SESSION ENDING EXAMINATION PHYSICS - CLASS XI

BLUE PRINT

Sl No

Units VSA 1 mark

SA I 2 marks

SA II 3 marks

VBQ 4 marks

Long Answer 5 marks

Total marks

Weightage marks

1 Physical world & Measurement

1(1) 2(1) 3(1) 6 23

2 Kinematics 1(1) 3(2) 5(1) 12

3 Laws of Motion

2(1) 3(1) 5

4 Work, energy, power

1(1) 4(1) 5 17

5 Motion of system of particles & rotational motion

3(2) 6

6 Gravitation 3(2) 6

7 Properties of matter

1(1) 2(1) 3(1) 5(1) 11 20

8 Thermodynamics

2(1) 3(1) 5

9 Kinetic theory of gases

1(1) 3(1) 4

10 Oscillations and waves

2(1) 3(1) 5(1) 10 10

Total 1x5=5 2x5=10 3x12=36 4x1=4 5x3=15 70 70

KENDRIYA VIDYALAYA SANGATHAN

PHYSICS Class XI Time : 3 hours Maximum marks : 70General Instructions

i) All questions are compulsoryii) There are 26 questions in total .Questions 1 to 5 carry one mark each, questions 6 to

10 carry two marks each, questions 11 to 22 carry three marks each, and questions 24 to 26 carry five marks each

iii) Question number 23 is a value based question which carries four marks iv) There is no overall choice. However, an internal choice has been provided in one

question of two marks, one question of three marks and all three questions of five marks. You have to attempt only one of the choices in such questions

v) Use of calculators is not permitted.vi) You may use the following values of physical constants wherever necessary

c = 3 x 108 m/s h = 6.626 x10 -34 JsBoltzman constant k = 1.381 x 10-23 J/K G = 6.67 x 10 -11 Nm2kg -2

Avogadros Number NA = 6.022 x 10 23 mole-1

Radius of earth = 6400 km Mass of earth= 6 x 10 24 kg.

1. How many parsec are there in one metre? 1

2. The position coordinate of a moving particle is given by y = 3 + 9t + 3t 2 (y is in metres and t in seconds), what is its velocity at t = 2 seconds.

13. What is the magnitude and direction of acceleration in uniform circular motion?

14. Define angle of contact

15. A narrow uniform glass tube contains air enclosed by 15cm long thread of mercury. When the

tube is vertical with the open end uppermost, the air column is 30cm long, when the tube is inverted the length of air column becomes 45 cm. Calculate the atmospheric pressure.

16. what is the maximum permissible error in the density of the material, if the errors involved in

the measurements of a side and mass of a cube are 3% and 2% respectively,?2

7. A ball is rolled on ice with a velocity of 5.6m/s and comes to rest after travelling 8m. Find the coefficient of friction. Given g=9.8m/s .

2

8. With the help of a diagram explain the working of Hydraulic lift. 2

9. State Newton’s law of cooling. Draw the graph showing the cooling of hot water with time. 2

10. A spring of force constant k is cut into two pieces, such that one piece is double the length of the other. What is the force constant of the longer piece of the spring

ORShow that the total energy of a particle executing simple harmonic motion is directly proportional to the square of the amplitude and frequency

11. The frequency υ of vibration of a stretched string depends upon its length lits mass per unit length ‘m’ and the tension ‘T’ in the string. Obtain dimensionally an expression for frequency υ. 3

12. On a foggy day two drivers spot each when they are just 80 meters apart. They are travelling at 72km/h and 60km/h respectively. Both of them applied brakes retarding their cars at the rate of 5m/s2 Determine whether they avert collision or not. 3

13 An object is in uniform motion along a straight line.Draw the position –time graph with (a) xo is positive and v is negative(b) xo is negative and v is positive(c) xo is negative and v is negative

3 14.(a)On what factors does the position of a center of mass of a rigid body depends

(b)Show that the center of mass of a system moves with a constant velocity when no external force acts on the system. 15. .It is easier to pull a lawn roller than to push it explain 3

16. (a) State the theorems of parallel axis and perpendicular axis

(b) Find the moment of inertia of a sphere about a tangent to the sphere, given moment of

inertia of the sphere about any of its diameters is MR2 where M is the mass and R, the radius

of the sphere. OR

A wheel has a moment of inertia 200kg/m2. It is to be accelerated uniformly from rest to an angular velocity of 300 rad/sec in 25 sec. Find the angular acceleration, torque and kinetic energy at the final angular velocity.

317. (a) Obtain expression for the acceleration due to gravity at an altitude above the surface of the

earth and below the surface of the earth. (b) If a man goes from the surface of the earth to a height equal to the radius of the earth, then what will be his weight relative to that on the earth .

318. (a) What are geostationary satellite .

(b) State the necessary conditions for a satellite to be stationary. 3

19. (a) Define the terms (i) Surface tension (ii) Surface energy. (b) Calculate the amount of energy released when 8 droplets of water of radius 0.5mm each combine to form one. Surface Tension of water is 0.072 N/m.

320. Explain the various operations for Carnot’s heat engine working between two temperatures.

Hence derive from it the efficiency of the engine .3

21. State the law of equipartition of energy. Find the ratio of Cp and Cv for a mono-atomic gas molecule using degrees for freedom.

322. A simple harmonic motion is represented by y t =10 Sin (20t +0.5). Write down its amplitude,

angular frequency, frequency, time period and initial phase, if displacement is measured in meters and time in seconds.

323. Hari went to railway station to receive his uncle .At platform, he saw that an old coolie was

carrying heavy load on his head. Suddenly the coolie tripped and the baggage was about to fell off his head . Hari went to the old man helped him and offered to carry some load for him

(a) What are the qualities of Hari?(b)A man weighing 55kg support a body of 20kg on his head. Calculate work done by him if he moves a distance of 20m.

(1) On horizontal road (2) Upon a smooth inclined plane of 1/5(g=10m/s2).

424. State parallelogram law of vector addition Derive equations for the magnitude and direction of the resultant.

ORDefine uniform circular motion. Derive an expression for the centripetal acceleration. Also

show that centripetal acceleration is directed towards the centre of the circular path. 5

25. (a) State and prove Bernoulli’s theorem. (b) A horizontal pipeline carries water in a streamline flow. At a point along pipe where the cross sectional area is 10cm2, the water velocity is 1m/s and the pressure is 2000Pa. What is the pressure at another point where the where the cross sectional area is 5cm2

OR(a) Define terminal velocity. Show that terminal velocity of a sphere of radius ‘r’ density ‘’

falling vertically through a viscous fluid of density ‘’ and coefficient of viscosity ‘’ is directly proportional to (-)

(b) A rain drop of radius 0.3mm falls through air with a terminal velocity of 1m/s. The viscosity of air is 18 x10-6 Pa.s Find the viscous force on the rain drop.

5

26. (a) What is Doppler Effect? (b) Derive an expression for the apparent frequency of sound, when

both the source and the observer are in motion. (c) Two engines pass each other in opposite directions with a velocity

of 60Kmh-1 each. One of them is emitting a note of freq 540 Hz. Calculate the frequencies heard in the other engine before and after they have passed each other. (Given velocity of sound = 316.67 m/s)

OR(a) What are standing waves (b) Obtain expressions for the normal modes of vibration in a stretched string. (c) Distinguish between harmonics & overtones.

--------------------------------- 5

KENDRIYA VIDYALAYA SANGATHAN

SESSION ENDING EXAMINATION

CLASS X1 ANSWER KEY SUBJECT - PHYSICS (THEORY )

1 3.25x10-17parsec½ + ½

29 +6t, at t= 2sec , v= 21 m/s

1

3 V2/r and towards the centre 14 Definition 15 , When the tube is inverted P2 = P – 15.

Therefore, (P + 15)× A × 30 = (P -15) × A × 45.Hence, P = 75 cm of Hg

1

6=

1+6=7

1

7 μ = a/g , , 0- (5.6 ) 2 = 2 x a x 8 Hence a= -1.96 m/s2

Hence μ = a/g = = 0.2

½ +1+ ½

8 Diagram, Working 1+1

9 Statement , Graph 1+1

10Force constant k=

The length of the longer part is .

So it’s force constant is k’= k

OR

E= m =2 m

E

E

1+1

1+1

11 Proportionality relation, dimensional equation, ½ +1 +

Finding dimensions =

½ +½+ ½

12 for the first car

U=72km/h=20m/s ,v=0, a= -5m/

s1=40m

s2 = =27.78m

Total distance covered = s1 + s2 =67.78mAs this distance is less than the initial distance(=80m) between the two cars, so the collision will be averted.

1+1+1

13 3 diagrams , 1 mark each 1+1+114 (a)C.M depends on

() geometrical shape of the body() distribution of mass in the body(b) = M

If = 0 so MA=0 or A=0 or V=a constant

1½+1½

15 Diagram , Explanation 1½+1½16

17

(a) Statement of parallel axis & Perpendicular axis, (b) 2/5 MR2 + MR2 = 7/5 MR2

OR α = 300/25 = 12 rad /s2

τ = I α = 200 x 12 = 2400 Nm , K E =½ I ω2 = ½ x 200 x 3002 =9x 10 6 J

a) Derivation of g’ =g ( 1 )

b) /g= = =

c) = mg

1+11

1+1+1

1½

1½

18(a) Definition(b) Necessary conditions(4 points)

1=2

19 a) i) Definition of Surface tension ii) Definition of Surface energy b) Work done = Surface tension x Change in surface area

Volume remains the same , R= 2rDecrease in surface area = 4π x 10 -6 m2

½+ ½½½½

Work done = 0.072x 4x3.14x 10-6 = 9.0432 x 10 -7 J ½

20 Isothermal expansionAdiabatic expansion Isothermal compressionAdiabatic compression

Efficiency =1-

=2

1

21 Statement of law For a monoatomic gas there are three degrees of freedom E = 3 x ½ KB T x N = 3/2 R T

CV = 3/2 R , CP = CV + R = 5/2 R Cp/Cv = 5/3

1

½+ ½

½+ ½

22 Comparing with the standard equation y = a sin ( ωt +φ)a = 10 m Angular frequency ω = 20 rad /sFrequency γ = ω/2π = 20/2π = 3.18 Hz Time period T = 2π/ω = 0.314 SInitial Phase φ = 0.5 rad

½½½½½½

231) Helpful,kind hearted 2) M=55+20=75kg,s=20m

a)on horizontal road =90o

w=fs cos =fscos90o =0

b)upon inclined plane, sin =

w=fs cos =mg sin

=75

=3000j

1+1

1

1

24 Statement, diagram, derivation of R , direction OR Definition, diagram,equation for ac, direction

1+1+2+11+1+2+1

25

a) Statement , Proof

= or 10cm2 1m/s =5cm2 = 2m/s

1+2

Using Bernoulli’s theorem ,

+ = +

+ = 2000+

=500Pa

½½½ +½

26 a) Definition

b) derivation γ’ = γ ( )

c) VL = VS = 60 kmh-1 = 16.67m/sγ = 540 , γ’ = ?

d) before crossing γ’= γ ( ) = 600 Hz

after crossing γ’ = 486 Hz ORa) Definitionb) Derivation of expression with all diagramsc) explanation

12

11

122