m’ hK = 3h + 2K 3x + 2y = xy

48. The value of dx21

xsin2

2

x

2

is:

(1) 4

(3) 8

Sol. Ans. (2)

2/

2/x

2

dx21

xsin

Using property fdx)x(fb

a

b

a

dx21

xsinI

2/

2/x

2

dx21

xsin2I

2/

2/x

2x

(i) + (ii)

2/

2/

2 dxxsinI2

2I2

49. Let x)x(f,xcos)x(g 2

roots of the quadratic equation Then the area (in sq. units) bounded by the curve

)x)(f0g(y and the lines x

(1) )23(2

1

(3) )13(2

1

Sol Ans. (3)

0x9x18 22

3/,6/

3/

6/

dxxcosA 3(2

1A

50. For each ,Rt let [t] be the greatest integer less than

or equal to t. Then

2314896

10

………(i)

………(ii)

is:

(2) 4

(4) 2

………(i)

dx)xba(f

………(ii)

2/

0

2 dxxsin 4/I

,x and )(, be the

roots of the quadratic equation .0x9x18 22 Then the area (in sq. units) bounded by the curve

,0yandx,x is:

(2) )12(2

1

(4) )13(2

1

)13

let [t] be the greatest integer less than





x

15....

x

2

x

1xlim

0x

(1) is equal to 120 (2) does not exist (in R) (3) is equal to 0 (4) is equal to 15 Sol. Ans. (1)

x

1xlim

0x

x

1

x

11

x

1

1x

1xx1

1limx

1xlimx1lim

0x0x0x

1x

1xlim

0x

Similarly 2x

2xlim

0x

Ans. 1 +2+3+….+ 15 = 120

51. If

9

1i

2i

9

1ii ,45)5x(and9)5x( then the standard

deviation of the 9 items x1, x2,..,x9 is: (1) 2 (2) 3 (3) 9 (4) 4

Sol Ans. (1)

2

i2

iD

9

)5x(

9

)5x(S

29

36

52. The integral

xcosxsinxsinxcosx(sin

xcosxsin23235

22

is equal to:

(1) Cxcot1

13

(2) cot1

13

(3) C)xtan1(3

13

(4) tan1(3

1

(where C is a constant of integration ) Sol. Ans. (4)

xcosxsinxsinxcosx(sin

xcosxsin23235

22

= dx)1xtanxtanx(tan

xsecxcosxsin2325

1022

dx)xtan1()xtan1(

xsecxsin2322

82


(2) does not exist (in R) (4) is equal to 15

then the standard

2

9

9

9

45

dx

)xcos 25

Cx3

C)xtan

13

dx)xcos 25

dx)xtan1(xsec

xsecxtan234

62

dx)xtan1(

xsecxtan23

22

txtan1 3

dtxdxsec.xtan3 22

2t

dt

3

1c

)xtan1(3

13

53. Let x)x(f:Rt{S

differentiable at t}. Then the set S is equal to:(1) }{

(3) (an empty set)

Sol. Ans (3)

xsin)1e(x)x(fx

We check differentiability at

xat

R.H.D. = e(h

lim0h

= 0

L.H.D. = e(h

lim0h

= 0 RHD = LHD so function is differentiable at x at x = 0

h

)1e(hlimD.H.R

h

0h

h

e(hlimD.H.L

h

0h

RHD = LHD so function is differentiable at x = 0 set S is empty set,

54. Let y= y(x) be the solution of the differentiable equation

0(x,x4xcosydx

dyxsin

If ,02

y

then

6y is equal to:

(1) 2

9

8

(3) 2

39

4

Sol. Ans. (1) xdx4dxcosyxdysin

dxx4)xsiny( '

2314896

11

c

xsin)1e(.x

is not

differentiable at t}. Then the set S is equal to: (2) },0{

(4) {0}

We check differentiability at 0xandx

h

0hsin)1h

h

0hsin)1h

RHD = LHD so function is differentiable

00hsin)

0h

0hsin)1

RHD = LHD so function is differentiable at x = 0

Let y= y(x) be the solution of the differentiable equation

).,0

is equal to:

(2) 2

9

4

(4) 2

39

8

),0(x





dxx4)'xsiny(

cx2xsiny 2 …………..(1)

02

y

2

c2

From (1) 2

x2xsiny2

2

6y

55. Let u

be a vector coplanar with the vectors

.kjbandkj3i2a

If u

is perpendicular to

,24b.uanda

then 2

u

is equal to:

(1) 256 (2) 84 (3) 336 (4) 315

Sol. Ans. (3)

)kj(y)kj3i2(xu

k)xy(j)yx3(ix2u

0a.u

7x + y = 0 ………..(i)

24b.u

x + y = 12 ………(ii) on solving (i) and (ii) x = -2, y = 14

336u2

56. 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) 3

1 (2)

3

2

(3) 3

2 (4)

3

2

Sol. Ans. (2) Foot of perpendicular drawn from (5, 1, 4)

222

11

111

)7415(

1

4

1

1

1

5

3

11,

3

4,

3

14111

Foot of perpendicular drawn from (4, -1, 3)

222

222

111

314(4

1

3

1

1

1

4

3

10,

3

2,

3

13222

Length of projection

221

221

221 )()()(

57. PQR is a triangular park with PQ = PR = 200m. A T.V. tower stands at the mid-point of QR. If the angle of


…………..(1)

9

8 2

be a vector coplanar with the vectors

is perpendicular to

………..(i)

………(ii)

The length of the projection of the line segment joining 1, 3) on the plane , x + y + z

Foot of perpendicular drawn from (5, 1, 4)

)

1, 3)

2

)7

3

2

PQR is a triangular park with PQ = PR = 200m. A T.V. point of QR. If the angle 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:

(1) 3100

(3) 100

Sol. Ans. (3)

Let height of tower MN is ‘h’ in QMN

030tanQM

MN

MRh3QM

Now in MNP MN = PM In PMQ

22 )h3()200(MP

From (2)

hh)h3()200( 22

58. 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 arrangement is: (a) at least 500 but less than 750(b) at least 750 but less than 1000(c) at least 1000 (d) less than 500

Sol. Ans. (3)

1CC 13

46 4 = 1080

Answer will be at least 1000 59. Let A be the sum of the first 20 terms and B be the sum

of the first 40 terms of the series

254.232.21 22222 If B – 2A = 100 , then is equal to:

(1) 464 (3) 232 Sol. Ans. (4)

54.232.21 22222

)19....31(SA 22220

)20....321( 2222

= 6

211110(2

6

412120 2

= 20774170

2314896

12

elevation of the top of the tower at P, Q and R are respectively 45º, 30º and 30º, then the height of the

(2) 250

(4) 50

of tower MN is ‘h’

………….(1)

…………..(2)

m100

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 arrangement is: (a) at least 500 but less than 750 (b) at least 750 but less than 1000

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

....6.2 2 is equal to:

(2) 496 (4) 248

....6.2 2

)20...42.(2 222

)20......42( 222

)21





42(.2)39.....31(SB 222240

41740274120 70241740274120A2B

)77287(40)140540(41 = 16400 + 8400

= 24810024800

60. Let the orthocentre and centroid of a triangle be A(

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:

(1) 2

53 (2)

2

53

(3) 10 (4) 102

Sol. Ans. (1)

612

323

12

523

A(-3, 5), C : (6, 2)

Radius = 2

53)25()36(

2

1 22

CHEMISTRY

61. Total number of lone pair of electrons in

(1) 9 (b) 12 (3) 3 (d) 6

Sol. Ans. (1) Total number of lone pair of electrons

in 3I is 9

62. Which of the following salts is the most basic in

aqueous solution: (1) FeCl3 (2) Pb(CH3

(3) Al(CN)3 (4) CH3COOKSol. Ans. (4)

63. Phenol reacts with methyl chloroformate in the presence

of NaOH to form product A. A reacts with Brproduct B. A and B are respectively:

(1) (2)


)40...4 22

77404170 = 16400 + 8400

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

2

Total number of lone pair of electrons in 3I is:

Which of the following salts is the most basic in

3COO)2 COOK

reacts with methyl chloroformate in the presence

of NaOH to form product A. A reacts with Br2 to form

(3)

(4)

Sol. Ans. (1) 64. The increasing order of basicity of the following

compounds is:

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

Sol. Ans. (1)

65. An alkali is titrated against an acid with methyl orange

an indicator, which of the following is a correct combination: Base Acid (1) Weak Strong (2) Strong Strong (3) Weak Strong (4) Strong Strong

Sol. (1)

2314896

13

The increasing order of basicity of the following

(b) (d) < (b) < (a) < (c) (d) (b) < (a) < (c) < (d)

An alkali is titrated against an acid with methyl orange an indicator, which of the following is a correct

End point Yellow to pinkish red Pink to colourless

Colourless to pink Pinkish red to yellow





66. The trans-alkenes are formed by the reduction of

alkynes with: (1) Na/liq. NH3 (2) Sn – HCl (3) H2 – Pd/C, BaSO4 (4) NaBH4

Sol. Ans. (1)

67. 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 CXHY completely to CO2 and H2O. The empirical formula of compound CXHYOZ is: (1) C3H4O2 (2) C2H4O3

(3) C3H6O3 (4) C2H4O Sol. Ans (2)


alkenes are formed by the reduction of

HCl

The ratio of mass percent of C and H of an organic

) is 6 : 1. If one molecule of the ) contains half as much

oxygen as required to burn one molecule of compound O. The empirical

3

68. Hydrogen peroxide oxidizes

36 ])CN(Fe[ in acidic medium but reduces

36 ])CN(Fe[ to 4

6 ])CN(Fe[

other products formed are respectively:(1) H2O and (H2O + O2)

(2) H2O and (H2O + OH ) (3) (H2O + O2) and H2O

(4) (H2O + O2) and (H2O + OH

Sol. Ans. (1)

69. The major product formed in the following reaction is:

(1) (3) Sol. Ans. (2)

2314896

14

Hydrogen peroxide oxidizes 4

6 ])CN(Fe[ to

in acidic medium but reduces

in alkaline medium. The

other products formed are respectively:

OH )

The major product formed in the following reaction is:

(2)

(4)





70. How long (approximate) should water be electrolysed

by passing through 100 amperes current so that the oxygen released can completely burn 27.66g of diborane (Atomic weight of B = 10.8u): (1) 3.2 hours (2) 1.6 hours(3) 6.4 hours (4) 0.8 hours

Sol. Ans. (1)

71. Which of the following lines correctly show the

temperature dependence of equilibrium an exothermic reaction:

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

Sol. Ans. (3)

72. At 518ºC, 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


How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66g of

(2) 1.6 hours (4) 0.8 hours

Which of the following lines correctly show the temperature dependence of equilibrium constant, K, for

(2) A and D (4) B and C

At 518ºC, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363

when 5% had reacted and 0.5

Torr s—1 when 33% had reacted. The order of the reaction is: (1) 1 (3) 2

Sol. Ans. (3)

73. Glucose on prolonged heating with HI gives:

(1) Hexanoic acid (3) n-Hexane

Sol. Ans. (3)

74. Consider the following reaction and statements:

243 Co[Br]Br)NH(Co[

(1) III and IV (3) I and II Sol. Ans. (4)

2314896

15

when 33% had reacted. The order of the

(2) 0 (4) 3

Glucose on prolonged heating with HI gives:

(2) 6-iodohexanal (4) 1-Hexane

Consider the following reaction and statements:

3333 NH]Br)NH(Co

(2) II and IV

(4) I and III





75. The major product of the following reaction is:

(1) (2)

(3) (4) Sol. Ans. (4)

76. Phenol on treatment with CO2 in the presence of NaOH

followed by acidification produces compound X as the major product. X on treatment with (CHpresence of catalytic amount of H2SO4 produces:

(1) (2)


The major product of the following reaction is:

in the presence of NaOH

followed by acidification produces compound X as the major product. X on treatment with (CH3CO)2O in the

produces:

(3)

Sol. Ans. (3)

77. An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1M solution Na2SO4 is added, BaSO4 just beings to final volume is 500 mL. The solubility product of BaSO4 is 1 10—10. What is the original concentrationof Ba2+: (1) 1.1 10—9 M (3) 5 10—9

Sol. Ans. (1)

78. Which of the following compounds will be suitable for

Kjeldahl’s method for nitrogen estimation:

2314896

16

(4)

An aqueous solution contains an unknown

. When 50 mL of a 1M solution just beings to precipitate. The

final volume is 500 mL. The solubility product of . What is the original concentration

(2) 1.0 10—10M (4) 2 10—9 M

Which of the following compounds will be suitable for Kjeldahl’s method for nitrogen estimation:





(1) (2) (3) (4)

Sol. Ans. (4)

79. 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:(1) Al (2) Fe (3) Zn (4) Ca

Sol. Ans. (1)

80. An aqueous solution contains 0.10 M H

HCl. If the equilibrium constants for the formation of HS from H2S is 1.0 10—7 and that of

ions is 1.2 10—23 then the concentration of

aqueous solution is:

(1) 6 10—21 (2) 5 10—

(3) 5 10—8 (4) 3 10—

Sol. Ans. (4)


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:

An aqueous solution contains 0.10 M H2S and 0.20 M

HCl. If the equilibrium constants for the formation of

and that of 2S from HS

entration of 2S ions in

—19 —20

81. The recommended concentration of fluoride ion in

drinking water is up to 1ppm as fluoride ion is required

to make teeth enamel harder by converting

])OH(Ca.)PO(Ca3[ 2243 to:

(1) ]CaF.)PO(Ca3[ 2243

(3) ]CaF[ 2

Sol. Ans. (1)

82. The compound that does not produce nitrogen gas by

the thermal decomposition is:

(1) NH4NO2

(3) Ba(N3)2

Sol. Ans. (2)

83. The predominant form of histamine present in human

blood in (pKa, Histidine = 6.0) (1) (3)

2314896

17

The recommended concentration of fluoride ion in

drinking water is up to 1ppm as fluoride ion is required

to make teeth enamel harder by converting

(2) ]CaF].)OH(Ca(3[ 22

(4) ])OH(Ca).CaF(3[ 22

The compound that does not produce nitrogen gas by

the thermal decomposition is:

(2) (NH4)2SO4

(4) (NH4)2Cr2O7

The predominant form of histamine present in human , Histidine = 6.0)

(2)

(4)





Sol. Ans. (2)

84. The oxidation state of Cr in [

266 )HC(Cr[ and ()CN(Cr[K 22

respectively are: (1) +3, 0, and +6 (2) +3, 0 and +4(4) +3, +4 and +6 (4) +3, +2 and +4

Sol. Ans. (1)

85. Which types of ‘defect’ has the presence of cations in

the interstitial sites: (1) Frenkel defect (2) Metal deficiency defect(3) Schottky defect (4) Vacancy defect

Sol. Ans. (1)


,Cl])OH(Cr[ 362

)NH)(O()O( 322

(2) +3, 0 and +4 +2 and +4

Which types of ‘defect’ has the presence of cations in

(2) Metal deficiency defect (4) Vacancy defect

86. The combustion of benzene (I) gives COH2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ molcombustion (in kJ mol—1) of benzene at constant pressure will be (R = 8.314 JK(1) 3260 (3) 4152.6

Sol. Ans. (2)

87. Which of the following are Lewis acids:

(1) PH3 and SiCl4 (3) PH3 and BCl3

Sol. (2)

88. Which of the following compounds contain(s) no

covalent bond(s): KCl, PH3, O2, B2H6, H2SO4 (1) KCl (3) KCl, B2H6, PH3

Sol. Ans. (1)

89. For 1 molal aqueous solution of the following

compounds, which one will show the highest freezing point: (1) OH2.Cl]Cl)OH(Co[ 2242

(2) OH3].Cl)OH(Co[ 2332

(3) 362 Cl])OH(Co[

(4) OH.Cl]Cl)OH(Co[ 2252

Sol. Ans. (2)

2314896

18

The combustion of benzene (I) gives CO2(g) and O(l). Given that heat of combustion of benzene at

3263.9 kJ mol—1 at 25ºC, heat of ) of benzene at constant

pressure will be (R = 8.314 JK—1 mol—1) : (2) –3267.6

(4) –452.46

Which of the following are Lewis acids:

(2) BCl3 and AlCl3 (4) AlCl3 and SiCl4

of the following compounds contain(s) no

(2) KCl, B2H6 (4) KCl, H2SO4

For 1 molal aqueous solution of the following compounds, which one will show the highest freezing

O





90. According to molecular orbital theory, which of the

following will not be a viable molecule:

(1) 2H (2) 2

2H

(3) 22He (4)

2He

Sol. Ans. (2)


According to molecular orbital theory, which of the following will not be a viable molecule:

2314896

19




Documents

Solution Jee-Main (2018) - Prime Classes