MELUHA INTERNATIONAL SCHOOL..., then the value of 1 is _____ 24. IfP P 1 2, denote the length of the...

MELUHA INTERNATIONAL SCHOOL

INDIA SR MPC Date: 27-03-2020 Time: 3 Hours

SECTION – I (SINGLE CORRECT ANSWER TYPE)

This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and -1 if not correct.

MATHS

SYLLABUS : Algebra of Matrices (up to adjoint of matrices), Determinants & Pair of straight lines & Limits and continuity

1. The values of x, so that 1 3 2 1

1 1 0 5 1 1 00 3 2

xx

is /are

1) 2 2) 0 3) 7 352

4) 9 532

2. Matrix X satisfying the equation 3 3 0

23 3 2

X Y

, 4 1 5

21 4 4

Y X

is

1 2 1 0

21 2 1

2) 2 1 1

21 2 0

3) 2 1 11 2 0

4) 3 1 21 2 0

3. If a p

A b qc r

,then det TAA is equal to

1) 0 2) 7 3) 2 4) 3

4. If cos sinsin cos

E

, then .E E is equal to

1) E 2) E 3) E 4) E

5. If cos sinsin cos

A

and A (adj A) = I , then the value of is

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

6. If A

is such that 2A I , then

1) 21 0 2) 21 0 3) 21 0 4) 21 0

7. If 0 11 1

A

and 0 11 0

B

, then which of the following is correct?

1) 2 2.A B A B A B 2) 2 2.A B A B A B

3) . 1A B A B 4) None of these

JEE MAINS MODEL

M E L U H A I N T E R N A T I O N A L S C H O O L

Page | 2 :: JEE MAINS MODEL 27 MAR 2020

8. If 0 tan

2

tan 02

A

and I is the identity matrix of order2, I Acos sinsin cos

is equal

to 1) A 2) 1 3) 1 A 4) None of these

9. If 2 3

2 34 0

X Y

and

2 23 2

1 5x y

, then the value of x and y is

1) 5 01 4

and 2 01 1

2) 5 021 0

and 2 10 1

3) 4 13 22

and 1 1

512

4)

2 125 511 35

and

2 135 5

14 25

10. If 0

00

a bA a c

b c

, if '112

Q A A and '212

Q A A , then 1 2,Q Q is equal to

1) 3l 2) 30 3) A 4) 2A 11. If 2 22 5 2 0x xy y represents two sides of a triangle whose centroid is 1,1 then the equation

of the third side is 1) 3 0x y 2) 3 0x y 3) 3 0x y 4) 3 0x y 12. If the pair of lines which joins the origin to the point of intersection

of 2 22 2 0ax hxy by gx , 2 21 1 1 12 2 0a x h xy b y g x are at right angles then

1) 1 1

1

g a bg a b

2)

1 1 1

g a bg a b

3)

1 1 1

h a bh a b

4) 1 1

1

h a bh a b

13. If a and b are positive numbers a b , then the range of values of K , for which a real be

found such that the equation 2 22 2 1 0ax xy by k x y represents a pair of straight lines is

1) 2a K b 2) 2a K b 3) 2K a or 2K b 4) 2K a or 2K b 14. Two of the lines represented by 3 2 2 33 3 0ax bx y cxy dy will be perpendicular if 1) 2 2 0a ac db d 2) 2 23 0a ac bd d

3) 2 23 0a ac bd d 4) 2 2 0a ac bd d

15. If the line 3y x cuts the curve 3 3 2 23 5 3 4 5 1 0x y xy x y x y at the point , ,A B C then . .OA OB OC is

1) 4 3 3 113

2) 2 73 3) 3 3 1 4) 4 3 3 1

13

16. Area enclosed by the pair of lines 2 22 2 2 0S ax hxy by gx fy c and 1 2 22 2 2 0S ax hxy by gx fy c

1) 2

c

h ab 2)

2

2 c

h ab 3)

2

3 c

h ab 4)

2

4 c

h ab



17. The lines through origin which make angles of 045 with the line 0lx my n are 1) 2 2 2 2 4 0l m x y lmxy

2) 2 2 2 2 4 0l m x y lmxy

3) 2 2 2 2 4 0l m x y lmxy

4) 2 2 2 2 4 0l m x y lmxy

18. The equation of image of pair of lines 1y x in y axis is

1) 2 2 5 0x y 2) 2 2 2 1 0x y x 3) 2 2 2 1 0x y x 4) 2 2 2 1 0x y x

19.

3sin44tan3

x

x

x

Lt

1) 0 2) 1 3) 2 4) 3

20. 3

33

88

n

rn

rLtr

1) 17

2) 27

3) 34

4) 72

SECTION-II

(Numerical Value Answer Type)

21. If b c c b

c c a a kabcb a a b

, then the value of k is ______

22. If 32

22

x y z x yz y z x y k x y zz x z x y

, then the value of k is ______

23. If

2

2

2

111

x xy yz z

and 1

1 1 1yz zx xyx y z

, then the value of 1 is ______

24. If 1 2,P P denote the length of the perpendiculars from the point 2,3 on the lines given by the

equation 2 215 31 14 0x xy y if 1 2P P then 2 21 2

1 7413 ____5 31

P P

25. If the angle between the pair of straight lines 2 2 2 2 2 1 0y Sin xySin x Cos is then the

value of 2 25 15 20 ____Sin Sin Cos



SECTION – I (SINGLE CORRECT ANSWER TYPE)

This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and -1 if not correct.

PHYSICS

SYLLABUS: Impulse and Momentum,Work, Energy 26. Choose the correct alternative among the following: 1) When a conservative force does positive work on a body, the potential energy of the body

increases 2) Work done by a body against friction always results in a loss of its kinetic energy 3) The rate of change of total momentum of a many particle system is proportional to the sum of the

internal forces on the system 4) In an elastic collision kinetic energy of the colliding bodies before collision is equal to kinetic

during collision 27. A spring of spring constant k is stretched by a length x from its natural length and released.

What about its potential energy: 1) Remain stored in spring in any form 2) Converted into heat 3) Transfer to air molecules 4) None of the above 28. Alight and a heavy body have equal momenta. Which one has greater K.E.? 1) The light body 2) The heavy body 3) The K.E. are equal 4) data is incomplete 29. The graph of kinetic energy (K) of a body versus velocity v is represented as: 1) 2) K K v v 3) 4) K K v v 30. The only force acting on a 2kg body as it moves along the positive x axis has component

6xF x N, where x is in metre. The velocity of the body at 3x m is 8 m/s. the velocity of the body at 4x m is:

1) 10m/s 2) 9 m/s 3) 7.4 m/s 4) 6.6 m/s 31. A toy gun has a spring of force constant k . After charging the spring by compressing it

through a distance x , the toy releases off shot of mass m vertically upwards. Then the shot will travel a vertical height of

1) 2mgkx

2) 2kx

mg 3) kx

mg 4)

2

2kxmg



32. The body of mass m is pulling, vertically up with constant speed, by applying force P. the free end of the string is pulled by l meter, the increase in potential energy of the block is:

1)

2mgl 2) mgl 3) 2mgl 4)

4mgl

33. A block of mass M is through over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force F. the kinetic energy of the block increases by 20 J in 1 s:

1) The tension of the string is Mg 2) The tension in the string is F.

3) The work done by the tension on the block is 20 J in the above 1 s. 4) The work done by the force of gravity is – 20 J in the above 1 s. 34. A force acts on a 30g particle in such a way that the position of the particle as a function of

time is given by 2 33 4x t t t , where x is in metres and t is in seconds. The work done during the first 4 seconds is

1) 5.28 J 2) 450 J 3) 490 J 4) 530 J 35. A particle of mass 100g is thrown vertically upwards with a speed of 5m/s. The work done by

the force of gravity during the time particle goes up is 1) -1.25 J 2) 5.17 kJ 3) 0.5 J 4) -0.5 J 36. A spherical ball of mass 20 kg is stationary at the top of a hill of height 100m. it slides down a

smooth surface to the ground, them climb up another hill of height 30m and finally slides down to a horizontal base at a height of 20mabove the ground. the velocity attained by the ball is

1) 10 /m s 2) 10 30 /m s 3) 40 /m s 4) 20 /m s 37. Two blocks, each of mass m moving with speed v collide with the spring of force constant k as

shown in figure. The maximum compression of the spring is:

1) 2mv

k 2)

22mvk

3) 2

2mv

k 4) Zero

38. Calculate the work done, if a wire is loaded by ‘Mg’ weight and the increase in length is ‘ l ’:

1) Mgl 2) 2

Mgl 3) 2Mg l 4) Zero

39. A 60g bullet is fired through a stack of fibre board sheers 200mm thick. If the bullet approaches the stack with a velocity of 600m/s, and emerges out with a velocity of 300 m/s, the average resistance offered to the bullet is:

1) 40.5 kN 2) 2 k N 3) 20.25 kN 4) 10 kN

40. A particle in a conservative force field has a potential energy given by 20xyUz

. The force

exerted on it is



1) 2

20 20 20y x xyi j kz z z

2) 2


3) 2


4) 2


41. A locomotive of mass m starts moving so that its velocity varies according to law v a s , where a is a constant, and s is the distance covered. The total work done performed by all the forces which are acting on the locomotive during first t seconds after the beginning of motion is:

1) 2ma t 2) 4

2ma t 3)

4 2

8ma t 4) None of these

42. A 10 kg block is pulled in the vertical plane along a frictionless surface in the form of an arc of a circle of radius 10m. the applied force is of 200N as shown in figure. If the block had started from rest at A, the velocity at B would be:

1) 1.7 m/s 2) 17 m/s 3) 27 m/s 4) 34 m/s

43. A ball of mass ‘m’ is thrown at an angle is ‘ ’ with the horizontal with an initial velocity ‘u’ .

The change in its momentum during its flight in a time interval of ‘t’ is

1) mgt 2) cosmgt 3) sinmgt 4) 12

mgt

44. An impulse “I” given to a body change its velocity from “ 1v to 2v ”. The increase in the kinetic

energy of the body is given by 1) 1 2I v v 2) 1 2 / 2I v v 3) 1 2I v v 4) 1 2 / 2I v v 45. An impulse is supplied to a moving object with the force at an angle 0120 with the velocityand

the change in momentum vector is 1) 0120 2) 00 3) 060 4) 0240

SECTION- II (Numerical Value Answer Type)

46. A force 6 2 3F i j k

acts on a particle and produces a displacement of 2 3s i j xk

. If the work done is zero, the value of x is: ____

47. A body of mass 0.5kg travels in a straight line with velocity 3/ 25v x . The work done by the net force during the displacement from 0x to 2x m is (J) _____

48. Figure gives the acceleration of ,a 2.0 kg body as it moves from res along x axis while a variable force acts on it from 0x m rest to 9x m . The work done by the force on the body when it reaches 7x m is given by (J) ______

49. At 0t a 1.0 kg ball is thrown from the top of a tall tower with velocity 8 9 /v i j m s

. The

change in the potential energy of the ball – earth system between 0t to 5t s :

210 /g m s (J) ______

50. A body is acted on by a force given by 10 2F t N . The impulse received by the body during the first four seconds is (Ns) ______



SECTION – I

(SINGLE CORRECT ANSWER TYPE) This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and -1 if not correct.

CHEMISTRY

SYLLABUS :Chemical Bonding & Stoichiometry 51. Given that g-atomic mass of H is 1.008 g. Calculate the mass of one H-atom in g. 1) 241.67 10 g 2) 241.90 10 g 3) 241.76 10 g 4) 60g 52. Hydrated barium iodide molecule contains x molecules of water 2 2.BaI xH O . A sample of it

weighing10.407g when heated strongly gives off water completely and anhydrous residue left weighs9.520g . Find the value of x .

1) 2 2) 3 3) 6 4) 4 53. Two samples of lead oxide were separately reduced to metallic lead by heating on a current of

hydrogen the weight of lead from one oxide was half the weight of lead obtained from the other oxide, the data illustrates 1) Law of reciprocal proportions 2) Law of constant proportions 3) Law of multiple proportions 4) Law of equivalent proportions

54. 50 ml of an aqueous solution of Glucose contains 226.02 10 molecules. The concentration of solution is

1) 0.1 M 2) 1.0 M 3) 0.2 M 4) 2.0 M 55. The molality of 15% /mass Vol solution of 2 4H SO of density 31.1 /g cm is approximately 1) 1.2 2) 1.4 3) 1.8 4) 1.6 56. Two elements X (atomic mass =75) and Y (atomic mass =16) combine to give a compound

having 75.8% of X. The formula of the compound is 1) XY 2) 2X Y 3) 2 2X Y 4) 2 3X Y 57. The reaction 22 2C O CO is carried out by taking 24g carbon and 96g 2O . Which one is

limiting reagent 1) C 2) 2O 3) 2CO 4) 2NO 58. What volume of hydrogen gas at 273 K and 1 atm. pressure will be consumed in obtaining

21.6 g of elemental boron (at mass =10.8). From the reduction of boron trichloride with hydrogen 1) 44.5 L 2) 22.4 L 3) 89.6 L 4) 67.2 L

59. Which reaction involves neither oxidation nor reduction? 1) 2 2

4 2 7CrO Cr O 2) 3Cr CrCl 3) Na Na 4) 2 22 3 2 62S O S O

60. How many gram of 4KMnO are contained in 4 litre of 0.05N solution? The 4KMnO is to be used

as an oxidant in acid medium 1) 1.58 g 2) 15.8 g 3) 6.32 g 4) 31.6 g

61. 0.455g sample of Plaster of Paris 4 21.2

CaSO H O

. When treated with excess of

3 2Ba NO gave 0.617 g anhydrous 4BaSO , Calculate the % 4CaSO in the sample 1) 79.15 % 2) 76.49% 3) 70.65% 4) 97.15%



62. Calculate the amount of MgS formed when 2g of Mg reacts with 2g of S 1) 4.5 g 2) 3.5 g 3) 2.5 g 4) 0.5 g 63. Addition of Zinc powder to 4CuSO solution precipitates copper due to 1) reduction of 2Cu 2) reduction of 2

4SO 3) reduction of Zn 4) hydrolysis of 4CuSO 64. Same amount of metal carbines with 0.1 g of oxygen and 1g of a halogen. Hence the equivalent

mass of halogen is 1) 9 2) 35.5 3) 80 4) 127 65. An oxide of sulphur contains 50% of sulphur. What will be its empirical formula? 1) SO 2) 2SO 3) 3SO 4) 2 3S O 66. A solution contains one mole of alcohol and four moles of water. What are the fractions of

water and alcohol? 1) 1/4, 4/1 2) 4/1, 1/4 3) 4/5, 1/5 4) 1/5, 4/5 67. 2Li O is one of the most efficient absorbents for 2CO in space crafts in terms of absorbing

capacity per unit mass. If the reaction is 2 2 2 3Li O CO Li CO .What if the absorption efficiency of pure 2Li O in litre 2CO (STP) per kg 1) 746.66 L/kg 2) 7466.6 L/kg 3) 74.66 L/kg 4) None of these

68. What is the charge involved when 0.1 mol of 6 5 2C H NO is reduced to 6 5C H NHOH 1) 0.3 F 2) 0.4 F 3) 0.6 F 4) 0.8 F 69. A gaseous hydro carbon gives upon combustion 0.72g of water and 3.08 g of 2CO . The

empirical formula of the hydrocarbon is 1) 3 4C H 2) 6 5C H 3) 7 8C H 4) 2 4C H 70. The volume strength of 1.15 N 2 4H SO solution is

1) 4.8 2) 8.4 3) 3.0 4) 8.0 SECTION- II

(Numerical Value Answer Type) 71. The density of a gaseous element is 5 times that of oxygen under similar condition. If the

molecule of the element is Triatomic. What will be its atomic mass? 72. What volume of 75% alcohol by mass 30.8 /d g cm must be used to prepare 150 cc of 30%

alcohol by mass 30.9 /d g cm ?

73. Given that the abundances of isotopes 54 56Fe Fe and

57Fe are 50%, 90% and 5% respectively the

atomic mass of Fe is 74. A saturated solution is prepared at 070 C containing 32.0 g 4CuSO . 25H O per 100 g solution. A

335 g sample of this solution is then cooled to 00 C so that 4 2.5CuSO H O crystallizes out. If the concentration of saturated solution at 00 C is 12.5 g 4 2.5CuSO H O per 100 g solution. How much 4 2.5CuSO H O is crystallized.

75. How many grams 4KMnO are contained in 4 litres of 0.05 N solution the 4KMnO is to be used as an oxidant in acid medium



MELUHA INTERNATIONAL SCHOOL

INDIA SR MPC Date: 27-03-2020 Time: 3 Hours

KEY SHEET MATHS

1) 4 2) 3 3) 1 4) 2 5) 1 6) 3 7) 2 8) 3 9) 4 10) 2

11) 1 12) 2 13) 4 14) 2 15) 4 16) 2 17) 1 18) 3 19) 1 20) 2

21) 4 22) 2 23) 0 24) 35 25) 20

PHYSICS

26) 2 27) 3 28) 1 29) 3 30) 4 31) 4 32) 1 33) 2 34) 1 35) 1 36) 3 37) 2 38) 2 39) 1 40) 2 41) 3 42) 2 43) 1 44) 2 45) 2 46) 2 47) 50 48) 30 49) 800 50) 56

CHEMISTRY

51) 1 52) 1 53) 3 54) 4 55) 4 56) 4 57) 1 58) 4 59) 1 60) 3 61) 1 62) 2 63) 1 64) 3 65) 2 66) 3 67) 1 68) 2 69) 3 70) 2 71) 53.33 72) 67.5 73) 55.95 74) 74.47 75) 6.32

JEE MAINS MODEL



HINTS & SOLUTIONS

MATHS - A

1. Take 1 3 2 1

1 1 0 5 1 1 00 3 2

xx

1

1 5 6 4 1 0x xx

21 5 6 4 0x x x 2 9 7 0x x 9 53

2x

2. Given 3 3 02

3 3 2x y

4 1 52

1 4 4x y

Multiply 2 with 6 6 04 2

6 6 4y

10 5 55

5 10 0X

2 1 11 2 0

X

3. Since, a p

A b qc r

, and T a b cA

p q r

T

a pa b c

AA b qp q r

c r

2 2

2 2

2 2

a p ab pq ac prab qp b q bc qrac pr bc qr c r

4. Since, cos sinsin cos

E

cos sinsin cos

E

and cos sinsin cos

E

cos sin cos sin.

sin cos sin cosE E

cos cos sin sin cos sin sin cossin cos cos sin sin sin cos cos

cos sinsin cos

.E E E

5. Since, cos sinsin cos

A

Adj (A) = cos sinsin cos

cos sin cos sin.

sin cos sin cosA adjA

2 2

2 2

cos sin sin cos sin cossin cos sin cos sin cos

1 00 1

. 1A adjA

But . 1A adjA 1 6. 1AA

1 00 1

2

2

1 00 1

On comparing the corresponding elements, we have

2 21 1 0 21 1

7. 0 1 0 1 0 01 1 1 0 2 1

A B

0 1 0 1 0 21 1 1 0 0 1

A B

2 0 1 0 1 0 1 0 1 1 1

.1 1 1 1 0 1 1 1 1 2

A A A

and 2 0 1 0 1 0 1 0 0 1 0

.1 0 1 0 0 0 1 0 0 1

B B B


Page | 11 JEE MAINS MODEL 27 MAR 2020

2 2 1 1 1 0 2 11 2 0 1 1 3

A B

and

0 0 0 2 0 0 0 0 0 02 1 0 1 0 0 4 1 0 5

A B A B

hence, 2 2A B A B A B

8.

22

22

1 tan12cos11 tan

2

tt

22

2 tan22sin

11 tan2

tt

22 2

2 2

2 2 2

2 2

2 11 21 1

1 2 2 11 1

t t tt tt t

t t t t tt t

22

2 2

2 2

2 2

1111 1

11 11 1

t tttt t

tt t tt t

0 1 0 1

0 0 1 1t t

t t

On putting the value of t in both equations, we get

1 tan 1 tan2 2

tan 1 tan 12 2

cos sinsin cos

I A I A

9. 2 3

2 34 0

x y

2 2

3 21 5

x y

On multiplying Eq. (i) by 2, Eq. (ii) by 3 and then subtracting, we get

4 6 6 6

4 6 9 68 0 3 15

x y x y

2 125 511 35

4 24 6 392 35 5 5 522 424 0 6 65 5

6 39 2 131 5 5 5 5

42 143 6 25 5

Y

10. 0

00

a bA a c

b c

' '1 12 2

A A A A A

1

0 01 0 02

0 0

a b a bQ a c a c

b c b c

'2

01 02

0

a bQ A A a c

b c

1 2 3Q Q Q 11. 2 22 5 2 2 2x xy y x y x y

G dividesOD in 2:1

3 3,2 2

D

Slope of 1OD m rOD AB equation of AB is

3 1 32 1 2

y x

3x y 3 0x y 12. Eliminate ' 'x in two equations

2 21 1 1 1 1 12 2 0ag a g x hg h g xy bg b g y

and 0a b

2 2 0Coeff of x Coeff of y

13. We have h , g K , 2c K , f K



Apply 0 , we get a quadratic equation in for real 2 4 0b ac

14. Let 1 2 3am m m

d

and ay xd

in the given

equation

15. 3Tan , the line is 1 1x x y y rCos Sin

0 01 32 2

x y r any point on the

Line is 3,2 2r r

, where ' 'r is the distance

from 0,0 Substituting in the curve

3 21 3 3 .... ...... 1 08

r r r

It is a cubic equation in ' 'r

1 48 3 3 1131 3 3

OA OB OC

16. 1 1 1 2 2 2S l x m y n l x m y n

11 1 1 2 2 2S l x m y n l x m y n

Above four lines form a parallelogram

1 2 1 2

1 2 2 1

c c d dArea

a b a b

17. OP

andOQ

be the bisectors of the angles between OA andOB

Equations ofOB andOA are 0lx my and 0mx ly and 0lx my mx ly

2 2 2 2 0lmx l m my lmy

Combined equation ofOP andOQ is

2 2

2 2122

x y xylm l m

2 2 2 2 4 0l m x y lmxy

18. 221 1y x y x

Replace x by x 22 1y x

2 2 2 1 0x y x 19.

3 3sin3 3 34 4 . . 0 0

4 4 4 4 4tan3 3

xx

x x x

x

xLt Lt Lt

x

20.

23

3 23 3

2 2 488 2 2 4

n n

r rn n

r r rrLt Ltr r r r

2

23 3

2 2 4.2 2 4

n n

r rn

r r rLtr r r

2

23 3

1 32 .2 1 3

n n

r rn

rrLtr r

2 23 2 41.2.3.4

1 1 2 7.12n

n n nLt

n n n n

2 2

3 2 41 12 2

1 1 27 71 1 1n

n n nLt

n n n

NUMERICAL VALUE HINTS & SOLUTIONS

21. We have b c c b

c c a ab a a b

0 2 2a ac c a ab a a b

0 1 1

2a c c a ab a a b

2a bc bc 22.

2 22 2 2

2 2 2

x y z x y x y z x yz y z x y x y z y z x yz x z x y x y z x z x y



1

2 1 00 0

x yx y z y z x

z x y

1

2 0 1 00 0 1

x yx y z x y z x y z

3 32 1 1 0 2x y z x y z

3k x y z 2k

23. We have, 1

1 1 1yz zx xyx y z

2

2

2

111

x xxyz y yxyz

z z

2

21

2

11 1 0

1

x xy yz z

24. 2 215 31 14 0x xy y 2 215 21 10 14 0x xy xy y 3 5 7 2 5 7 0x x y y x y

5 7 3 2 0x y x y

5 7 0x y , 3 2 0x y

16 6 129 4 13

P

; 2

10 21 3125 49 74

P

2 2

1 21 74 1 144 74 31 3113 135 31 5 13 31 74

P P

1 175 355

25 2 2 1 0 90a b Sin Cos

PHYSICS 26. (2) 27. Potential energy of the spring will transfer

to the kinetic energy of air molecules.

28. As 2

;2PKm

1Km

.

29. We know that 212

K mv

It represent a parabola about y axis 30. f iW K K .

or 4

2 2

3

16 2 82 fx dx v

or 42

2

3

6 642 fx v

or 2 2 23 4 3 64fv

6.6 /fv m s

31. 212

kx mgh

or 2

2kxhmg

32. In the device when free end of the string is

pulled by l , the block will rise by 2l . So

increase in potential energy / 2U mgl . 33. The tension in the string, T = F. Also gravity tensionW W K tension gravityW K W

34. 2 3

23 4

3 8 3d t t tdxv t t

dt dt

0t , 3 /m s 4t , 23 8 4 3 4 19 /fv m s

Now, 2 212 f iW m v v

3 2 21 30 10 19 3 5.282

J

35. 2 212 f iW m v v

3 2 21 100 10 0 52

1.25J

36. 21100 202

mg mv

40 /v m s

37. 2 21 122 2

kx mv

38. 2 2

Mg l MglW

39. 2 212 f iFx m v v

or

3 3 2 21200 10 60 10 300 6002

F



340.5 10F N 40. Given, 120U xyz

20x

U xFx z

20y

U yFy z

and 2

20z

U xyFz z

x y zF F i F j F k

41. See examples 42. gravity FW W K

0 010 10 10cos60 200 10sin 60g

21 10 02 fv

17 /fv m s 43. J Ft P

44. 2 22 1

12

KE m v v , J p m v u

45. Both the impulse and change in momentum are in same direction. 46. . 0F S

or 6 2 3 . 2 3 0i j k i j xk

or 6 2 2 3 3 0x 2x

47. At 320, 5 0 0ix v

and 2x m , 325 2 5 8 /fv m s

W K

2 212 f im v v

2 21 0.5 5 8 02

= 50 J 48. W area of F x

m area of a x

1 12 5 3 6 2 1 6 302 2

j

49. 8 9 /iv i j m s

and 8 9 5 8 41 /fv i g j i j m s

so 2 28 9 147iv

and 2 28 41 1745fv

1 1 145 17452

U K

800J

2


50. .J F dt

CHEMISTRY 51. One g atom = mass of AN atoms 231.008 6.022 10g H-atoms

2423

1.008 1.67 106.022 10

Mass of one H atom g

52. Mass of water escaped = 10.407-9.520=

0.887 g 2 2.BaI xH O

2 9.520BaI g 2 0.887xH O g 2

2

moles of H O xmoles of BaI

0.887 /18 29.520 / 391

x x

53. Law of multiple proportions

54. 22

23

6.02 10cos 0.16.02 10

Moleof Glu e

cos0.1 100 2

50glu eM

55.

15 1.6100 1.1 15

981000

m

56. g-atoms of 75.875

X and

g-atom of 24.216

Y

57. Limiting reagent is one which is

completely consumed in reaction 58. 3 22 3 2 6BCl H B HCl 22 10.8 3 22.4g Boron LH

3 22.4 21.621.62 10.8

g Boron

267.2L H 59. Oxidation number remains unchanged. 60. Meq. Of 4KMnO

4 1000 0.5 100031.6

6.32 g 61.



4 3 4 32 2aqCaSO Ba NO BaSO Ca NO

1mole 1mole 136 g 233 g Mass of 4CaSO required to get 0.617 g 4BaSO

136 0.617 0.360233

g 4CaSO

% Purity of 4CaSO sample

0.360 100 79.15%0.455

62 Mg +S Mg S 1 mole 1 mole Mass ratio = 24 g 32 g

When 2g S 2/32mole reacts with Mg. It acts as limiting reagent while Mg is excess reagent. Thus amount produced will be decided by moles of S not by moles of Mg, thus

1 1mole S mole MgS

232

Moles of Mg S formed

And mass of

Mg S formed 2 56 3.532

g

63. Reduction of 2Cu 64. Eq. metal = eq. of 2O = eq. of halogen

Eq. of 20.1 .1 0.1

32 84

OE

Eq. of halogen 1 0.18E

80E 65. Conceptual

66. 1 11 4 5

2 0Hx =4/5

67. 2Li O reacts with 2CO as 2 2 2 3Li O CO Li CO (Mass of 2 30Li O g ) 1 mole of 2Li O reacts with 22.4 L 2CO at NTP

21000 g Li O absorbs 22.4 100030

=746.66 litre 2CO Absorption efficiency is 746.66 L/kg 68. 6 5 2 6 54C H NO e C H NHOH 4 6 5 4 0x 24 7 2 0x 25x 29x 25 29 4n

1 mole = 4F (or) 0.1 mole = 0.4F

69. 2 2 24

4 2X YYC H x O XCO H O

3.08 g 0.72 g 0.07 moles 0.04 moles

0.07 7/ 2 0.04 8

X XY Y

70. 2 2 2 22 2H O H O O 68 g 22.4 l ( at STP) Mass of 2 2H O in 1.5 N solution = E. wt. of 2 2 1.5H O N 17 1.5 25.5 g = Volume strength of1.5N 2 2H O solution

22.4 25.5 8.468.0l g l

NUMERICAL HINTS 71. Molecular mass of oxygen = 32

V.D of oxygen = 32 162

V.D of gaseous element 16 5 80 Molecular mass of gaseous element

=80 2 160 As molecule id Triatomic, Its atomic mass

molar massAtomicity

160 53.333

Ans: 53.33

72. 75 300.8 150 0.9100 100

V

67.5V m L 73. 5 90 554 56 57

100 100 100Fe

55.95amu 74. Mass of 4 2.5CuSO H O in 335 g solution

32 335 107.20100

g

Mass of water = 335-107.20 =227.80 g



At 00 C mass of 4 2.5CuSO H O in 100 g of solution = 12.5 g

Mass of water= 100-12.5 g =87.5 g

87.5 2 12.5H o 227.80 ?

12.5 227.8 32.7387.5

g

Amount of 4 2.5CuSO H O crystallised =107.20-32.73 = 74.47

75. Wt. of 4 4 1000 0.5 100031.6

wKMnO

6.32w g

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