solution sr 15 main test 4 full portion 08-02-15 for uploding

ANSWER SHEET

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

RAO ALL INDIA TEST SERIES

MAIN FULL TEST [4]Batch: Seniors-15 Date:

Rao All India Test Series_Sr-2015 Batch_MAIN Full Test - [4]_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= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =1. (A)

b = 2, c = 3 , 30A Now, 2 2 2 cosa b c bc A

34 3 2 2 3 1

2

( ) tan tan2 2 2A b c a Ar s a

3 1 tan152

3 1 3 12 3 1

3 12

Topic: S.O.T. (Incentre)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

2. (A)For x > 0 or x < 0

'( ) 2 1af x bxx

'(1) 0f a + 2b + 1 = 0 ..........(i)

and '(3) 0f 6 1 03a b ..........(ii)

On solving Eqs. (i) and (ii), we geta = 3 / 4, b = 1 / 8

Topic: A.O.D. (Maxima & Minima)_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

3. (C)On shifting standard deviation remains constant.On magnifying standard deviation gets magnified by same value.

2 7new old Topic: Statistics (Standard deviation)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

4. (B)p q p q p q ( ) ( )p q p q T T T F FT F F T FF T T F FF F T F F

Topic: Reasoning (Logic)_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.



5. (C) x y dx dy dx dy

dx dydx dyx y

on integration, ln lnx y x y c ( ln c = constant of integration)

ln x y c

x yc x y e

. x yx y k e 1ck

Topic: Differential Equation_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

6. (C)Let S = cot1 2 + cot1 8 + cot1 18 + cot1 32 + ..... Tn = cot

1 2n2

12

1tan2n

1 12

2 (2 1) (2 1)tan tan4 1 (2 1)(2 1)

n nn n n

= tan1 (2n + 1) tan1 (2n 1)

1 11{tan (2 1) tan (2 1)}n

nS n n

= 1 1tan tan 1

2 4 4

Topic: I.T.F. (Series)_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

7. (D)Let E denote the event that a six occurs and A the even that the man reports it is a six. We have

P(E) = 1/6, P(E ' ) = 5/6, P(A/E) =34 and P(A/E

' ) = 1/4

By Bayes theorem( ) ( / )( / )

( ) ( / ) ( ') ( / ')P E P A EP E A

P E P A E P E P A E

1 336 4

1 3 5 1 86 4 6 4

Topic: Probability (Bayes Theorem)_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.



8. (B)Given that A is a singular matrix. | A | = 0 A adj A = | A | I = 0 A adj A is a zero matrix.

Topic: Matrices (Singular Matrix)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

9.. (D)10 1 10

1 1 1( 1)

n

n i ni n

1 2 3 ..... 10 1 1 ......10 times

10(10 1) 10

2

55 1045

Topic: Progressions (Summation)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

10. (B) x y = 4 .....(i)and xy = 9 .....(ii)Eqs. (i) and (ii) are the equations of rectangular hyperbolas. 1 22 2,e and e then e1 e2 = 0

Topic: Hyperbola (Rectangular Hyperbola)_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

11. (D)Since, y-axis is major axis. f (4a) < f (a2 5) 4a > a2 5 ( f is decreasing) a2 4a 5 < 0 ( 1,5)a

Topic: Ellipse (Formation of ellipse)_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

12. (B)

We have, 1 2 3

1 1 1 1, , ,...,nx x x x

are in AP..

2 1 3 2 1

1 1 1 1 1 1...n n

dx x x x x x

(say)

11 2 2 3

1 2 2 3 1

... m nn n

x xx x x x dx x x x x x

Now, x1x2 + x2x3 + ... + xn + 1 xn

1 2 2 3 11 [ ... ]n nx x x x x xd

1 nx xd



But,1

1 1 ( 1)n

n dx x

1

1

( 1)nn

x x n dx x

or 1 1( 1)n nx x n x x

d

x1x2 + x2x3 + ... + xn1xn = (n 1)x1xn

Topic: Progressions (Harmonic Progression)_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

13. (A)sin x > sin3 x

/ 2 / 2 3

0 0sin sinx dx x dx

I1 > I2

Topic: Definite Integration (Trigonometric)_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

14. (D)

31dxI

x x

On multiplying the Nr and Dr by x2, we get2

3 31x dxI

x x

On putting 1 x3 = t2 31t x

3x2 dx = 2t dt

2 3 22 1

3x dx t dt and x t

223 (1 )

t dtIt t

2

2 2 1 1log3 1 3 2 1

dt t ct t

1 1log3 1

t ct

3

3

1 1 1log3 1 1

x cx

Topic: Indefinite Integration (Algebraic function)_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

15. (C)Let projection be x, then

( ) ( ) 2 2

x i j x i ja xk

2

2x ja xk



unit vector 2

3 3ka j

Topic: Vectors (Projection)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

16. (D)If p1 and p2 be the distance between parallel sides and be the angle between adjacent sides, thenRequired area = p1p2 cosec

Where, 1 22 21 1,

(1 ) (1 )p p

m n

(distance between parallel lines)

p1

p2y = mx

y = mx + 1y = mx

y = nx + 1 | m n |

1 + mn

and| |tan|1 |

m nmn

Required area2 2

2 2

(1 ) (1 )1 .| |(1 ) (1 )

m nm nm n

1| |m n

Topic: Straight line(Area of parallalogram)_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

17. (A)Equation of the plane through (5, 1, 2) is

a(x 5) + b(y 1) + c(z 2) = 0 ... (i)Given plane (i) is perpendicular to the line

2 4 51/ 2 1 1x y z

... (ii)

Equation of normal to plane of Eq. (i) and straight line (ii) are parallel

i.e.,1/ 2 1 1

a b c k (say)

, ,2ka b k c k

From Eq. (i).

( 5) ( 1) ( 2) 02k x k y k z

or x + 2y + 2z = 11 ... (iii)

Any point on Eq. (ii) is 2 ,4 ,52

Which lies on Eq. (iii), then 2 . Required point is (1, 2, 3).

Topic: 3-D(Plane)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.



18. (B)599 = 5(52)49 = 5(25)49 = 5(26 1)49 = 5 26 (Positive term) 5 5 26 (positive term) 5 13 13 5 26 (another positive term) 8 So, when it is divided by 13, it gives the remainder 8.

Topic: Mathematical Induction_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

19. (C)AB A

2ABA A 2AB A BA B

2A A AB A Similarly 2B B

Topic: Matrices( Simple algebraic operations )_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

20. (D)A

DB CSince AD is altitude of ,ABC D will always lie on circles having diameters as AB and AC.(angle in semi-circle is 90)

Topic: probability_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

21. (A)Given, u v u w and w u v

( )u v u u v

( )u v u v

( )v u v u v

( ) 0u v u

( ) 0u v

Now, [ ] ( )u v w u v w

( ( ))u v u v u

( ( ) )u v u v v u 2(| | ( ) )u v u u v v v u

2 2| | | | 1v u

Topic: Vectors(Scalar triple product)_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

22. (D)

arg (z 3i) = arg (x + iy 3i) = 34

x < 0, y 3 > 0 34

isin II quadrant



3 3tan 14

yx

y = x + 3 ...(i)0 3x and y

and arg (2z + 1 2i) = arg[(2x + 1) + i(2y 2)] 4

2x + 1 > 0, 2y 2 > 0 4is in I quadrant

2 2 tan 12 1 4yx

2y 2 = 2x + 1

3 1, , 12 2

y x x y ...(ii)

From Eqs. (i) and (ii)

3 /4

(0, 3)

/4(0, 1)

12

,0

It is clear from the graph that there is no point of intersection.Topic: Complex Numbers (Vector representation)_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

23. (B)52 32 2 2cos cos ....cos cos cos ...cos

65 65 65 65 65 65

6

6

2 64sin sin65 65

2 sin 64sin65 65

sin1656464sin

65

Topic: Trignometric Ratios (Product Series)_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

24. (B)cos cos 2 cos3 cos 4 cos5 5x x x x x which is possible only when cos cos 2 cos3 cos 4 cos5 1x x x x x and is satisfied by 0x only.Hence, number of solution = 1

Topic: Trigonometric equations_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.



25. (B)

Let 1/3( )f x x

Now, 2 / 3( ) ( ) '( ). 3xf x x f x f x x

x

we may write 0.007 as 0.008 0.001Taking 0.008 and 0.001x x we have

2 /30.001(0.007) (0.008)

3(0.008)f f

as 1(0.007) 0.2

120f

1/3 230.007120

Topic: Application of Derivative_(Approximation)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

26. (C)

Total number of rectangle 15.10(15 1)(10 1)

4

6600

Number of squares 10

1(11 )(16 )

rr r

= 660

Required number 6600 6605940

Topic: Permutation and Combination_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

27. (C)Given summation is equal to

2

3lim 1nn

n

3( 1)(2 1)lim

6 1nn n n

n

3

1 11 21 2 1lim

1 6 1 36 1nn n

n

Topic: Limits _ / _L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir..

28. (A)

Homogeneous equation is 2

2 2 34 y xx yc

2 2 2 2 2 24 9 6c x c y y x xy If they are perpendicular pair of lines,

2 236 4 0c c (coeff. of 2 2coeff. of 0x y )2 20c

Topic: Pair of st. lines_(Homogenisation)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.



29. (A)The event that the fifth toss results in a head is independent of previous results.

Probability of required event 12

Topic: Probability_(Matually independent events)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.

30. (C)Required area

1,4 2

3 1,4 2

( , 0)( , 0)

(0, 1)

(0, 1)4

34

(cos sin )x x dx

= / 4 / 4(sin cos )x x

1 1 1 1 2 22 2 2 2

Topic: Area under the curve_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_AHm sir.



PART-B: PHYSICS= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =31. (C)

The rotational inertia about the X axis ( XI ) , about the Y axis ( YI ), and about the Z axis ( ZI )may be written as

2 2

2 2

2 2

( )

( )

( )

X i i ii

Y i i ii

Z i i ii

I m y z

I m z x

I m x y

Wherein the point masses are multiplied by the squares of their distances from the respective axis. In thepresent case, we may substitute the coordinates of the four particles and masses in the above expressions toget (z = 0)

2

2

2

(2 3 1.5 2.5)(1.5 ) 20.25

(2 3 1.5 2.5)(1.0 ) 9.00

(2 3 1.5 2.5)(1 1.5 ) 29.25

X

Y

Z

III

Topic: Rotational Motion_(Moment of inertia)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

32. (A)Heat lost by steam of mass M is M [2300 + (4.2)(20)] kJ or [2384 M] kJ. This must be equal to the heatgained by the calorimeter system, which is [4.2(65)] kJ = 273 kJ. Equating heat lost with heat gained, weget 0.11 kg for M.

Topic: Heat and thermodynamics_(Method of mixtures experiment)_L-1_IIT-JEE_Sr-15_Full Test-4_MainLevel_DKp sir.

33. (A)

The volume of the material of the alloy is 4 300.800 1.0 108000

V m .

The volume of the displaced water is 4 30.800 0.698 1.02 10

1000V m .

This means the overall volume of the block is in excess by 4 30.02 10V m .Topic: Fluids_(Archimedis principle)_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

34. (B)Work energy theorem gives the work as equal to

2 21

1

12

0.2(10) 2

0.2(40) 8

f i

i

f

W m v v

v m sv m s

Topic: Work Power Energy _(Work)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.



35. (A)

The time period of a simple pendulum is given by 2lTg

. Therefore,

2' '

'

T g g RandT g g R h

Topic : Oscillations_(Simple pendulum)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

36. (C)

The resistance of a cable is given by 2 2L mR andr ne

.

Topic: DC Circuits_(Resistance)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

37. (A)

We know that 2 2n

hr n and r np

. 2nhp

r

Topic: Atomic Physics_(Bohr theory)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

38. (A)We have to solve the two equations:

0.54 1.53

h W

h W

Topic: Wave-particle duality_(Photoelectric effect)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

39. (C)

0.50 ; 0.251 1 1

1 2.0

u m v m

v u f

f

Topic: Geometrical optics_(eye defects)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

40. (B)

In the fundamental mode, the quarter wavelength equals the column length plus the end correction.The first overtone corresponds to three quarters of wavelength equaling the column length plus theend correction. Accordingly, we have



0.104

3 0.354

Solving these equations gives the result.

Topic: Wave-motion_(Air column resonance)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

41. (B)

Faradays law is applicable for self-induced emf, with the sense of the emf governed by the principle ofLenzs law.

Topic: Electromagnetics_(Self-induced emf)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

42. (C)

The self-inductance of a solenoid is proportional to 2

2 Nn lA Al

. N is the total number of turns and l is

the length, A being area of cross-section. As such the inductance is increased by the factor 8.Topic: Electromagnetism_(Self-inductance)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

43. (A)

2

2

6

6

61000(0.2) 200

PI R P VI

VII R

VIR

R

Substituting the data yields the result.

Topic: DC circuits_(Power transmission lines)_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

44. (A)

Remember the reactance of an inductor is proportional to the frequency, while that of a capacitor isinversely proportional to frequency. Resistance is independent of frequency.

Topic: AC circuits_(Impedance)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

45. (A)Consider the two linear oscillations:

sinmx x t

sinmy y t



When the phase difference is 2

, we can solve for the trajectory by

eliminating time, to find that it is a circle m my x . Unequal amplitudes give elliptical motion.Zero phase difference or 180 deg phase difference lead to linear motion inclined to the axes and aslope which is the ratio of the two amplitudes. Please work out these cases for yourselves.

Topic: SHM_(Combination of SHMs)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

46. (D)

From the equation of Newtons II law of motion, we get for the speed of a planet, GMv

R ,

where M is the mass of the sun. The magnitude of the angular momentum for the circular orbit isgiven by m v R, m being the planets mass. Hence the result.

Topic: Gravitation_(Planetary motion)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

47. (A)

The activity is given by dNR Ndt

.

Topic: Nuclear physics_(Radioactivity)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

48. (A)The electric field E is related to the potential by the equations:

, ,x y zV V VE E Ex y z

And the force components are obtained by multiplying these with the point charge. You may noticethat the field is uniform independent of position.

Topic: Electrostatics_(Potential and field)_L-2_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

49. (A)The gases used obviously do not approximate sufficiently well towards an ideal gas. As per the definedgas temperature, we have to reduce the data to zero gas pressure. Assuming a linear dependence, wemay try

0 1T T c P Solve for the true temperature after substituting the data on the argon thermometer. The true temperatureis 0T = 450.00 K. (for P = 0) and

11 0.02c K kPa

.Now substitute the true temperature in the equation for the nitrogen thermometer:

0 2T T c P Using the single reading given, we deduce that the constant for this meter turns out to be

12 0.03c K kPa

. Use this result along with the true temperature in the above equation for thepressure of 1.0 kPa.

Topic: Heat and Thermodynamics_(Temperature measurement_L-1_IIT-JEE_Sr-15_Full Test-4_MainLevel_DKp sir.



50. (B)

The statement I is correct by virtue of Newtons II law of motion; the book being in equilibrium. TheIII law refers to the mutual force between two bodies; weight force on the book is due to the earth andit is only the normal contact force of the book on the table that pairs (in accordance with the III law)with the normal contact force of the table on the book.

Topic: Dynamics_(Newtons III law of motion)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

51. (B)The momentum of the bead at A is iP mvi

mv = piA

Bmv/2 = pf

yx

The moment of the bead at B is 2f

mvP i

Therefore, the magnitude of the change in momentum between A and B is32f i

P P P mv

Average force exerted by the bead on the wire is32av

p mvFt T

Topic: Centre of mass and Collision_(Thrust force)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

52. (B)Rolling is rotation about point of contact. Applying impulse momentum equation about P.

PJ R h I ....(i)and J mv ....(ii)

As sphere rolls 2 2 22 7, and5 5

v R I mR mR mR

After solving, we get 25

hR

Topic: Rotational motion_(Rolling)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

53. (C)The air pressure is greater inside the smaller bubble (4 / )T r . Hence, air flows from the smaller to the largerbubble.

Topic: Properties of matter_(Surface tension)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

54. (B)Change in pressure due to placing of mass on piston is

MgpA

From bulk mudulus definition /dpK

dV V

dV p MgV K AK



From 343

V r

3 13 3

dV dR dR dV MgV R R V AK

Topic: Proportion of matter_(Elasticity)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

55. (A)

1 2,kQ kQ kQ kQV Va d b d

a bd

QQ

V1 V2

0

1 2

41 1 2

QCV V

a b d

Topic: Electrostatics_(Capacitors)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

56. (B)Let a current of x ampere pass through the voltmeter; then 4 x ampere passes through the resistance R.Therefore, voltmeter reading is

2020 4 , . ., , . ., 54

x R i e R i e Rx

Topic: Current electricity_(Meters)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

57. (C)By Lenzs law, clockwise current is induced in both loops. Greater the area, large will be the induced emf,outer loop has greater area.

Topic: EMI _(Induced emf)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

58. (A)Condition for maxima is

sind n 0.50sin 0.252.0

n n nd

As sin lies between 1 and 1, so we wish to find all values of n for which | 0.25 | 1n These values are 4, 3, 2, 1,0, 1, 2, 3, 4 . For each of these, there are two different values of except for 4 and +4. A single value of , 90 and 90 , is associated with 4n and 4n ,respectively. Thus, there are 16 different angles in all and therefore 16 maxima.

Topic: Wave optics _(Interforma)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

59. (B)

For an isothermal process, PV constant and for the given process 2PV constant.Therefore the gas is cooled because volume expands by greater exponent than in an isothermal process.

Topic: Heat and Thermo_(Thermodynamics)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.

60. (C)1 cos(2 2 )

2t kxy a

cos(2 2 )2 2a ay t kx

Topic: Waves_(Superposition)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_DKp sir.



PART-C: CHEMISTRY= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =61. (A)

Sol: Maximum capacity or volume of balloon 8 480 548.7mL7

Apply1 2

1 2

V VT T

2

480 548.57278 T

2T 317.71KTopic: Gaseous state_(Charls Law)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

62. (A)

Sol: E photon absorbed = hc

196.625 10 J

E photon re emitted out = 194 10 Joule E absorbed = EI photon + EII photon re emitted out

19 19IIE photon 6.625 10 4 10

192.6 10 Joule Topic: Atomic structure (Quantum theory of radiation )_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

63. (A)Sol: 238 m 492 A 2X X He

238 = m + 4m = 234

92 = A + 2A = 90234 90 144

Topic: Radioactivity_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

64. (B)

2 2 33 2g g gN H NH

23

32 2

C

NHQ

N H

Given, 38.13 1.570.4065 0.078520 20

NH M M M M

21.92 0.09620

H M M



23 2

3

0.40652.379 10

0.0785 0.096C

MQ M

M M

C CQ K , so the reaction mixture is not in equilibrium. C CQ K , it indicates that the reaction will proceedin the direction of reactants.

65. (B)In 2 2 23 2 x y z x y zsp d s p p p d d

Topic: Chemical bonding,(VBT)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

66. (D)

2 4K NiCl is paramagnetic with two unpaired electrons.

Cl ions are weak field ligands, so pairingnot possible.

Topic: Coordination compounds_(VBT)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

67. (D)Lewis acid strength 3 3 3 3BF BCl BBr BI . Greater the extent of back bonding lesser the Lewisacid strength and extent of backbonding follows order 3 3 3 3BF BCl BBr BI

Topic: p-block, Boron Family_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

68. (C)Conceptual.

Topic: Coordination Compounds_(VBT)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

69. (A)In alkyl halide degree of unsaturation is zero

Topic: Alkyl halides_(General formula)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_NGch sir.

70. (A)

Sol:

COOHOH

H3 2(CH CO) O

COOH3OCOCH

3CH COOH

AspirinSalicylic acid

+

Topic : Phenols_(Aspirin)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_NGch sir.

71. (B)Sol: Only 3 alcohols give alkene when heated with copper.Topic: Alcohols_(Dehydrogenation Cu/300C)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

72. (B)Sol: (A) & (C) will oxidise phenol due to high electron density in ring. Dil. 3HNO being weak oxidising agent will

nitrate phenol(D) will produce Nitrosophenol.

Topic: Phenols_(Nitration)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_NGch sir.



73. (D)canizaro reaction takes place in aldehydes having no hydrogen.

Topic: Aldehydes & ketones_(canizaro reaction)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_NGch sir.

74. (D)OH

BO

BHO O

BO

B OHO

O

OH

+2 Na . 8H O2

Water molecules are associated with the metal ions.Borax is usally written as option (C), so both (A) and (C) are correct.

Topic: P-block_(Boron family)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

75. (A)Diborane 2 6B H undergoes hydrolysis to give boric acid. Alkane doesnot undergo hydrolysis.

Topic: P-block _(Carbon family)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

76. (A)Bond energy (kJ/mol)C C 348Si Si 297Ge Ge 260Sn Sn 240down the group bond energy decreases.

Topic: Chemical bonding_(Bond energy)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

77. (D)

* 44 :SiO tetrahedral unit exists without any sharing of oxygen (ortho)

* 62 7 :Si O two tetrahedral units are joined by sharing the oxygen at one corner..

* All four corner of 4SiO tetrahedron are shared and 3-D lattice of formula 2SiO is formed.Topic: P-block_(Silicates)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

78. (C) 2 1 2 1 0 1x x

Topic: Mole concept-2_(Oxidation state)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

79. (C)H N C NH2 2

NHH +

Guanidine is very strong base



CH N2 NH2

:

NH2:

CH N2 NH2

NH2

:

CH N2 NH2

NH2+

+ +

Topic: GOC_(Acid-base)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_NGch sir.

80. (B)At constant T, PV = constant

& PV T (PV = nRT)so, 2 1T T .

Topic: Gaseous state_(Ideal gas)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

81. (A)Charge distance = dipole moment

18

81.2D 1.2 10Charge1 10

esu cmcm

101.2 10 esu The fraction of an electronic charge, e, is

10

19charge 1.2 10

charge of one electron 1.6 10esuc

19 19 9

10

Charge on an electron 1.6 10 1.6 10 3 10

4.8 10 Charge on an electron in esu.

c esu

esu

The fraction of an electronic charge 10

101.2 10 0.25 25% of4.8 10

e e

Topic: Chemical bonding_(Dipole moment)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

82. (A)

Bond order 1bond length

Bond order

12

32

NONO

Bond order No. of bonding electrons -No. of antiboning electrons2

12 2

2

23

NN

2

2

2.52

OO

22

2

1

1.5

O

O

Topic: Chemical bonding_(Molecular orbit)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

83. (D)Gas constant per molecule is Boltzmans constant.

Topic: Basic concept_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.



84. (A)

2 2 4 2 42

2 2 2 4

1 5 13434 1 0.2

mol H S mol H SO litre H SOgm H Sgm H S mol H S mol H SO

25 litre.

Topic: Mole concept-2_(Concentration terms)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

85. (A

OC

HO

H

+

Topic: GOC_(Hydrogen bonding)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_NGch sir.

86. (B)

Cl 3 4 4 2 2

10

12 16

10

2.8 MTopic: Mole concept -2_(Molarity)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

87. (D)

For MX, we have 2; ( )( )spMX M X K M X s s s

Hence 1/ 21/ 2 8 2 44.0 10 2 10sps K M M

For MX2, we have22 2 2 3

2 2 ; ( )(2 ) 4spMX M X K M X s s s

Hence 1/31/3 14 3 5( / 4) 3.2 10 / 4 2 10sps K M M For 33 3we have 3M X M X M X

3 3 3 4(3 ) ( ) 27spK M X s s s

Hence 1/ 4 15 4 1/ 4( / 27) (2.7 10 / 27)sps K M

41.0 10 M Hence, the solubility follows the order 3 2MX M X MX Therefore, the choice (D) is correct.

Topic: Ionic equilibrium_(Solubility product)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.



88. (B)Hydrogen is produced due to the reaction

2 22 ( ) 2 ( ) 2 ( )H O l e H g OH aq

Since ( / )( / ),em Q F M V we get1 1 ee e

nFVm Q ltn tM F V F V I

Substituting the given data, we get1

43

(0.01 )(96500 )(2) 19.3 10(10 10 )

mol C molt sA

Therefore, the choice (B) is correct.

Topic: Electrochemisty_(Faradays Law)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

89. (C)In 2,CuF Cu exist as 2Cu with one unpaired electron. Its colour is blue.

In other choices, Copper exists as Cu with no unpaired electrons and thus are colourless.Topic: D-block_(Colour)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_PDch sir.

90. (D)Electron-donating 3CH group stabilizes carbocation. Hence, (I) is more stable than (III).2 carbocation is more stable than 1 carbocation. Hence (II) is more stable than (IV)Of the species (II) and (III), the species (III) will be more stable due to the lone-pair of electrons on oxygenatom. Hence, the order of stability is I > III > II > IV

Topic: GOC_(Carbocation)_L-1_IIT-JEE_Sr-15_Full Test-4_Main Level_NGch sir.

solution sr 15 main test 4 full portion 08-02-15 for uploding

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