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Work, Energy, and PowerClass 11thPhysics NCERT Exemplar Solutions Class 11th Chapter Sixphysics ncert solutions class 11thPhysics

A bob of mass m suspended by a light string of length L is whirled into a vertical circle as shown in Fig. 6.11. What will be the trajectory of the particle if the string is cut at 

(a) Point B?

(b) Point C?

(c) Point X

28 Views|Posted 8 months ago
Asked by Shiksha User
1 Answer
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8 months ago

This is a short answer type question as classified in NCERT Exemplar

At B the velocity of B is vertically downward, therefore when string is cut at B then it fall downwards.

At C velocity along horizontally right, so when we cut it at C then it will move to right. But under the action of gravity its p

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An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of the object is d1 from C and the distance of the image formed is d2 from C, the radius of curvature of this mirror is:

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Vishal Baghel

Using Newton’s formula,

⇒ ( f + d 1 ) ( f − d 2 ) = f 2

⇒ f = d 1 d 2 d 1 − d 2 ⇒ R = 2 d 1 d 2 d 1 − d 2

Two blocks M1 and M2 having equal mass are free to move on a horizontal frictionless surface. M2 is attached to a massless spring as shown in Fig. 6.10. Initially M2 is at rest and M1 is moving toward M2 with speed v and collides head-on with M2.

(a) While spring is fully compressed all the KE of M1 is stored as PE of spring.

(b) While spring is fully compressed the system momentum is not conserved, though final momentum is equal to initial momentum.

(c) If spring is massless, the final state of the M1 is state of rest.

(d) If the surface on which blocks are moving has friction, then collision cannot be elastic.

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Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(c) When M1 comes in contact with the spring. M1 is retarded by the spring force and M2 is accelerated by the spring force. 
 
a) So spring will continue compress until the blocks moves with the same velocity. 
 
b) As

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A bullet of mass m fired at 30° to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target. Which of the following statements are correct in respect of bullet after it emerges out of the target?

(a) The velocity of the bullet will be reduced to half its initial value.

(b) The velocity of the bullet will be more than half of its earlier velocity.

(c) The bullet will continue to move along the same parabolic path.

(d) The bullet will move in a different parabolic path.

(e) The bullet will fall vertically downward after hitting the target.

(f) The internal energy of the particles of the target will increase.

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Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(b) conserving energy between “O” ans ”A”

Ui + Ki = Uf + Kf

0+1/2mv2= mgh + 1/2mv’

( v ' ) 2 2 = v 2 2 = - g h

(v’)2=v2-2gh = v’= v 2 - 2 g h ……….1

Let speed after emerging be v1 then

=1/2mv12=1/2[1/2mv’2]

1/2m(v1)2=1/4m(v’)2=1/4m[v2-2gh]

V1= v 2 2 - g h ………….2

From eqn 1 and 2

v ' v 1 = v 2 - 2 g h v 2 - 2 g h / 2 = 2

So v1 = v’/

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A man, of mass m, standing at the bottom of the staircase, of height L climbs it and stands at its top.

(a) Work done by all forces on man is equal to the rise in potential energy mgL.

(b) Work done by all forces on man is zero.

(c) Work done by the gravitational force on man is mgL.

(d) The reaction force from a step does not do work because the point of application of the force does not move while the force exists.

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Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.

Hence total work done =-

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A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001s, the force that the batsman had to apply to hold the bat firmly at its place would be

(a) 10.5 N

(b) 21 N

(c) 1.05 ×104 N

(d) 2.1 × 104 N

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Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(c) m =150g =3/20kg

Time of contact =0.001s

U=126km/h= 126 × 1000 60 × 60 = 35 m s

V= -35m/s

Change in momentum of the ball = m (v-u)= 3 20 - 35 - 35 k g m / s

=21/2

F= dp/dt=- 21 / 2 0.001 N = - 1.05 × 10 4 N

Here – negative sign indicates that force will be opposite to the direction of movement of the ball be

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Physics NCERT Exemplar Solutions Class 11th Chapter Six 2025

Physics NCERT Exemplar Solutions Class 11th Chapter Six 2025

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