CLASS XI

SAMPLE PAPERS

PHYSICS

-(1)-

1. In

bV

V

aP

2 constant, P and V refer to pressure and volume, while a and b

are constants. Find the dimensions of a and b.

2. A body moves with constant speed in a circle. Does it possess a constant or avariable acceleration?

3. Find the angle between the vectors ( k6j3i2 ) and ( k12j6i4 ).

4. Distinguish between Distance and Displacement.

5. At which point in the path of a projectile, is the velocity and accelerationperpendicular to each other.

6. Define one Steradian.

7. If the percentage error in the measurement of radius and length of a cylinderare 2% each, then find the percentage error in its surface area and Volume.

8. Why do action and reaction forces cannot cancel each other?

9. If acceleration a = k t, find the displacement in “t” second, considering the initialvelocity to be zero at the origin.

10. Find the component of ji ˆˆ along j-i2 .

11. If x = 6t and y = (8t + 6t2) represent the position of a projectile on the move, findthe angle and velocity of projection. Also find the time at which the velocity willbe along the X- axis only.

12. A car covers first half of the distance in a straight line with a speed of 40km/hand the second half it covers with speed 60km/h and 80km/h for equal length.Find the average velocity in the journey.

General Instructions:1 . All questions are compulsory.2 . Questions numbered 1 to 8 are very short answer questions and carry 1 mark each.3 . Questions numbered 9 to 18 are short answer questions and carry 2 marks each.4 . Questions numbered 19 to 27 are also short answer questions and carry 3 marks each.5 . Questions numbered 28 to 30 are long answer questions and carry 5 marks each.6 . Use of calculators is not allowed.

CLASS

XI

Half Yearly Examination 2013-14Date : 25-08-2013

Duration : 3 hrs.Max. Marks : 70

Physics(Theory)

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13. What are the limitations of dimensional analysis.

14. Ifdc

baP

1/3

1/22

is to be calculated when the % error in a, b, c and d are 1%, 2%, 3%

and 4% respectively, find the % error in P?

15. If the velocity of a car moving in a straight line is given by V= 4t2 +5t +7, find(i) The displacement in 2 sec and

(ii) Acceleration at t = 2

16. Why do we bank a curved track?

17. What is the significance of the following:

(i) Area under Force – displacement graph.

(ii) Area under the Force — time graph.

18. Give the location of the “centre of mass” of mass of a (i) sphere, (ii) cylinder, (iii)ring, and (iv) cube, each of uniform mass density. Does the centre of mass of abody necessarily lie inside the body?

19. If l = )1.00.2( cm and b = 0.2)12.3( cm are the length and breadth of a rectangle,

find the error involved in the measure of its (i) area (ii) Perimeter.

20. A car accelerating with 5 ms-2 along a straight high-way from rest begins todeaccelerate after 20s and comes to rest in another 100m length. Calculate

(i) the Maximum velocity

(ii) Displacement in the straight line.

(iii) Draw velocity-time graph for the same.

21. From the top of a tower 100m height, a ball of mass 2kg is dropped. Prove that theenergy possessed at any point during the fall is 2000joule.

22. (i) If x = 3tn, prove that2

2d xdt

is independent of “t” for n <= 2

(ii) Evaluate from x = 0 to x =/2, the integral of 2x3 + cos 2x

23. Four particles of mass m1 = 1kg, m2 = 2kg, m3 = 3kg, m4 = 4kg, are located at thecorners of a rectangle as shown in figure. Find the positon of centre of mass.

b

m1

m4 m3

m2aX

Y

24. Volume of water flowing through a capillary tube per second is dependent on thepressure gradient (p/l), radius (r) and the coefficient of viscosity of the liquid.Derive the relation.

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25. A ball is thrown up vertically with a speed of 25m/s. Simultaneously another ballis dropped from a height of 130m with the same speed and in the same Verticalline. When and where do they meet?

26. What is meant by elastic collision? Show that in case of two dimensional elasticcollision of two bodies of equal masses with one moving with a velocity v and theother at rest, after the collision they will move at right angles.

27. Three masses 5Kg, 4Kg and 2Kg are at the vertices of an equilateral triangle ofside “a” with the 5Kg at the origin. Locate the centre of mass.

28. What is the acceleration of the block and the trolley systemshown in fig, if the coefficient of kinetic friction between thetrolley and the surface is 0.04? What is the tension in thestring? Take g = 10 ms–2. Neglect the mass of the string.

20 kg

3kg

30 N

trolley

OR

Distinguish between uniform and non-uniform circular motion. Derive a relationof centripetal acceleration. For a body slowing down at the rate of 5 ms-2 when itsvelocity is 20ms–1 in a circle of radius 2m, find the acceleration.

29. (a) Proving that the path of a projectile is a parabola, find the time taken toreach the maximum height.

(b) Show that for two complementary angles of projection of a projectile thrownwith the same velocity, the horizontal ranges are equal.

OR

(a) A mass is projected horizontally from a tower of height h with a velocity u.As it traverses, prove that the path is a parabola.

(b) If t is the time taken to reach the ground, find the velocity at a time t/2.

(c) Find the velocity when it lands on the ground.

30. Derive the three equations of motion graphically or otherwise. Give the conditionunder which the equations of motion are applicable.

OR

(a) What is relative velocity? Write an expression for the same.

(b) Rain is falling vertically with a speed of 30 ms–1. A woman rides a bicyclewith a speed of 10 ms–1 in the north to south direction. What is the relativevelocity of rain with respect to the woman? What is the direction in whichshe should hold her umbrella to protect herself from the rain?

× · × · × · × · ×

-(4)-

1.

2a

V = [P]

[a] = [PV2]

= [ML–1T–2][L3]2 = [ML5T–2]

[b] = [V] =[L3]

2. It possesses variable acceleration.

3..cos

| || |A B

A B

= – 1

180

4. Distance is a scalar quantity whereas displacement is a vector quantity.

Displacement can be zero, positive or negative, whereas distance can never benegative.

5. At the topmost point of the projectile, velocity and acceleration are perpendicularto each other.

6. Solid angle subtended at the centre of a sphere by a surface of the sphere equalin area to that of a square, having each side equal to the radius of the sphere.

2surface area

radius

7. Volume of cylinder = r2h

2100 100 6%V r hV r h

Surface area = 2rh

100 100 4%S r hS r h

8. Action and reaction forces donot cancel each other because they do not act onthe same body.

CLASS

XI

Hints/Solutions to Half Yearly Examination 2013-14Date : 25-08-2013

Duration : 3 hrs.Max. Marks : 70

Physics(Theory)

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9. a = kt

v = adtv = ktdt

v =kt c

2

2

At t = 0, v = 0

So 0 =20

2k c

c = 0

v =kt 2

2

s = kt ktvdt dt

2 3

2 6

s =kt 3

6

10. Let ^^a = i – j

^^b = i – j2

a . b a b aprojection of on

a . b b a bprojection of on

.(2 ) 2 – projection of on^ ^ ^^ ^ ^i – j i -j i j a b

(2 + 1) = a b5 projection of on

a b 3Projection of on5

Component of ^ ^( )i j

along^ ^3 (2 )^ ^(2 )

5 5i ji j

3 ^ ^(2 )5

i j

11. x = 6t

Velocity along x-axis vx =dxdt = 6

y = 8t + 6t2

Velocity along y-axis vy =dydt = 8 + 12t

At t = 0,vx = 6 and vy = 8 + 12t = 8

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Velocity projection = x yv v2 2

= 2 26 8

= 36 64 = 100 = 10 m/s

tan =y

x

vv

86 =

43

= tan–143

12. Average velocitytotal displacement

total time taken

2 4 4 50.526 /

2 40 4 60 4 80

avg

l l l

V km hl l l

13. (i) The method doesn’t give any information about the dimensionless constant.

(ii) It fails to derive relationship which involve trignometric, logarithmic orexponential function.

(iii) It fails when a physical quantity is the sum or difference of two or morequantities.

14. % error in P = 2 × (1%) +12 × (2%) +

13 × (3%) + 1 × (4%)

= 2 + 1 + 1 + 4 = 8%

15. (i) s = vdt t t dt2(4 5 7) =t t3 24 53 2 +7t

=323 + 10 + 14 = 34.67 units

(ii) a =dv d t tdt dt

2(4 5 7)

= 8t + 5 = 8 × 2 + 5 = 21 units

16. The component of the normal reaction can provide the necessary centripetalforce and thus reduce the dependence on coefficient of friction.

17. (i) work done (ii) Impulse

18. (i) Geometrical center

(ii) Center of its axis

(iii) Center of ring

(iv) Point of intersection of the diagonals

No, it is not necessary that center of mass of the body lies inside the body liesinside the body.

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19. (i) A r e a = l × b = 2 × 1 2 . 3 = 2 4 . 6 c m

2

AA

=l bl b

A24.6

=0.1 0.22.0 12.3

A = 24.6 (0.05 + 0.01626)

A = 1.63 cm

(ii) P = l + b + l + b

P = l + b + l + b= 0.1 + 0.2 + 0.1 + 0.2 = 0.6 cm.

20. (i) Maximum velocity (vmax) = u + at

= 0 + 5 × 20 = 100 m/s

(ii) v2 = u2 + 2as

0 = (100)2 + 2a × (100)

2a =2100–

100

Time

Vel

ocit

y

100 m/s

20s 2st

2a = –100

a = –50 m/s2

Diplacement in acceleration,

s = ut +12 at 2= 0 × t +

12 × 5 × 202

= 1000 m

Total displacement = 1000 m + 100 m

= 1100 m

v = u + at

0 = 100 – 50 t

t =10050 = 2s

Hence, v – t graph is as shown.

21. Let the ball drop through a distance x from the top.

At point P,

v2 = u2 + 2gx [u = 0]

v2 = 2gx

K.E. at P =12 mv2 =

12 m22gx = mgx

AO

OP

x

O C

L = 100

P.E. of P =mg (100 – x)

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Total Energy at P = mgx + mg (100 – x)

= 100 mg= 100 × 2 × 10 = 2000 J

This proves that energy at any point P is 2000 J.

22. (i) 3 nx t

13 ndx n tdt

2

22 ( 1)3 nd x n n t

dt

2

2 0 for 2d x ndt

(ii)4

32

23. 0.5 0.7cmr a i b j

24.C

a b pV k rl

3 1 1 1 2 2a CbL T ML T L ML T

3 1

1 1

volumetime

V L T

ML T

0 3 1 2 2a c a b c a cM L T M L T

a = – 1, b = 4, c = 14kr pVl

25. Thrown Up: X = 25 t – ½ gt2

Dropped down: 130 – x = 25t + ½ gt2

Solve to get, 130 = 50t13050

t Sec = 2.6 s and x = 31.2 m from ground

26. Write the law and as discussed in class the derivation of velocity of rocket maybe shown.

27. 4 2 3,11 11cm cmax y a

Or

4kg

5kg 2kg

Y

X

2 kg

5kg 4kg

Y

X

5 3,11 11cm cmax y a

28. a = 0.96 m/s2, T = 27.12 N

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29. (i) The horizontal distance covered by the body in time t is

xx ut tu

The vertical distance travelled by the body in time t is given by

212

s ut at

or 2 21 102 2

y t gt gt [For vertical motion, u = 0]

or2

22

12 2

x gy g xu u

xtu

or 2y kx therefore, trajectory is parabola

(ii) 2 2x yv u u

22 2

2tu g

(because at time = t/2, vy = 0 + gt/2)

(iii) 2 2 2v u g t .

OR

(a) Do as discussed in class

(b) Do as discussed in class

(c) Centripetal acceleration ac =vr

2

Tangential acceleration at = 5 m/s2

ac =220

2 = 200 m/s2

anet = c ta a2 2

= 2200 25 = 200.062 m/s2

30. Derive as discussed in class

OR

Define relative velocity.

Here

velocity of woman cyclistWOA v

S N

B D

AC

Wv

Vertical

Wv10m/s 10m/sO

Rv

30

m/

s RW R Wv v v

= 10 ms–1, due south

velocity of rainROB v

= 30 ms–1, vertically downward

WOC v

= Opposite velocity of the woman cyclist.

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× · × · × · × · ×

R W RWOD v v v

2 2 2 210 30RWv OD OC OB

110 10 31.6 ms

If OD makes angle with the vertical, then

10tan 0.3333 or 18 26'30

BD OCOB OB

The woman should hold her umbrella at 18°26 ' with the vertical in the direciton

of her motion i.e., towards south.