PHYSICSorigenceonline.com/uploadedfiles/CBCT-1 (13th Jan) (Shift...1 PHYSICS 1. A 2 kg block is lying on a smooth table which is connected by a body of mass 1 kg by a string which

1

PHYSICS 1. A 2 kg block is lying on a smooth table which is connected by a body of mass 1 kg by a string which

passes through a pulley. The 1 kg mass is hanging vertically. The acceleration of block and tension in the string will be

(A) 23.27m / s ,6.54N (B) 24.38m / s ,6.54N

(C) 23.27m / s ,9.86N (D) 24.38m / s ,9.86N

Solution (A):

Acceleration 2

1 2

mg

m m

21

9.8 3.27 m/s2 1

and 1T m a 2 3.27 6.54 N

2. Two forces of magnitude F have a resultant of the same magnitude F. The angle between the two forces is (A) 45° (B) 120° (C) 150° (D) 60° Solution (B):

2 2F (F) (F) 2F.Fcos 120

3. Figure shows the displacement of a particle going along the X-axis as a function of time. The force acting

on the particle is zero in the region

(A) AB & CD (B) BC (C) CD & DE (D) DE

Solution (A): In region AB and CD, slope of the graph is constant i.e. velocity is constant. It means no force acting on the particle in this region.

4. The engine of a jet aircraft applies a thrust force of 510 N during take off and causes the plane to attain a

velocity of 1 km/sec in 10 sec. The mass of the plane is

(A) 210 kg (B) 310 kg (C) 410 kg (D) 510 kg

Solution (B):

Acceleration produced in jet = Change in velocity

Time

3(10 0)

a10

2100m / s

Mass = 5

32

Force 1010 kg

Acceleration 10 .

5. A body of mass 100 g is sliding from an inclined plane of inclination 30°. What is the frictional force

experienced if 1.7

(A) 1

1.7 2 N3

(B) 1

1.7 3 N2

(C) 1.7 3 N (D)

11.7 2 N

3

Solution (B):

k k kF R mgcos

kF 1.7 0.1 10 cos30 3 1

1.7 N2 2

2

6. A body B lies on a smooth horizontal table and another body A is placed on B. The coefficient of friction between A and B is . What acceleration given to B will cause slipping to occur between A and B

(A) g (B) g / (C) / g (D) g

Solution (A): There is no friction between the body B and surface of the table. If the body B is pulled with force F then

A BF (m m )a

Due to this force upper body A will feel the pseudo force in a backward direction.

Af m a

But due to friction between A and B, body will not move. The body A will start moving when pseudo force is more than friction force.

i.e. for slipping, A Am a m g a g

7. A uniform rope of length l lies on a table. If the coefficient of friction is , then the maximum length 1l of the

part of this rope which can overhang from the edge of the table without sliding down is

(A) l

(B)

l

l (C)

l

1

(D)

l

1

Solution (C): For given condition we can apply direct formula

1l l1

8. A block of mass 5 kg lies on a rough horizontal table. A force of 19.6 N is enough to keep the body sliding

at uniform velocity. The coefficient of sliding friction is (A) 0.5 (B) 0.2 (C) 0.4 (D) 0.8 Solution (C):

kF

R

19.6 20.4

5 9.8 5

9. A block of mass 2 kg rests on a rough inclined plane making an angle of 30° with the horizontal. The

coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is

(A) 9.8 N (B) 0.7 9.8 3 N (C) 9.8 3 N (D) 0.8 9.8N

Solution (A):

Limiting friction lF mgcos

lF 0.7 2 10 cos30 12N (approximately)

But when the block is lying on the inclined plane then component of weight down the plane mgsin

2 9.8 sin30 9.8 N

It means the body is stationary, so static friction will work on it Static friction = Applied force = 9.8 N

3

10. A force of 5 N acts on a 15 kg body initially at rest. The work done by the force during the first second of motion of the body is

(A) 5 J (B) 5

J6

(C) 6 J (D) 75J

Solution (B):

21W Fs F at

2 21

from s ut at2

21 FW F t

2 m

2 2 2F t 25 (1) 25 5J

2m 2 15 30 6

11. The potential energy of a certain spring when stretched through a distance ‘S’ is 10 joule. The amount of

work (in joule) that must be done on this spring to stretch it through an additional distance ‘S’ will be (A) 30 (B) 40 (C) 10 (D) 20

Solution (A):

21kS 10 J

2 (given in the problem)

2 2 21 1k (2S) (S) 3 kS

2 2

= 3 × 10 = 30 J

12. The potential energy of a body is given by, U = 2A Bx (Where x is the displacement). The magnitude of force acting on the particle is (A) Constant (B) Proportional to x

(C) Proportional to 2x (D) Inversely proportional to x

Solution (B):

2U A Bx dU

F 2Bxdx

F x

13. Two masses of 1 gm and 4 gm are moving with equal kinetic energies. The ratio of the magnitudes of their

linear momenta is

(A) 4 : 1 (B) 2 :1 (C) 1 : 2 (D) 1 : 16

Solution (C):

P 2mE. If E are same then P m

1 1

2 2

P m 1 1

P m 4 2

14. Power of a water pump is 2 kW. If 2g 10m / sec , the amount of water it can raise in one minute to a

height of 10 m is (A) 2000 litre (B) 1000 litre (C) 100 litre (D) 1200 litre

Solution (D):

P = mgh

t m =

3p t 2 10 60

gh 10 10

1200 kg

As volume = mass

density 3

3 3

1200kgv 1.2m

10 kg/m

Volume = 3 31.2m 1.2 10 litre 1200 litre

4

15. Two perfectly elastic particles P and Q of equal mass travelling along the line joining them with velocities 15 m/sec and 10 m/sec. After collision, their velocities respectively (in m/sec) will be (A) 0, 25 (B) 5, 20 (C) 10, 15 (D) 20, 5

Solution (C): For a collision between two identical perfectly elastic particles of equal mass, velocities after collision get interchanged.

16. A smooth sphere of mass M moving with velocity u directly collides elastically with another sphere of mass

m at rest. After collision their final velocities are V and v respectively. The value of v is

(A) 2uM

m (B)

2um

M (C)

2u

m1

M

(D) 2u

M1

m

Solution (C):

2 1 1 12 2

1 2 1 2

m m 2m uv u

m m m m

2Mu 2u

mM m1

M

17. A shell of mass m moving with velocity v suddenly breaks into 2 pieces. The part having mass m/4 remains

stationary. The velocity of the other shell will be

(A) v (B) 2v (C) 3

v4

(D) 4

v3

Solution (D):

According to conservation of momentum

1 2m 3m

mv v v4 4

24

v v3

18. A mass of 10 gm moving with a velocity of 100 cm/s strikes a pendulum bob of mass 10 gm. The two

masses stick together. The maximum height reached by the system now is 2(g 10m / s )

(A) Zero (B) 5 cm (C) 2.5 cm (D) 1.25 cm Solution (D):

Initially mass 10 gm moves with velocity 100 cm/s

Initial momentum = 10 × 100 = gm m

1000sec

After collision system moves with velocity sys.v then

Final momentum = sys.(10 10) v

By applying the conservation of momentum

10000 = sys.20 v sys.v 50 cm/s

If system rises upto height h then

2sys.v 50 50 2.5

h 1.25 cm2g 2 1000 2

5

19. The displacement x of a particle moving in one dimension under the action of a constant force is related to

the time t by the equation t x 3 , where x is in meters and t is in seconds. The work done by the force

in the first 6 seconds is (A) 9 J (B) 6 J (C) 0 J (D) 3 J

Solution (C):

(c) 2x (t 3) dx

v 2(t 3)dt

at t 0 ; 1v 6m / s and at t 6sec , 2v 6m/s

so, change in kinetic energy 2 22 1

1 1W mv mv 0

2 2

20. A shell is fired from a cannon with velocity v m/sec at an angle with the horizontal direction. At the

highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed in m/sec of the other piece immediately after the explosion is

(A) 3vcos (B) 2vcos (C) 3

v cos2

(D) 3

v cos2

Solution (A):

m m

mv cos vcos V2 2

V 3vcos

21. A body starts from rest with uniform acceleration. If its velocity after n second is , then its displacement in

the last two seconds is

(A) 2 n 1

n

(B)

n 1

n

(C)

n 1

n

(D)

2 n 1

n

Solution (D): v 0 na a v / n

Now, distance travelled in n sec. 2n

1S an

2 and distance travelled in (n 2)sec 2

n 21

S a(n 2)2

Distance travelled in last two seconds,

n n 2S S 2 21 1an a(n 2)

2 2

2 2an (n 2)

2

a[n (n 2)][n (n 2)]

2

= a(2n 2) v

(2n 2)n

2v(n 1)

n

22. A particle is projected with velocity 0 along x axis . The deceleration on the particle is proportional to the

square of the distance from the origin i.e., 2a x . The distance at which the particle stops is

(A) 03

2

(B)

1

3o3v

2

(C)

20

3

2

(D)

12 30

3

2

Solution (D):

dv dv dxa

dt dx dt 2dv

v xdx

(given)

0 S2

v 00

vdv x dx

0 S2 3

v 00

v x

2 3

2 30

v S

2 3

12 30

3vS

2

6

23. A ball of mass 0.1 kg is suspended by a string. It is displaced through an angle of o60 and left. When

the ball passes through the mean position, the tension in the string is (A) 19.6 N (B) 1.96 N (C) 9.8 N (D) Zero

Solution (B): 2mv

T mgl

= m

mg 2gl(1 cos )l

=mg 2mg(1 cos60 ) = 2mg =2× 0.1 × 9.8=1.96N

24. A body is projected horizontally from a height with speed 20 metres/sec. What will be its speed after 5

seconds ( 2g 10 metres / sec )

(A) 54 metres/sec (B) 20 metres/sec (C) 50 metres/sec (D) 70 metres/sec

Solution (A):

Horizontal velocity xv 20m/s

Vertical velocity yv u gt 0 10 5 50m/sec

Net velocity 2 2 2 2x yv v v (20) (50) = 54 m/s.

25. A man standing on the roof of a house of height h throws one particle vertically downwards and

another particle horizontally with the same velocity u . The ratio of their velocities when they reach the

earth's surface will be

(A) 22gh u : u (B) 1: 2

(C) 1:1 (D) 22gh u : 2gh

Solution (C): When particle thrown in vertical downward direction with velocity u then final velocity at the ground level

2 2v u 2gh 2v u 2gh

Another particle is thrown horizontally with same velocity then at the surface of earth.

Horizontal component of velocity xv u

Resultant velocity, 2v u 2gh

For both the particle final velocities when they reach the earth's surface are equal.

7

26. Objects A and B each of mass m are connected by light inextensible cord. They are constrained to move on a frictionless ring in a vertical plane as shown in figure. The objects are released from rest at the positions shown. The tension in the cord just after release will be:

(A) mg 2 (B) mg/ 2 (C) mg/2 (D) mg/4

Solution (B):

From length constraint on AB a cos 45° = b cos 45° a = b T sin 45° = m(a) mg – T sin 45° = mb mg – ma = ma

2ma = mg g

a = 2

T mg

22

mgT =

2

27. A circular curve of highway is designed for traffic moving at 72 km/h. If the radius of the curved path is 100

m, the correct angle of banking of the road should be given by

(A) 1 2tan

3

(B) 1 3tan

5

(C) 1 2tan

5

(D) 1 1tan

4

Solution (C): 2V = gRtanθ (20) 10 100 tanθ

14 2 tan θ = θ = tan (2/5)

10 5

28. A ball of mass m is attached to the end of a thread fastened to the top of a vertical rod which is secured to

a horizontally revolving round table. If l = 6 cm, r = 10 cm and 45 , in equilibrium, the angular speed ,

is (rounding off to one significant digit)

axis of rotation

l

r

(A) 11 (B) 12 (C) 8 (D) 5

8

Solution (C):

r sin x l

T

T cos

m ( )2 x T sin

mg

For equilibrium, 2m (r sin ) Tsin l

mg Tcos

1gtan(8)s

r sin

l 29. The moment of inertia of a triangular lamina of mass M about edge AB is equal to

b

A B

C

60°

M

30°c

a

(A) 2 2 2Ma b c

12 (B)

2Ma

24

(C) 2 2 2M(c a b )

24

(D)

2Mb

12

Solution (B):

2

ABmh

I6

I

h

h asin30 a / 2

2

ABMa

I24

30. The ratio of the time taken by a solid sphere and that by a disc of the same mass and radius to roll down a

smooth inclined plane from rest from the same height is

(A) 15 : 14 (B) 15 : 14 (C)14 : 15 (D) 14 : 15

Solution (D): Conceptual

9

CHEMISTRY 31. Rearrange the following (I to IV) in the order of increasing masses and choose the correct answer. (atomic

mass; O=16, Cu=63, N=14) (I) 1 molecule of oxygen (II) 1 atom of nitrogen

(III) 101 10 g molecular weight of oxygen

(IV) 101 10 g atomic weight of copper

(A) II<I<III<IV (B) IV<III<II<I (C) II<III<I<IV (D) III<IV<I<II Solution (A):

(I) 1 molecule of 232 23

32O 5.3 10 g

6.022 10 g

(II) 1 atom of 2323

14N 2.3 10 g

6.022 10 g

(III) 1010 g mol. wt. of oxygen 10 910 32 3.2 10 g

(IV) 1010 g atomic weight of copper 10 910 63.5 6.35 10 g

Order of increasing mass is II<I<III<IV.

32. A gaseous hydrocarbon gives upon combustion 0.72g of water and 3.08 g of 2CO . The empirical formula

of the hydrocarbon is:

(A) 2 4C H (B) 3 4C H (C) 6 5C H (D) 7 8C H

Solution (D):

218g H O contains 2 g of H 20.72g H Ocontains 0.08g of H.

244g CO contains 12 g of C

23.08gCO contains 0.84 g of C.

0.84 0.08C :H : 0.07 : 0.08 7 : 8

12 1

Empirical formula 7 8C H

33. At STP, 0.48 g of 2O diffuse through a porous partition in 1200s. What volume of 2CO will diffuse in the

same time and under the same conditions? (A) 286.6 mL (B) 346.7 mL (C) 112.2 mL (D) 224.8 mL Solution (A):

Volume of 2O diffused22400 0.48

336mL32

Let the volume of 2CO diffused be x mL

Rate of diffusion of 2336

O1200

mL 1s

2 2 2

2 2 2

O O CO

CO CO O

r V /t M

r V /t M

or 336 44

1200 32x

1200

x 286.6mL

10

34. For a monoatomic gas, kinetic energy E . The relation with rms velocity is:

(A)

1

22Eu

M

(B)

1

23Eu

2M

(C)

1

2Eu

2M

(D)

1

2Eu

3M

Solution (A):

rms3pV

uM

pV nKT(K Boltzmann's constant)

For a molecule, n 1

rms3KT

pV KT uM

Kinetic energy 3 2

(E) KT or KT E2 3

rms

23 E

2E3uM M

35. As the temperature is raised from o o20 C to 40 C, the average kinetic energy of neon atoms changes by a

factor of which of the following?

(A) 1

2 (B)

313

293 (C)

313

293 (D) 2

Solution (C):

Average 03

KE RT/N2

313

293

(KE) 313KE T

(KE) 293

36. Which hydrogen like species will have same radius as that of Bohr orbit of hydrogen atom?

(A) 2n 2, Li (B) 3n 2, Be (C) n 2,He (D) 3n 3, Li

Solution (B):

Radius of orbit 2 2

22 2

n h 1 0.529(r) n

Z Z4 me

Å

For H atom 2

H0.528n

r1

Å

3 2rBe 0.529n /4 Å and ' 3nr rBe n 2

37. The wave number of the first emission line in the Balmer series of H-spectrum is (R=Rydberg constant)

(A) 5R

36 (B)

9R

400 (C)

7R

6 (D)

3R

4

Solution (A:

For Balmer series, 1 2 1n 2and n (n 1) for first emission line (2 1) 3

We know that,

Wave number, H 2 2 2 21 2

1 1 1 1 1 1 5Rv R R R

4 9 36n n (2) (3)

11

38. Which of the following is the energy of a possible excited state of hydrogen? (A) 13.6eV (B) 6.8eV (C) 3.4eV (D) 6.8eV

Solution (C):

Since, at n 1, the population of electrons is maximum, i.e. at ground state. So, maximum excitation will

take place from n 1to n 2 .

Hence, n=2 is the possible excited state. Now, we have the formula for energy of H-atom 2

n H 2

Z(E ) 13.6 eV

n where, Z=atomic number, Z for H-atom=1

n H 2

1 13.6(E ) 13.6 eV eV 3.4eV

42

39. The orbital diagram in which Aufbau principle is violated is:

(A) (B)

(C) (D) Solution (B): Aufbau principle states that electrons are filled in the increasing order of energy. Hence, 2s-orbital should be filled first before filling 2p-orbitals.

40. Which of the following pairs has zero dipole moment?

(A) 2 2 3CH Cl and NF (B) 4 3SiF and BF

(C) 3PCl and ClF (D) 3 3BF and NF

Solution (B):

4SiF is a symmetrical tetrahedral molecule and 3BF is a triangular planar (symmetrical) structure and

hence, have zero dipole moment.

41. Bond distance in HF is 119.17 10 m. Dipole moment of HF is 306.104 10 cm. The per cent ionic

character in HF will be (electron charge=1.60 1910 C)

(A) 61.0% (B) 38.0% (C) 35.5% (D) 41.6% Solution (D):

% ionic character observed

calculated

100

’

calculated e d

19 11 291.6 10 C 9.17 10 m 1.467 10 cm

% ionic character30

29

6.104 10100 41.6%

1.467 10

42. Which of the following compounds has the smallest bond angle in its molecule?

(A) 2H O (B) 2H S (C) 3NH (D) 2SO

Solution (D):

Bond angle of 2H S is smallest because S-atom is larger in size and has low electronegativity.

n 4

n 3

n 2

n 1

Maximum population

of e of H atom

12

43. The species in which the N-atom is in a state of sp-hybridisation is:

(A) 2NO (B) 3NO (C) 2NO (D) 2NO

Solution (D):

44. Which of the following species is not paramagnetic?

(A) NO (B) CO (C) 2O (D) 2B

Solution (B): To identify the magnetic nature we need to check the molecular orbital configuration. If all orbitals are fully occupied, species is diamagnetic while when one or more molecular orbitals is/are singly occupied, species is paramagnetic.

(A) NO(7 8 15) 2 * 2 2 * 2 2 2 2 * 1 * 0x y z x y1s , 1s , 2s , 2s , 2p 2p , 2p , 2p 2p

One unpaired electron is present. Hence, it is paramagnetic.

(B) 2 * 2 2 * 2 2 2 2x y zCO(6 8 14) 1s , 1s , 2s , 2s , 2p 2p , 2p

No unpaired electron is present. Hence, it is diamagnetic.

(C) 2 * 2 2 * 2 2 2 2 * 1 * 12 z x y x yO (8 8 16) 1s , 1s , 2s , 2s , 2p , 2p 2p , 2p 2p

Two unpaired electron are present. Hence, it is paramagnetic.

(D) 2 * 2 2 * 2 1 12 x yB (5 5) 1s , 1s , 2s , 2s , 2p 2p

Two unpaired electrons are present. Hence, it is paramagnetic.

45. The stability of the species 2 2 2Li ,Li and Li increases in the order of:

(A) 2 2 2Li Li Li (B) 2 2 2Li Li Li (C) 2 2 2Li Li Li (D) 2 2 2Li Li Li

Solution (A):

Stability of a molecule bond order 2 * 2 22Li (6) 1s , 1s , 2s

Bond order 4 2

12

2 * 2 2 * 12Li (7) 1s , 1s 2s , 2s bond order

4 30.5

2

2 * 2 12Li (5) 1s , 2s , 2s bond order

3 20.5

2

As both 2 2Li and Li has 0.5 bond order but 2Li is less stable because its valence electron is present in

antiorbital. The stability order is 2 2 2Li Li Li

13

46. The compound with longest bond length is:

(A) NO (B) NO (C) CN (D) CN

Solution (A):

Bone length is inversely proportional to their bond order. The bond order for NO ,NO ,CN and CN are

2,3,3,2.5. Therefore, NO has highest bond length.

47. An ideal gas expands in volume from 3 3 2 31 10 m to1 10 m at 300K against a constant pressure of

5 21 10 N/m . The work done is:

(A) 900J (B) 900kJ (C) 270kJ (D) 900kJ

Solution (A): Work done due to change in volume against constant pressure is

2 1W p(V V ) 5 2 2 3 31 10 N/m (1 10 1 10 )m 900Nm 900J

48. A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant

temperature of 37. o0 C . As it does so, it absorbs 208 J of heat. The values of q and W for the process will

be [R=8.314 J/mol K (in 7.5=2.01)] (A) q 208 J,W 208J (B) q 208 J,W 208J

(C) q 208J,W 208J (D) q 208J,W 208J

Solution (B): As the process is carrying out at constant temperature, therefore this type of expansion is called isothermal reversible expansion, for which, U 0.Hence, q W

i.e. heat absorbed by the system is equal to the work done by the system. Thus, q 208J

W 208J

49. For two moles of an ideal gas:

(A) p v(C C ) 2R (B) p v(C C ) 0

(C) p v(C C ) R (D) p v(C C ) R/2

Solution (A):

The equation, p vC C nR

2 moles of an ideal gas is equal to

p vC C 2R

50. The direct conversion of A to B is difficult, hence it is carried out by the following shown path:

Given (A C)S 50 eu

(C D)S 30 eu

(B D) 20 euS

Where, eu is entropy unit, then (A B)S is:

(A) + 100 eu (B) + 60 eu (C) -100 eu (D) -60 eu

C D

A B

14

Solution (B): S A C 50 eu

C D 30 eu

D B 20 eu

A B 60 eu

51. For a particular reversible reaction at temperature T, Hand S were found to be both positive. If eT is

the temperature at equilibrium, the reaction would be spontaneous when.

(A) eT T (B) eT T (C) eT is 5 times T (D) eT T

Solution (B): G H T S

At equilibrium, G 0

eT H/ S

For a reaction to be spontaneous G should be negative, so this implies H T S.

eH

T;T TS

Therefore, T should be greater than eT .

52. 2.1g of Fe combines with S evolving 3.77 kJ. The heat of formation of FeS in kJ/mol is:

(A) -1.79 (B) -100.5 (C) -3.77 (D) None of these Solution (B): When 2.1g Fe combines with S heat evolved=377kJ When 56g (atomic mass) combines with S, then heat evolved.

H/mole of 3.77 56

FeS 100.52.1

Since, the heat is evolved in the formation of FeS, thus the heat of formation of FeS is 1100.5 kJmol

53. Heat of combustion of 12H (g) 241.8 kJmol

1 12 5C(s) 393.5kJmol , C H OH( ) 1234.7 kJmol l

Hence, heat of formation of 2 5C H OH( )l is:

(A) 12747.1kJmol (B) 1277.7 kJmol

(C) 1277.7kJmol (D) 12747.1kJmol

Solution (B):

2 2 21

H (g) O (g) H O( )2

l ;

o 1H 241.8 kJmol

2

o 1t(H O)

H 241.8 kJmol

2 2C(s) O (g) CO (g),

o 1H 393.5 kJmol

2

o 1f (CO )H 393.5 kJmol

o 12 5 2 2 2C H OH( ) 3O (g) 2CO (g) 3H O( ); H 1234.7kJmol l l

15

54. Consider the reaction,

2 2 2 5 r4NO (g) O (g) 2N O (g); H 111kJ.

If 2 5N O (s) is formed instead of 2 5N O (g) in the above reaction, the rH value will be:

(Given, H of sublimation for 2 5N O is 54 kJ 1mol )

(A) 165kJ (B) 54 kJ (C) 219kJ (D) 219kJ

Solution (A):

From Hess law,

f sub reactionH H H

f reaction subH H H

111kJ (54 kJ)

111 54kJ 165kJ

Thus, the enthalpy of formation, fH for 2 5N O (s) is 165 kJ

55. Which of the following represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se

and Ar? (A) S Se Ca Ba Ar (B) Ba Ca Se S Ar

(C) Ca Ba S Se Ar (D) Ca S Ba Se Ar

Solution (B): Ionisation energy increases along a period from left to right and decreases down a group. Therefore, increasing order of first ionisation enthalpy is Ba Ca Se S Ar .

56. The successive ionization energy values for a element are given below?

(I) 1st ionization energy 1410 kJmol (II) 2nd ionization energy 1820 kJmol

(III) 3rd ionization energy 11100 kJmol (IV) 4th ionization energy 11500kJmol

(V) 5th ionization energy 13200 kJmol

Find out the number of valence electron for the atom X. (A) 4 (B) 3 (C) 5 (D) 2 Solution (A):

In the given values, there is a biggest jump between 4IE and 5IE . Hence, there are 4 valence electrons

for the atom X.

57. The IUPAC name of neopentane is: (A) 2-methylbutane (B) 2,2-dimethylpropane (C) 2-methylropane (D) 2,2-dimethylbutane Solution (B): The neopentane

IUPAC name: 2,2-dimethylpropane

H 111kJ2 2 2 54NO (g) O (g) 2N O (g),

H ? 54kJ

2 52N O (s)

3CH

3 3CH C CH

3CH

3 2 1

16

58. The IUPAC name of the compound shown below is:

(A) 2-bromo-6-chlorocyclohex-1-ene (B) 6-bromo-2-chlorocyclohexene (C) 3-bromo-1-chlorocyclohexene (D 1-bromo-3-chlorocyclohexene Solution (C):

Unsaturation, i.e. double bond is at priority over halogen, So, the correct IUPAC name is 3-bromo-1-chlorocyclohexene.

59. Which of the following compounds represents 2,2,3-trimethyl hexane?

(A) 3 3 2 2 2 3 2CH C(CH ) CH CH CH(CH ) (B) 3 3 2 2 3 2 2 3CH C(CH ) CH CH(CH ) CH CH

(C) 3 3 2 3 2 2 3CH C(CH ) CH(CH )CH CH CH (D) 3 3 2 2 3 2 3CH C(CH ) CH C(CH ) CH

Solution (C): 2.2,3-trimethylhexane is represented by

3 3

3 2 2 3

3

CH CH| |

CH C CH CH CH CH|CH

60. The structural formula of 2-oxo-3-methyl-(N-bromo) butanamide is:

(A) 3 2CH CH CO CO NH Br (B) 3

3

CH|

CH CH CO CO NH Br

(C) 3

3

CH|

CH CH CO CO NOBr (D) 3 3(CH ) C CO CO NHBr

Solution (B): 2-oxo-3-methyl-(N-bromo) butanamide is:

Cl

Br

Cl

Br

12

34

5

6

3H C HC C C N Br 34 2 1

H

3H C O O

17

MATHEMATICS 61. The value of 5 2 3 2log log log log 512 is

(A) 0 (B) 1 (C) 2 (D) 4 Solution (A):

5 2 3 2log log log log 512 95 2 3 2log log log log 2

5 2 3 2log log log 9log 2 25 2 3 5 2 3log log log 9 log log log 3

35 2 3log log 2log 5 2 5log log 2 log 1 0

62. The value of

7 7

7

log log 7 7 7

71 3log 2

is equal to

(A) 5 (B) 7 (C) 9 (D) 11 Solution (B):

3/2 3/47 7 7 7 7 7 7

31 1/2

7/4 7/847 7 7

Hence 7/87 7 7 7log log 7 7 7 log log 7 7 7

7log 1 log 8

8 71 3log 2

7 7

7

log log 7 7 7

7 7 1 7.1 3log 2

63. If 2 3 then the degree of the polynomial equation of least degree with rational coefficients whose

one of the roots is is

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

2 3 2 2 3 2 2. 3

2 5 2 6 2

2 5 24

4 210 25 24 4 210 1 0

is roots of 4 2x 10x 1 0

Hence degree of the equation is 4.

64. If

xy yz 1 zx 11, ,

x y y z 2 z x 3 then the value of z is

(A) 1 (B) 2 (C) 1

2 (D)

1

3

Solution (C):

xy1

x y

x y 1

xy 1

1 11

x y …(i)

yz 1

y z 2

y z2

yz

1 12

y z …(ii)

zx 1

z x 3

z x3

zx

1 13

z x …(iii)

Adding the equations (i), (ii) and (iii):

18

1 1 12 6

x y z

1 1 13

x y z

1

1 3z

1

2z

1

z2

65. If o0 45 then the solution to

22x 2 2 sin x sin2 04

is

(A) sin x cos (B) sin x cos

(C) cos x sin (D) cos x sin

Solution (B):

22x 2 2 sin x sin2 04

2 1 12x 2 2 sin . cos . x 2sin cos 0

2 2 2x sin cos x sin cos 0

x sin x cos 0

22x 2 2 sin x sin2 04

x sin x cos 0 sin x cos

NOTE: o0 45 cos sin

66. If tan tan x and cot cot y then value of tan is equal to

(A) 1 1

x y (B) x y (C)

xy

x y (D) xy x y

Solution (C):

cot cot y

1 1y

tan tan

tan tany

tan .tan

xy

tan .tan, tan tan x

x

tan .tany

tan tan x xytan

x1 tan tan x y1

y

67. The general solution of the equation tan2 .tan 1 is

(A)

n6

(B)

n

3 4 (C)

n

3 6 (D) none of these

Solution (A):

tan2 .tan 1

2

2tantan 1

1 tan

2

2

2 tan1

1 tan 2 22tan 1 tan

19

23tan 1 2 1tan

3

22 1

tan3

2 2tan tan6

n

6

68. The solution to 2x x 1 x 2 1 0 is

(A)

1 2

x2

(B)

1 5

x2

(C)

1 3

x2

(D) none of these

Solution (B):

x x 1 x 1 x 2 1 0

2 2x x x x 2 1 0

y y 2 1 0 , where 2x x y

2

y 1 0 y = 1 2x x 1

2x x 1 0

1 5

x2

69. The equation, sin2x cos2x sinx cosx 1 0 has

(A) One solution in 0, (B) Two solutions in 0,

(C) Three solutions in 0, (D) No solution in 0,

Solution (B): sin2x cos2x sinx cosx 1 0

22sinxcosx 2cos x 1 sinx cosx 1 0

sinx 2cosx 1 cosx 2cosx 1 0

sinx cosx 2cosx 1 0

sinx cosx 0 or 2cosx 1 0

1

tanx 1 or cosx2

x ,2 or x ,4 4 3 3

3 7 2 4

x , , ,4 4 3 3

Hence there are two solutions viz.

3 2

x ,4 3

in 0,

70. If

1 sin2 cos2f

2cos2 then value of o of 11 .f 34 equals

(A) 1

2 (B)

3

4 (C)

1

4 (D) 1

Solution (A):

1 sin2 cos2f

2cos2

1 cos2 sin2

2cos2

2

2 2

2cos 2sin cos

2 cos sin

cos cos sin

cos sin cos sin

cos

cos sin

20

So,

o oo o

o oo o

cos11 cos34f 11 .f 34

cos34 sin34cos11 sin11

o o

o o o o

o o o o

1 2cos11 cos34

2 cos11 cos34 cos11 sin34

sin11 cos34 sin11 sin34

o o

o o o o

o o o o

1 2cos11 cos34

2 cos11 cos34 sin11 sin34

cos11 sin34 sin11 cos34

o o

o o o

cos 45 cos 231

2 cos 34 11 sin 34 11

o o

o o

cos 45 cos 231

2 cos 23 sin 45

1

2

71. If & are the solutions of the equation

log acot2 bcosec222 c , then tan 3 3 is

(A)

2 2

2 2

a a 3c

c c 3a (B)

2 2

2 2

a c 3a

c a 3c (C)

2 2

2 2

c c 3a

a a 3c (D) none of these

Solution (C):

log acot2 bcosec222 c

acot2 bcosec2 c

a bc

tan2 sin2

2 2

a bc

2 tan 2 tan

1 tan 1 tan

2 21 tan 1 tana b c

2tan 2tan

2 2a 1 tan b 1 tan c 2tan

2b a tan 2c tan b a 0

Let tan x

2b a x 2cx b a 0

Now roots of above equation are tan & tan .

2ctan tan

b a &

b atan tan

b a

tan tantan

1 tan tan

2c2c cb atan

b a 2a a1

b a

tan 3 3 tan3

3

2

3 tan tan

1 3 tan

21

3 3 2

3 3

2 2 2

2

3c c c 3a c

a a a

3c a 3c1

a2 a

3 2 2

3 2 2

c 3a c a

a a 3c

2 2

2 2

c c 3a

a a 3c

72. In a triangle ABC, if tanA tanB tanC 6 then the value of cot A.cotB.cotC is:

(A) 6 (B) 1 (C) 1

6 (D) –6

Solution (C):

In a ABC, A B C

A B C tan A B tan C

tanA tanBtanC

1 tanA tanB

tanA tanB tanC tanA tanBtanC

tanA tanB tanC tanA.tanB.tanC

tanA tanBtanC 6 1

cot A cotBcotC6

73. The value of sin .cos tan cot sin2 .cos2 tan2 cot2 sin3 .cos3 tan3 cot3

sin4 .cos4 tan4 cot 4 .... upto n number of terms is:

(A) 1 (B) n (C) n n 1

2 (D) cannot be found

Solution (B):

sin .cos tan cot

sin cossin .cos

cos sin

2 2sin cossin .cos .

sin .cos 1

Hence each term of the given series 1. Hence sum of the series:

n terms

1 1 .... 1 n

74. Let A, B and C are the angles of a triangle and A 1 B 2

tan ,tan2 3 2 3

. Then C

tan2

is equal to

(A) 7

9 (B)

2

9 (C)

1

3 (D)

2

3

Solution (A):

A B C

A B C

2 2 2 2

A B Ctan tan

2 2 2 2

A Btan tan

C2 2 cotA B 2

1 tan .tan2 2

1 213 3

1 2 C1 . tan

3 3 2

9 / 9 1

C7 / 9tan

2

C 7

tan2 9

22

75. In triangle ABC, if 2 2 2a c 2011b then cot A cotC

cotB is equal to

(A) 1

2010 (B)

1

1005 (C)

3

2010 (D)

2

1005

Solution (B):

cot A cotC

cotB

cos A cosC

sin A sinCcosB

sinB

2 2 2 2 2 2

2 2 2

b c a 2R a b c 2R

2bc a 2ab c

c a b 2R

2ca b

,

a b cusing sine rule 2R

sinA sinB sinC

2 2 2 2 2 2

2 2 2

b c a a b c

c a b

2 2

2 2 2 2 2

2b 2b

c a b 2011b b, 2 2 2a c 2011b

2

2

2b 1

10052010b

76. In a ABC, the perimeter = 2s and the ex-radii are 1 2r ,r & 3r . Then 12 2 3 3 1r r r r r r is equal to

(A) 2s (B) 22s (C) 23s (D) 24s

Solution (A):

1 2 3r ,r & r

s a s b s c

12 2 3 3 1r r r r r r

. . .

s a s b s b s c s c s a

2 1 1 1

s a s b s b s c s c s a

2 s c s a s b

s a s b s c

2 23s a b c 3s 2s

s a s b s c s a s b s c

2 s

s a s b s c

2 22 2 2

2

s ss

s s a s b s c

23

77. The range of the function f x 5 3x 7 , 2 x 1

(A) 8,1 (B) 8,1 (C) 8,1 (D) 8,1

Solution (B):

2 x 1 6 3x 3

6 7 3x 7 3 7 13 3x 7 4

4 3x 7 13 13 3x 7 4

5 13 5 3x 7 5 4 8 f x 1

Range of f x is 8,1

78. If 1/3 1/3

3x 2 2 x then x

(A) ,1 (B) 1, (C) ,1 1, (D) none of these

Solution (B):

1/3 1/3

3x 2 2 x cubing both sides

3x 2 2 x 4x 4 x 1 x 1,

79. If

2x 17 x 0 then x , where [ ] denotes greatest integer

(A) , 1 3, (B) 1,3

(C)

, 2 10 , (D) none of these

Solution (C):

Given

2x 17 x 0

As

17 4 & 3 thus the given inequality is equivalent to 2x 4x 3 0 x 1 x 3 0

x 1 or x 3 x ,1 3,

Since

2 1& 10 3 therefore c is the correct answer.

80. Which of the following is correct: where [ ] denotes greatest integer

(A) 2 3

log 3 log 2 (B)

ee

(C)

1 1

2 2 (D) all are correct

Solution (D):

81. If

1

x0 3 9 then x

(A)

1 1, ,

2 2 (B)

1,0 0,

2

(C) ,0 0, (D) none of these

Solution (A):

1

2x

1

0 2x

1

x2

1x

2

1

x2

or 1

x2 x

1 1, ,

2 2

24

82. If the roots of the quadratic equation 2 2x 2kx k k 5 0 are less than 5, then k lies in the interval

(A) 5,6 (B) 6, (C) ,4 (D) 4,5

Solution (C):

Let 2 2f x x 2kx k k 5 .

For both the roots to be less than 5 the graph of y f x must be as shown below:

Hence the following conditions must hold simultaneously:

(i) D 0 2 24k 4 k k 5 0

4k 20 0 k 5

(ii) b

52a

k 5

(iii) f 5 0 225 10k k k 5 0

2k 9k 20 0 k 4 k 5 0

k 4 or k 5

Taking intersection of (i), (ii) & (iii) k ,4 .

83. The number of quadratic equations of the form 2ax bx c 0 , where a,b,c 2,3,6,7 and a b c

such that they all have real roots is (A) 2 (B) 4 (C) 3 (D) 6 Solution (D):

Discriminant 2D b 4ac

For b 2 or b 3 discriminant is always negative.

Hence for b 2 or b 3 number of quadratic equations having real roots is zero.

CASE-I: b 6

2b 36 . For roots to be real, discriminant must be greater than or equal to zero.

36 4ac 0 ac 9

As a b c , therefore possible values of a and c are:

a 2& c 3 or a 3 & c 2 .

Hence for b 6 number of quadratic equations having real roots is two.

CASE-II: b 7

2b 49 . For roots to be real, discriminant must be greater than or equal to zero.

49 4ac 0 ac 12.25

As a b c , therefore possible values of a and c are:

a 2& c 3 or a 3 & c 2 or a 2& c 6 or a 6 & c 2

Hence for b 7 number of quadratic equations having real roots is four. Therefore total number of

quadratic equations having real roots is 6.

84. The number of ordered triplet x,y,z satisfying the system of equations

10 10 10log 2000xy log x.log y 4 , 10 10 10log 2yz log y.log z 1 & 10 10 10log zx log z.log x 0 is:

(A) 0 (B) 1 (C) 2 (D) 3

25

Solution (C):

Let 10 10 10log x a,log y b & log z c

10 10 10log 2000xy log xlog y 4

log2000 a b ab 4log10

4a b ab log10 log2000

410a b ab log

2000 a b ab log5 …(i)

10 10 10log 2yz log ylog z 1

log2 b c bc log10

b c bc log10 log2

10b c bc log

2

b c bc log5 …(ii)

10 10 10log zx log zlog x 0

c a ca 0 …(iii)

From (i) & (ii): a b ab b c bc

a c b a c a c 1 b 0 a c or b 1

CASE I: a c

From (iii):

22a a 0

a 2 a 0 a 0 or a 2

10 10log x 0 or log x 2

x 1 or x 100

(a) For x 1: As a c

10 10log x log z x z z 1

From (i):

10a b ab log 5

100 b 0 b log 5 10b log 5

10 10log y log 5 y 5

x,y,z 1,5,1

(b) For x 100

As a c

10 10log x log z x z z 100

From (i):

10a b ab log 5

102 b 2b log 5 102 b log 5

102 log 5 b 10 10b 2log 10 log 5

10100

b log5

10b log 20

10 10log y log 20 y 20

x,y,z 100,20,100

26

CASE II: b 1

10log y 1 y 10

From (i):

10a 1 a 1 log 5 101 log 5 a

Similarly b & c

Hence real solution to the given system of equations is x 1,y 5,z 1or x 100,y 20, z 100 .

Therefore the number of ordered triplet x,y,z satisfying the system of equations is 2.

85. If x/2xlog a,a and blog x are in GP, then x is equal to:

(A) a blog log a (B) a alog log b

(C) a e a elog log b log log a (D) none of these

Solution (A): x/2

xlog a,a & blog x are in GP.

2

x/2x ba log a.log x x

ba log a

a bx log log a

86. The range of the function sinxcos3x

ysin3xcos x

, x R is

(A) 1

y 33

(B) 1

y or y 33

(C) 1

y or y 42

(D) 1

y 42

Solution (B):

Let

3

2

sinxcos3x tanx tanxy

sin3xcos x tan3x 3 tanx tan x

1 3 tan x

2

3

tanx 1 3 tan x

3 tanx tan x

2

2

1 3 tan x

3 tan x, tanx 0

2 23y tan x y 1 3 tan x 23y 1 y 3 tan x

2 3y 1tan x

y 3

1 3y 1x tan

y 3.

For x to be real:

3y 10

y 3

1y or y 3

3

But when tanx 0 :

2

2

1 3 0 1y

33 0.

Also let 1

y3

:

2

2

1 1 3 tan x

3 3 tan x

2 23 tan x 3 9tan x

28tan x 0 tanx 0

tanx 0 1

y3

1

y or y 33

27

87. In ABC, If A – B = 120° and R = 8r, then the value of equals

(All symbols used have their usual meaning in a triangle) (A) 12 (B) 15 (C) 21 (D) 31 Solution (B):

88. Let be a real valued function satisfying . Where

and such that . Then the condition that will have real roots is

(A) (B)

(C) (D)

Solution (D):

Using hypothesis we get

89. If for any real , we have then the value of is

(A) (B)

(C) (D)

Solution (A):

, true then

--- (1)

Similarly

--- (2)

Combined (1) & (2) we get

90. 10 10 1016 25 81

7log 5log 3log15 24 80

equals to

(A) 10log 2 (B) 10log 3 (C) 10log 5 (D) zero

Solution (A):

16 25 817log 5log 3log

15 24 80

7 5 3

7 5 3

16 25 81log

15 24 80

28 10 12

7 7 15 5 12 3

2 5 3log

3 5 2 3 2 5

28 10 12

27 10 12

2 5 3log log2

2 5 3

C

C

cos1

cos1

cos cos cos 1r

A B CR

12cos cos 1 cos

8 2 2

A B A BC

21sin 2sin

8 2 2

C C

1sin

2 4

C

f x 2. ,a f x b f x px qx r x R

, , 0p q r R ,a b R a b 0f x

2 2

4

a b q

a b pr

2

2

4a b pr

a b q

2 2

4

a b q

a b pr

2

2

4a b pr

a b q

2( ) ( )

qxf x f x

a b

x2

2

21 2

3 4

x nx

x x

n

1, 40 6n

[ 1,3)n

40 6, 1n

1, 40 6n

2

2

22 0

3 4

x nx

x x

2 ( 6) 10 0x n x x R

20 ( 6) 40 0 40 6 40 6D n n 2

2

2

21 0 2 ( 3) 2 0

3 4

x nxx x x

x x

20 ( 3) 16 0 1 7D n n

1, 40 6n