Work Power & Energy — Page 4 | NEET Physics

Q31

A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity 4 m s

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1 . It collides with a horizontal spring of force constant 200 N m

-

1 . The maximum compression produced in the spring will be

A 0.5 m

B 0.6 m

C 0.7 m

D 0.2 m

Correct Answer

Option B

Solution

At maximum compression the solid cylinder will stop so loss in K.E. of cylinder = gain in P.E. of spring

\Rightarrow {1 \over 2}m{v^2} + {1 \over 2}I{\omega ^2} = {1 \over 2}k{x^2}

\Rightarrow {1 \over 2}m{v^2} + {1 \over 2}{{m{R^2}} \over 2}{\left( {{v \over R}} \right)^2} = {1 \over 2}k{x^2}

\Rightarrow {3 \over 4}m{v^2} = {1 \over 2}k{x^2}

\Rightarrow {3 \over 4} \times 3 \times {\left( 4 \right)^2} = {1 \over 2} \times 200{x^2}

\Rightarrow {{36} \over {100}} = {x^2} \Rightarrow x = 0.6\,m

Q32

The potential energy of a particle in a force field is

U = {A \over {{r^2}}} - {B \over r}

where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is

A

{B \over {2A}}

B

{{2A} \over B}

C

{A \over B}

D

{B \over A}

Correct Answer

Option B

Solution

Here,

U = {A \over {{r^2}}} - {B \over r}

For equilibrium,

{{dU} \over {dr}} = 0

\therefore - {{2A} \over {{r^3}}} + {B \over {{r^2}}} = 0

\Rightarrow {{2A} \over {{r^3}}} = {B \over {{r^2}}} \Rightarrow r = {{2A} \over B}

For stable equilibrium,

{{{d^2}U} \over {d{r^2}}} > 0

{{{d^2}U} \over {d{r^2}}} = {{6A} \over {{r^4}}} - {{2B} \over {{r^3}}}

{\left. {{{{d^2}U} \over {d{r^2}}}} \right|_{r = \left( {2A/B} \right)}} = {{6A{B^4}} \over {16{A^4}}} - {{2{B^4}} \over {8{A^3}}} = {{{B^4}} \over {8{A^3}}} > 0

So for stable equilibrium, the distance of the particle is

{{2A} \over B}

.

Q33

A mass m moving horizontally (along the x-axis) with velocity

v

collides and sticks to a mass of 3m moving vertically upwards (along the y-axis) with velocity 2

v

. The final velocity of the combination is

A

{3 \over 2}v\widehat i + {1 \over 4}v\widehat j

B

{1 \over 4}v\widehat i + {3 \over 2}v\widehat j

C

{1 \over 3}v\widehat i + {2 \over 3}v\widehat j

D

{2 \over 3}v\widehat i + {1 \over 3}v\widehat j

Correct Answer

Option B

Solution

According to conservation of momentum, we get

mv\widehat i + \left( {3m} \right)2v\widehat j = \left( {m + 3m} \right)\overrightarrow {v'}

where

\overrightarrow {v'}

is the final velocity after collision

\overrightarrow {v'} = {1 \over 4}v\widehat i + {6 \over 4}v\widehat j = {1 \over 4}v\widehat i + {3 \over 2}v\widehat j

Q34

A body projected vertically from the earth reaches a height equal to earth's radius before returning to the earth. The power exerted by the gravitational force is greatest

A at the highest position of the body

B at the instant just before the body hits the earth.

C it remains constant all through.

D at the instant just after the body is projected.

Correct Answer

Option B

Solution

Power, P =

\overrightarrow F .\overrightarrow v = Fv\cos \theta

Just before hitting the earth $\theta$ = 0°.

Hence, the power exerted by the gravitational force is greatest at the instant just before the body hits the earth.

Q35

Force F on a particle moving in a straight line varies with distance d as shown in figure. The work done on the particle during its displacement of 12 m is

A 18 J

B 21 J

C 26 J

D 13 J

Correct Answer

Option D

Solution

Work done = Area under (F-d) graph = Area of rectangle ABCD + Area of triangle DCE =

2 \times \left( {7 - 3} \right) + {1 \over 2} \times 2 \times \left( {12 - 7} \right) = 8 + 5 = 13J

Q36

A particle of mass M, starting from rest, undergoes uniform acceleration. If the speed acquired in time T is V, the power delivered to the particle is

A

{{M{V^2}} \over T}

B

{1 \over 2}{{M{V^2}} \over {{T^2}}}

C

{{M{V^2}} \over {{T^2}}}

D

{1 \over 2}{{M{V^2}} \over T}

Correct Answer

Option D

Solution

Power delivered in time T is P = F·V = MaV or

P = MV{{dV} \over {dT}} \Rightarrow PdT = MVdV

\Rightarrow PT = {{M{V^2}} \over 2}

or

P = {1 \over 2}{{M{V^2}} \over T}

Q37

An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of 2 m/s. The mass per unit length of water in the pipe is 100 kg/m. What is the power of the engine?

A 400 W

B 200 W

C 100 W

D 800 W

Correct Answer

Option D

Solution

Here, Mass per unit length of water, $\mu$ = 100 kg/m Velocity of water, v = 2 m/s Power of the engine, P = $\mu$ v 3 = (100 kg/m) (2 m/s) 3 = 800 W

Q38

A ball moving with velocity 2 m/s collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in m/s) after collision will be

A 0, 1

B 1, 1

C 1, 0.5

D 0, 2

Correct Answer

Option A

Solution

Here, m 1 = m, m 2 = 2m, u 1 = 2 m/s, u 2 = 0 Coefficient of restitution, e = 0.5 Let v 1 and v 2 be their respective velocities after collision.

Applying the law of conservation of linear momentum, we get m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 $\therefore$ m × 2 + 2m × 0 = m × v 1 + 2m × v 2 or 2m = mv 1 + 2mv 2 or 2 = (v 1 + 2v 2 ) ...(i) By definition of coefficient of restitution,

e = {{{v_2} - {v_1}} \over {{u_1} - {u_2}}}

or e(u 1 – u 2 ) = v 2 – v 1 $\Rightarrow$ 0.5(2 – 0) = v 2 – v 1 ...(ii) Solving equations (i) and (ii), we get v 1 = 0 m/s, v 2 = 1 m/s

Q39

A block of mass M is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value k. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be

A 2Mg/k

B 4Mg/k

C Mg/2k

D Mg/k

Correct Answer

Option A

Solution

When the mass attached to a spring fixed at the other end is allowed to fall suddenly, it extends the spring by x.

Potential energy lost by the mass is gained by the spring.

Mgx = {1 \over 2}k{x^2} \Rightarrow x = {{2Mg} \over k}

Q40

A body of mass 1 kg is thrown upwards with a velocity 20 m/s. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? (g = 10 m/s 2 )

A 30 J

B 40 J

C 10 J

D 20 J

Correct Answer

Option D

Solution

Initial velocity u = 20 m/s; m = 1 kg Kinetic energy = maximum potential energy Initial kinetic energy =

{1 \over 2} \times 1 \times {20^2} = 200\,J

Mgh (max) = 200 J $\therefore$ h = 20 m The height travelled by the body, h' = 18 m $\therefore$ Loss of energy due to air friction = mgh – mgh' $\Rightarrow$ Energy lost = 200 J – 1 × 10 × 18 J = 20 J.