Solution Sr 15 Main Test 5

Batch: Seniors-15 Date:

ANSWERKEY

MATHEMATICS1. (C) 2. (A) 3. (A) 4. (D) 5. (C)

6. (C) 7. (A) 8. (A) 9. (B) 10. (A)

11. (C) 12. (D) 13. (C) 14. (D) 15. (B)

16. (A) 17. (B) 18. (A) 19. (D) 20. (C)

21. (B) 22. (B) 23. (A) 24. (A) 25. (C)

26. (C) 27. (A) 28. (C) 29. (D) 30. (D)

PHYSICS31. (A) 32. (A) 33. (A) 34. (B) 35. (A)

36. (D) 37. (B) 38. (B) 39. (D) 40. (C)

41. (A) 42. (B) 43. (C) 44. (D) 45. (B)

46. (A) 47. (B) 48. (B) 49. (A) 50. (C)

51. (C) 52. (C) 53. (B) 54. (A) 55. (A)

56. (D) 57. (D) 58. (B) 59. (A) 60. (D)

CHEMISTRY61. (A) 62. (A) 63. (B) 64. (D) 65. (D)

66. (C) 67. (B) 68. (D) 69. (D) 70. (B)

71. (B) 72. (C) 73. (A) 74. (D) 75. (B)

76. (A) 77. (B) 78. (B) 79. (A) 80. (B)

81. (C) 82. (D) 83. (A) 84. (C) 85. (C)

86. (D) 87. (A) 88. (B) 89. (A) 90. (D)

RAO ALL INDIA TEST SERIES

MAIN FULL TEST [5]

Rao All India Test Series_Sr-2015 Batch_MAIN Full Test - [5]_Solutions

2SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR |SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

PART-A: MATHEMATICS [SOLUTION]= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =1. (C)

aRb a b is not transitive for(2,3) R but (3,2) R

Topic:- Sets, Relation, Functions_(Relations)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

2. (A)

2 113 3

xf xx x

2

1' 03

f x x Ax

f is one - one

and 2 3 13 1

x yy xx y

Clearly 3 1,1

yy B x Ay

f is onto

Topic:- Sets, Relation, Functions_(Function)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

3. (A)iee = cos sin cos cos sin sin sinie e i

Topic:- Complex number_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

4. (D)

1 12 23 3

1 13 3

33 2 188 3 427

8

Topic:- Quadratic Equations_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

5. (C)

2det 4sin 0, 4A Topic:- Matrices & Determinants _(Determinants)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

6. (C)Let n be the no. of sides of the polygon.n 160° = (n–2) 180° 20°n = 360 n = 18The number of diagonals = 18

2 18 135C Topic:- Permutations & combinations_L-2_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.



7. (A), , , , , , , , , , ,T T T O O I I N N C U S

( , , , , , , )T O I N C U S

4 distinct : 74 840P

3 same, 1 distinct : 6 4 24

2 same, 2 same :

42

4! 362!2!

C

2 same, 2 distinct : 4 61 2

4! 14402!

C C

Total no. of words = 840 +24 +36 +1440 = 2340Topic:- Permutations & combinations_L-2_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

8. (A)

Gen. term in 81 1

3 512

y y

is 81 1

8 3 512

r r

rC y y

83 5

81

2

r r

r rC y y

40 815

81

2

r

r rC y

For independent of y, r = 5

The value of the term independent of 85 3

12

y C

38 7 6 11 2 3 2

7

Middle term in

7 11 13 51

2y y

is

4 41 18 3 5

412

C y y

Coefficient 84 4

1 8 7 6 5 1 351 2 3 4 16 82

C

Topic:- Binomal Theorem_L-2_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

9. (B)

1 2 1

1 1

1cot cot 1 cot tan1 1n

n nn nn n



1 1cos lim tan 1 tan 1n

n

cot 12 4

Topic:- Inverse trignometric functions_L-2_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

10. (A)

1 5 1 5 1 1sum 19 3 2 9

aa a

Topic:- Sequence & Series_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

11. (C)

2

3, 1 33 , 3

3 13 , 14 2 4

x xf x x x

x x x

3 3

3 0, lim lim 0x x

f f x f x

1 1

1 2, lim lim 2x x

f f x f x

Topic:- Continuity_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

12. (D)

tan , 04

cot ,4 2

3tan ,2 43cot ,4

x x

x xf x

x x

x x

Since tan x, cot x are periodic with period , f (x) is periodic with period

f (x) is clearly not continuous at 30, , ,

2 2x

f is not differentiable at 3 5, , .

4 4 4x etc

Topic:- Differentiabiltiy_L-2_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.



13. (C)

1 1 1'( ) 2 0 ,0 ,2 2

f x x xx

Topic:- Monotonic functions_L-2_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

14. (D)

2/32

2 1/ 32

2 /3

323 32 6

2

ydy dy x yy xdx dx y y

1 1

16 8y x

Topic:- Definite Integration_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

15. (B)

3

1 3/23

cos cos 2cos cos31 cos

x x dx x cx

Topic:- Indefinite Integration_L-2_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

16. (A)

/ 4

20

sec1 2sin

x dxx

/ 4

2 20

cos(1 sin )(1 2sin )

x dxx x

12

2 20 (1 )(1 2 )

dtt t

and 22 21 1/ 6 1/ 6 2 / 3

1 1 1 21 1 2 t t tt t

Topic:- Definite Integration_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

17. (B)

Required area 312

0

1 9 26 2

m mx x mx m

Topic:- Areas_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.18. (A)

2 21 1 0x y dy y x dx

2 21 1 0y xdy dx

y x

1 1log logy x cy x



1 1log x cy x y

Topic:- Differential equation_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

19. (D)

| 2 | 4 22,189 16c c

Topic:- Straight lines_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

20. (C)

Family of circle 2 2 2 22 4 1 ( 1) 0x y x y x y

2 2(1 ) (1 ) 2 4 (1 ) 0x y x y

Centre 1 2,

1 1

radius 24

1

Since it touches 2

1 441 12 011 4

x y

1 Topic:- Circles_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

21. (B)

2 21 1 2 2, 2 , , 2P at at Q at at be the end points of a normal chord.

Then 2 11

2t tt

....(1)

Slope of 1

2OPt

, slope of 2

2OQt

1 21 2

4 1 4t tt t

....(2)

(1) and (2) 1 2t Topic:- Conics_(Parabola)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

22. (B)

8 89 9

y x m

tangent 2 2 2y mx a m b

8 59 9

y x



8 9 5 0x y

If 1 1,x y is point of contact then 1 14 9 1xx yy

1 11 1

4 9 1 2 1, ,8 9 5 5 5x y x y

Topic:- Conics_(Ellipse)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

23. (A)

Tangent to parabola 2y mxm

. It is tangent to 2

2 13yx , then

222 3 2m m

m

Topic:- Conics_(Hyperbola)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

24. (A)

1 1 1 13 1sin cos sin sin cos sin2 3 2 6

Topic:- Inverse Trigonometric fuctions_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

25. (C)

Given 24, 5 5.2x n and

If the remaining observations are 1 2,x x then 2 5.2 2| | 5.2ix x

n

2 2 2 2 21 24 4 1 4 2 4 6 4

5.25

x x

2 21 24 4 9x x ....(1)

But 4x

1 2 1 2 6 45

x x

1 2 11x x ....(2)

Solving (1) and (2) 1 24, 7x x Topic:- Statistics_(Mean and Variance)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

26. (C)

1ˆ ˆˆ ˆ( )2

a b c b

1ˆ ˆ ˆˆ ˆ ˆ ˆ( . ) ( . )2

a c b a b c b



1 ˆ ˆˆ ˆ ˆ ˆ. ( . ) 02

a c b a b c

1 ˆ ˆˆ ˆ ˆ ˆ. 0, . 0 (2

a c a b b and c are non parallel)

angle between ˆ ˆ, 60ºa c

angle between ˆˆ, 90ºa b Topic:- Vectors_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

27. (A)

. 0 ( ). 0

. . 02 0

c xa yba c xa yb a

x a a y b ax y

( 2 )ˆˆ( 3 3 )1| | 1

3 2ˆˆ

2

c x a b

x j k

c x

j kc

Topic:- Vectors_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

28. (C)

Gen point on the line , 2 1,3 2r r r

Dr’s of (1, 6, 3) and , 2 1,3 2r r r

1,2 5,3 1r r r

Now 1 2 2 5 3 3 1 0r r r

1r foot of perpendicular (1, 3, 5)the perpendicular distance 9 4 13

Topic:- 3-D_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

29. (D)

Clearly 1 ( ) ( )8

P A B P A P B

Topic:- Probability_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.

30. (D)

Required probability 51 5! 2 2 2 3 3 12

3!2!2

Topic:- Probability_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_RKm sir.



PART-B: PHYSICS [Solution]= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =31. (A)

The correct choice is (A)The maximum possible error in is given by the relation

4

R l P QR l P Q

It is clear that the error in the measurement of R is magnified four times on account of the occurrence of R4

in the formula. Hence the radius (R) of the capillary tube must be measured most accurately. Thus thequantity which is raised to the highest power needs the most accurate measurement.

Topic : Units & dimensions,(errors)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

32. (A)Since the initial velocity of the stone is zero, the total time taken by the stone to hit the ground is given by

21 2 2 125 52 10

hh gt or t sg

During the first four seconds, the stone falls a distance h given by

21 (4) 8 80 m2

h g g

Distance h2 through which the stone falls in the last (i.e. fifth) second= 125 – 80 = 45 m.

Now, 1 2 5 45 1 9h / h / / . Hence the correct choice is (A).Topic : Kinematics,(free fall under gravity)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

33. (A)The acceleration of the block while it is sliding down the upper half of the inclined plane is sin g . If is thecoefficient of kinetic friction between the block and the lower half of the plane, the retardation of the blockwhile it is sliding down the lower half sin g cos ( g ) . For the block to come to rest at the bottomof the inclined plane, the acceleration in the first half must be equal to the retardation in the second half, i.e.

sin sin cos g g g

or cos 2sin

2 tan Hence the correct choice is (A).

Topic : NLM,(friction) _L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

34. (B)Total distance 2 42 2h e h e h ....

2 22 (1 )h e h e ...

2 2

2 2

2 11 1

e h eh he e



2

2

1153411

3

hh .

which is choice (B).Topic : System of Particles,(collisions)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

35. (A)When the string D is cut, the rod will rotate about point A. Let a be the linear acceleration of the centre ofmass and T the tension in string C, Then (see Fig.)

mg – T = ma ......(1)

Torque of force mg about A is 2LMg .

B

mg

L/2 Centre of

mass

T

A

Moment of inertia of the rod about A is I = ML2/3. Therefore, the angular acceleration of the rod about A is32

gI L

.....(2)

Also 2

La

Using Eq.(2), we get 34ga . Then from Equation (1).

34 4mg mgmg T T

So the correct choice is (A).Topic : Rotation,(angular and linear acceleration)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

36. (D)

If is the density of the material of each sphere, then the mass of the sphere of radius r is 31

43

M r and

the mass of the sphere of radius 2r is 32

4 (2 )3

M r .

Distance between their centres is 2 3d r r r.



Now

3 3

1 22 2

4 4 (2 )3 3

9

G r rG M MF

d r

which gives 4F r , which is choice (D).Topic : Gravitation,(gravity force)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

37. (B)Refer to Fig. For vertical equilibrium

cosT mg

or cosmgT

If L is the original length of the wire, the increase in length is

TLlAY

Straincos

l T mgL AY AY

m

mg

T cos

T sin

T

O

Topic : Elasticity,(normal strain)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

38. (B)Work is done

2 1 3 24 1 4 1 / RW R n Rr

(1)

1 3( )/R n r

Heat produced 343

Q ms T R S T (2)

Equating (1) and (2) we find that choice (B) gives the correct expression for T .Topic : Fluids/Subtopic : Statics_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

39. (D)Let h1 and h2 be the depths of the holes below the surface of water.

The volume of water flowing out from the holes per second is 1 2 1 1and where 2 a a , g h

2and 22g h . Hence mass of water flowing per unit time is 1 1 2 2and m a m a . Therefore,momentum per second carried by water is

1 1 1p m in one direction

and 2 2 2p m in the opposite directionHence force required to keep the tank at rest is

F rate of change of momentum

1 1 2 2 m m 2 21 2 a a



1 22 2 a gh a gh

1 22 ( )ag h h

2aghTopic : Fluids/Subtopic : Dynamics_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

40. (C)If k is the force constant, we have

mg = kx

orm xk g

2 2 m xTk g

Topic : Simple Harmonic Motion,(spring control motion)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

41. (A)340 0.8425

m

for the first harmonic, 4Xl

= 20 cm.Topic : Waves,(resonance of air columns)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

42. (B)Mass of 2 litres of water = 2 kg. Heat energy needed to raise to temperature of 2 kg of water from 20°C to75°C is 3 32 (4 2 10 ) 55 462 10 JQ . It t is the time taken heat energy supplied by the heater in time t is Q1 = (power × time) = 1000 t jouleHeat energy lost in time t is Q2 = 160 t jouleHeat energy available for heating water is 1 2 840 JQ Q Q t

Equating Q Q , we get 550s.tTopic : Heat,(specific heat capacity)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

43. (C)The two adiabatic paths ad and bc for the gas intersect the two isothermals ab and cd at temperatures T1and T2 (see Fig.). Since point a and d lie on the same adiabatic path, we have

1 11 2

a dTV T V

or 1

2

1

a

d

V TV T

(1)

Since points b and c also lie on the same adiabatic path,



1 11 2b cT V T V

or 1

2

1

b

c

V TV T

(2)

From (1) and (2), we get 1 1

a b

d c

V VV V

ora b

d c

V VV V

Thus the correct choice is (C).Topic : Thermodynamics,(adiabatics & isothermals)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

44. (D)The root mean square speed is given by

3r ms

kTm

The escape velocity is given by 2e gR

For rms e , we require

3 22 or3

kT mgRgR T ,m k

which is choice (D)

Topic : Kinetic Theory of Gases,(rms speed, escape speed)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

45. (B)U Uˆ ˆF i jx y

ˆ ˆF 6i 8j

At the time of crossing Y-axis, x-coordinate will be zero. Suppose position vectors is y j then

21S ut at2

2

ˆ ˆ6i 8j1ˆ ˆ0 6 i y 4 j t2 2

236 t2

2t 4 t 2sec.

Topic : Work, Power, Energy,(potential energy & force)_L-2_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.



46. (A)

Electric flux through the cubical surface 0

q

where q is the net charge enclosed in the surface. Since

half the disc lies inside the surface, one-fourth of the rod lies inside the surface, point charge –7C liesinside the surface and point charge 3C lies outside the surface, the net charge enclosed in the surfaceis

6 8 7 02 4C Cq C

3 2 7 2C C C C

0

2Electric flux C .

So the correct choice is (A)

Topic : Electrostatics,(electric flux)_L-2_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

47. (B)Total energy = KE + PE = K + UIn the region 10 1x ; U E. Therefore, kinetic energy is K1 = Total energy – U1 = nE – E = (n – 1) E

112 2 1

h hmK m n E

(1)

In the region x > 1 ; U2 = 0./ Therefore kinetic energy isK2 = nE – 0 = nE

222 2

h hmK mnE

(2)

Dividing (1) by (2), we get

1

2 1n .

n

So the correct choice is (B)

Topic : Modern Physics(de Broglie wavelength)_L-2_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

48. (B)Till then m remains on B, B will continue to accelerate due to contact force between m and B. Hence,velocity of B will be maximum when finally m leaves B.

Topic: Centre of mass & Collision_(Centre of mass)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

49. (A)Velocity of m when string gets tight, 1 2 2 2v g l gl

Combined velocity, 12 3

mvvm

1 23 3v gl

Topic: Centre of mass & Collision_(Impulse)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.



50. (C)Let the velocity components of point B along x and y directions are andx yv v . As length of the rod isconstant, along the length of the rod, velocity of each point must be the same (rigid body constraint).

Bl 45º

45º45º

vC

Av=0Fixed

1 2vy

Along rod ,2 2 2

yx vv VAC

Along rod , 02 2

yx vvAB

2x yVv v

Topic: Kinematics_(Constraints)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

51. (C)

Shearing stress | viscous force|

area

dvA dvdx

A dx

3 2 3 2510 / 0.5 10 /10

N m N m

Topic: Properties of matter_(Viscosity)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

52. (C)Let the bar be rotated through a small angle .The restoring torque of the forces 1 2,mg k x and k x about O can be given as

1 2sin cos cos2lmg k x l k x l

Since is small, sin , cos 1x l and Putting 1 2 ,k k k we obtain

22lkl mg

or2

2lI kl mg

2

. 2( / 2) 3 3

2( / 3osckl mg l k g

m lml

Topic: SHM_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.



53. (B)y

xDotted shape shows pulse position after a short time interval. Direction of the velocities are decided ac-cording to direction of displacements of the particle.

Topic: Waves_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

54. (A)

vC C V

vdQ dU dV PdVP C CdT dT dT dT

comparing, dVP VdT

2RT dV dV dTVV dT R TV

On integration

1 ln ln RT T

V R V

constantVIRT Topic: Thermodynamics_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

55. (A)The current density at P is higher than that at Q. For the same current flowing through the metallic conductorPQ, the cross-sectional area at P is narrower than that at Q. The resistance per unit length r is given by

1 ,rA

where is the resistivity and A is the cross-sectional area of the conductor PQ. Thus, r is

inversely proportional to the cross-sectional area A of the conductor.Topic: Current electricity_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

56. (D)

Current density : 2 2 24

( / 2) 3I I

R R R

R/2BA

R

Current in smaller cylinder (if there were): 2

1 2 3R II p

For wholecylinder small cylinder: AA B B B

0 0( / 3)02 ( / 2) 3A

I IBR R



For wholecylinder small cylinder: BB B B B

0 02

( / 3)( / 2) 032

I I R IRR

Topic: Magnetic effect_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

57. (D)According to Lenz’s law, emf of same megnitude in clockwise direction is induced in the two loops intowhich the figure is divided. So, current is induced in the clockwise direction in the outer boundary but nocurrent is there in wire AB.

Topic: EMI_(Induced emf)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

58. (B)Kinetic energy gained by a charge q after being accelerated through a potential difference V volt, is given by

212

qV mv

2mV MqV

As, 2bh h

mV qmV

For cut-off wavelength of X-rays, we have m

hcqV

or mhcqV

Now, 2b

m

qVm

c

As 11 11.8 10q C kgm

for electron,

we have11 3

81.8 10 10 10 / 2 0.1

3 10b

m

Therefore, the answer is (B).

Topic: Modern physics_(de Broglie)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

59. (A)Maximum number of nuclei will be present when rate of decay = rate of formationor N

N

Topic: Mordern physics_(Radioactivity)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.



60. (D)

Using 1 1 2 ,v u R we get

1 1 1 30

15 10v cm

v

Now, using 2

2im OMvV Vu

2

02i m mvV v V Vu

2

230

(1) 10 115

iV

145iV cms

So, the image will move with velocity 145 .cms

Topic: Optics_(Refflection by mirror)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_DKp sir.

PART-C: CHEMISTRY [Solution]= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =61. (A)

In (2) and (4) components cancell out. In (1) add up. In (3) there will be some net dipolemoment butless than (1)

Topic: GOC _(Dipole moment)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

62. (A)

F

F F

FF P

sp d3

PF5 - trigonal bipyramidal

BrF

F F

FF

..

sp d3 2

BrF5 – square pyramidalHence PF5 and BrF5 are not isostructrual.

Topic: Chemical bonding_(VSEPRT)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

63. (B)

Rate 12 5k N O (first order reaction)

12 5 Rate /N O k



4

51.02 10 33.4 10

Topic: Chemical Kinetics_(Ist order reaction)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.64. (D)

Order of elements in Electro chemical series is Zn, Fe, Ni (in decreasing reducing power)Zn can reduce Fe or Ni

2 2( ),

Zn s Ni Zn NiX Zn Y Ni

2

2

2

2 º 0.76

2 º 0.44

2 º 0.23

Zn Zn e E

Fe Fe e E

Ni Ni e E

Reducing power decreasesTopic: Electrochemistry_(Electrochemical series)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.65. (D)

As for B.C.C.

33 44

a r so r a

3 3514

r

152 pmTopic: Solid state_(Packing of crystal)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

66. (C)Moles at Elquilibrium A Bx y xA + yBy+ x–

1 – x y1i x y

( 1) ( 1)i x y

1( 1)

ix y

Topic: Colligative properties_(van’t Hoff factor)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

67. (B)

f fT iK m

1i for glucose.

0 02.8fT O C C 02.8 2.8C K

10002.8 1.86 B

B A

WM W

10002.8 1.86

62 1000BW

= 93.3 gTopic: Colligative Properties_(Freezing point depression)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.



68. (D)Let the bond dissociation energy of XY, X2 and Y2 be x kJ mol–1, x kJ mol–1 and 0.5 x kJ mol–1

respectively.11 1

2 22 2 ; 200fX Y XY H kJ mol

H reaction = (sum of bond dissociation energy of all reactants) – (sum of bond dissociation energy ofproduct)

2 2

1 12 2X Y XYH H H

0.5 2002 2x x x kJ

1200 8000.25

x kJ mol

Topic: Themochemistry_(Bond energy)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

69. (D)Compounds with methyl group give positive Iodo form test.

Topic: Carbonyl_(Iodoform reaction)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

70. (B)Due to aromatic character of phenol. So it is very stable and hence equilibrium will be shiftedforward to give phenol.

Topic: GOC_(Tautomerism)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

71. (B)

3 2 2:CHCl NaOH CCl NaCl H O Topic: Electrophilic subs reaction_(Reimertieman)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

72. (C)Here the complex involves d2sp3 hybridisation, and hence it is an inner orbital complex.

Topic: Coordination chemistry_(Mixed)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

73. (A)Ring activating power of 3OH CH ; 2,4,6 positions are available only in (A). In (C) and (D) onlytwo positions available.

Topic: Phenol_(Reactions)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

74. (D)Loss of two moles of HCl, followed by tautomerism.

Topic: Elimination reactions_(E2)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

75. (B)Density at constant mass will be highest if volume is lowest.Volume decreases at low temperature and high pressure.

Topic: Gaseous state_(Intensive properties)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

76. (A)These are reactive metals and react with water, so they are electrolysed in molten state.

Topic: Metallurgy_(Electro chemical principle)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.



77. (B)Both (A) and (B) converts

C = O to CH2 but in presence of acid sensitive group (–OH)

(B) is most suitable.Topic: Reduction_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.78. (B)

O O –H

+

O O

Hmost stable

conjugate anion (resonance stabilised)Topic: GOC_(Acid base)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.79. (A)

CHO NaOHHCHO

CH OH2

+ H – C – O Na +

O–

Cannizzaro reactionTopic: Carbonyl _(Cannnizzaro reaction)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

80. (B)

2 3 3 2 3Fe O C Fe CO Limiting reagent is carbon.So 1.20 mol C reacts with 0.40 mol of 2 3Fe O to give 0.80 mol of Fe and 1.20 mol of CO (according tostoichiometric coeficient of balanced chemical reaction)

Topic: Mole concept-1_(Limiting reagent)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

81. (C)The temperature and pressure at which all the three states of water i.e. ice, water, vapour exist atequilibrium is called triple point of water.Temp = 0.0098ºCPressure = 0.006 atm or 4.58 mm Hg

2 2 2( ) ( ) ( )H O s H O l H O v Topic: Solution and Colligative properties_(Phase diagram of water)_L-1_IIT-JEE_Sr-15_Full Test-5_Main

Level_PDch sir.

82. (D)

2CO deviates from ide al behaviour maximum and H2 minimum among above gases. Lighter gsesalways deviate positively (Z > 1)

Topic: Gaseous state_(Real gases)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

83. (A)126 0U W q

vC T U

VU nC T ; 1n



so VU C T 1 1 112.6 126T J k mol J mol

126 1012.6

T k k

Topic: Themodynamics_(Processes)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

84. (C)

2

1ln VS nR

V

8.314 2.303log 2 5.76 JKmol–1.

Topic: Themodynamics_(Entropy)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

85. (C)1 1

t O

KtC C

tC concentration at any time t.

oC initial concentration.

for half life 0

2tCC

so, 1/ 20 0

2 1 ktC C

1/ 20

1ktC

1/ 2 1/ 20 0

1 1t tkC C

Topic: Chemical Kinetics_(IInd order reaction)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

86. (D)Only urea undergo hydrolysis to give CO2 and NH3 .

Topic: GOC_(Physical and Chemical properties)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.87. (A)

Cl

:

O OO

SO

O: –

OC

–F

:F

FO

H

:H

H

+

Cl – F

:

:

F

F

BF F

F

Planer

CH H

H

Planer

+

–

Topic: Chemical bonding_(VSEPRT)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.



88. (B)

Cr

OH2

OH2

OH2

OH2

Cl

Cl

(cis)

CrOH2

OH2

Cl

(Trans)

H O2

H O2

ClTopic: Coordination Chemistry_(Isomerism)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

89. (A)As metallic character of element attached to oxygen decreases (electronegativity increases) acidity of oxideincreases.

Topic: P-block_(Oxides)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

90. (D)Fe has (III) oxidation state not (II)

Topic: Coordination Compound_(Isomerism)_L-1_IIT-JEE_Sr-15_Full Test-5_Main Level_PDch sir.

Documents

Solution Sr 15 Main Test 5