Class- XI (Physics)

Marking Scheme (2015

Class- XI (Physics)

1. Gravitational force 1

2. Yes, In circular motion

3. S α t

S = kt

ds/dt =k

d2s/dt2 =0 ½

So acting force F=ma =0 ½

4. work done is zero 1

5. τ α d , So we prefer to use a wrench of longer arm. 1

6.V=43πr 3 ½

%error= ΔV/V =3(Δr/r)×100 ½

Correct Value= 5.7% 1

7. Taking both side self dot product ½

4AB Cosθ =0 1

Θ = 900 ½

8. y = ut +1/2(gt2)

dy/dt = v = u+ gt ½

a = d2y = g ½

F = ma ½

So F= mg ½

9. dW = Fds = m(dv/dt).ds ½

dW= mv(dv/dt) ½

W = m u∫vvdv =1/2(mv2-mu2) ½

W = Kf –Ki Change in K.E. of the body ½

10. Definition ½

Derivation 1½

Factor ½

11. Units of

A= ms-3 1

B= ms-2 1

C= ms-1 1

12. S= 4.29 light year = 4.29×9.46×1015/3.08*1016 =1.32 parsec 1

b= diameter of earth’s orbit = 2AU 1

θ = b/s = 2/1.32 = 1.515 s 1

13. ds =(u+at)dt 1

S = n-1∫n(u+at)dt 1

Snth= u+ ½ a(2n-1) 1

14. Case-1

Average velocity = 5kmh-1 ½

Average speed = 5kmh-1 ½

Case-2

Average velocity = 0 kmh-1 ½

Average speed = 6kmh-1 ½

Case-3

Average velocity = 1.875kmh-1 ½

Average speed = 5.625kmh-1 ½

15. Friction is necessary ( disadvantage)- any three 1 ½

Friction is an evil (disadvantage)- any three 1 ½

16. Proof of first law from second law 1 ½

Proof of first law from second law 1 ½

17. mu1+0 = mv1 + mv2

u1 = v1 + v2 ½

½ mu12 = ½ mv12 + ½ mv22

u12 = v12 + v22 ½

taking self dot product of velocity and from above equation

2V1V2Cos θ= 0 1

Angle between V1 & V2 is 900 ½

Θ1 = 900 – 370 = 530 ½

18. radus of well = 3/2 m

V=πr2h = 99m3 ½

Average height = (6+20)/2 = 13m ½

P=hp = 5×746

t = W/P =mgh/P ½

t=3381.6 s 1

19. Parallel axis theorem

Definition 1

Formula I= ICM + Md2 ½

Perpendicular axis theorem

Definition 1

Formula I= Ix + Iy ½

20. Definition 1

g’=g(1-h/r)2 1

using binomial theorem g’=g(1-2h/r) 1

21. gh =g(1-2h/R)

gd =g(1-d/R) ½

gh/gd =(R-2h)/(R-d) ½

Wh/Wd =(R-2h)/(R-d) ½

Wh/Wd =61/62 = 0.984 1

22. Definition 1

Derivation 1 ½

Correct unit ½

23. (a) Scientific thinking, determination and curiosity 2

(b) Input power = Vρgh

Output power = 40W ½

Efficiency ƞ= Output power/Input power ½

90/100 = 40/ Vρgh ½

V = 0.453×10-3 m3/s

V = 0.453 litre/s ½

24 (a)Analytical treatment- Use on of them triangle law or parallelogram law of vector addition

β θ

O B D Correct fig. 1

Vector OB =Vector P , Vector BC =Vector B and suppose Vector OC =Vector R

BD = QCosθ & CD = QSinθ ½

Using Pythagoras theorem in right angle triangle ΔOCD

OD2 = OD2 + OD2 =( OB2 + BD2 ) OD2 ½

R2 = P2 + Q2Cos2θ + Q2Sin2θ+2PQCosθ

R2 = P2 + Q2 +2PQCosθ ½

Tanβ =CD/OD = QSinθ /(P+QCosθ) ½

(b) Given R2 = P2 + Q2

R2 = P2 + Q2 +2PQCosθ

From above both equation

2PQCosθ =0 1

Cosθ= 0

θ= 900 1

(a) Definition of projectile motion 1

(b) Rmax =2/√3R

U2/g =2/√3 U2Sin2θ /g` 1

Sin2θ=√3/2

2θ=600

θ=300 1

h2 =u2 Sin2 (90-θ)/2g 1

h2 =u2 Cos2θ)/2g

h1 :h2 = Tan2θ:1 1

25. Angle of friction- The angle which the resultant of the limiting friction and the normal reaction makes with the normal reaction 1

tan θ =µs where θ is the angle of friction ½

Angle of repose-The minimum angle that an inclined plane makes with horizontal when a body placed on it just to slide down. 1

mgSinɸ mg mgCosɸ

ɸ ½

tan ɸ =µs where ɸ is the angle of repose ½

Relation between angle of friction & angle of repose

tan θ =µs (i)

tan ɸ =µs (ii) ½

θ= ɸ

or Angle of friction = Angle of repose ½

Banking of curve road- The system of raising the outer edge of a curved road above its inner edge is called banking of curved road 1

Needs of banking of curved road- To avoid the large amount of friction between the tyres and the road ½

Correct fig. 1

Equating the force along horizontal and vertical direction respectively

RSinθ + fCosθ= mv2/r (f =µR ) (i) ½

RCosθ - fSinθ= mg (ii) ½

V2 = rg[(µ+ tan θ)/(1- µtan θ}] ½

Optimum velocity- µ=0 ½

V2 = rgtanθ ½

26.(a)Orbital velocity- Orbital velocity is the velocity required to put the satellite in to its orbit around the earth. 1

Expression for orbital velocity-

M= mass of the earth , R= Radius of earth, m= Mass of the satellite ,

vo = Orbital velocity of satellite , h = Height of the satellite above the earth’s surface

& R+h = orbital radius of satellite

Fg =GMm/(R+h)2 (i)

Fc = mvo2/(R+h) (ii) ½

In equilibrium

Fg = Fc

mvo2/(R+h) = GMm/(R+h)2 ½

vo2 = GM/(R+h)

vo =[ GM/(R+h)]1/2

Put GM=gR2

Vo = R[g/(R+h)]1/2 ½

When satellite revolves close to the surface of the earth then vo=(gr)1/2 ½

(b) R=6400km Me = 6×1024kg & h = 1000km

R+h= 6400+1000 =7400km ½

Vo =[ GM/(R+h)]1/2

= [ 6.67×10-11 ×6×1024 /7400×103] ½

Vo = (54.08×106)1/2 =7352ms-1 1

Geostationary satellite-A satellite which revolves around the earth in its equatorial plane with the same angular speed and in the same direction as the earth rotates about its own axis is called geostationary or synchronous satellite. 1

Necessary condition- (four) 4×1/2

(i) It should revolves in an orbit concentric and coplanar with the equatorial plane of the earth.

(ii) Its sense of rotation should be same as that of the earth i.e. from west to east.

(iii) Its period of revolution around the earth should be exactly same as that of the earth about

its own axis i.e. 24 hours.

(iv) It should revolve at a height of nearly 36,000km above the earth’s surface.

Use- (any four) 4×1/2

(i) It communicating radio, TV and telephone signals across the world.

Geostationary satellites act as reflectors of such signals.

(ii) In studying upper regions of the atmosphere.

(iii) In forecasting weather.

(iv) In determining the exact shape and dimensions of the earth.

(v) In studying meteorites

(vi) In studying solar radiations and cosmic rays.