Physics previous year question Of Work , Power and Energy for NDA

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previous year question Of Work , Power and Energy for NDA

previous year question

Q 2878078806

An elastic spring has a force constant k. It is cut into three
equal parts. The force constant of each part is

(A)

k/3

(B)

(C)

`k^3`

(D)

Solution:

Correct Answer is `=>` (B) 3k

Q 2863178945

A unloaded car moving with velocity `u` on a frictionless road can be stopped in a distance `s`. If the passengers add `40%` to its weight and breaking force remains the same, then the stopping distance at velocity `u` is now

(A)

`(1.4)^3 s`

(B)

`1/(1.4)^2 s`

(C)

`1.4 s`

(D)

`( 1.4)^2 s`

Solution:

Let initial mass of the passengers be `m kg` and breaking force be `F.`

`therefore 1/2 m u^2 = F xx s` .................(1)

On increasing mass by `40 % `

`m' = 1.4 m`

Now `1/2 ( 1.4 m) u^2 = F xx s'` ................(2)

On dividing Eq. (i) by Eq. (ii), we get

`1.4 = ( s')/s` or `s' = 1.4 s `

Correct Answer is `=>` (C) `1.4 s`

Q 2853878744

The work done in time `t` on a body of mass `M` which is accelerated from rest to speed `v` in time `t_1` as a
function of time `t` is given

(A)

`(mv t^2)/(2t_1) `

(B)

`(mv t^2)/t_1`

(C)

`( mv^2 t^2)/(2t_1^2) `

(D)

`( m v^2 t_1^2)/(2t^2)`

Solution:

Using `v = u +at ` we have

`v = 0+a t_1 \ \ \ \ [ because u = 0 ] `

or `a = v/t_1` .............(1)

`therefore F = ma = ( mv)/t_1 ` [from Eq.(1)]

`therefore` Distance traveled, `s = u t + 1/2 a t^2`

` = 0+1/2 (v/t_1)t^2` [from Eq.(i)]

` = ( v t^2)/(2 t_1) `

`therefore` Work done ` W = F s `

` = ( mv)/t_1 xx ( v t^2)/(2t_1) = 1/2 ( m v^2 t^2)/t_1^2`

Correct Answer is `=>` (C) `( mv^2 t^2)/(2t_1^2) `

Q 2813778649

A small ball is released from a height `H` as shown in the figure in a vertical circular track. so that it loops inside of the circular track of radius `R`. The value of `H` must he

(A)

`(2R)/3`

(B)

`(3R)/2`

(C)

`R`

(D)

`(5R)/2`

Solution:

At the lowest point

`mgH = 1/2 mv^2`

`mgH = 1/2 m ( sqrt(5R g))^2 \ \ \ \ [ because v = sqrt(5Rg) ]`

`gH = 1/2 xx 5 Rg`

`=> H = (5R)/2`

Correct Answer is `=>` (D) `(5R)/2`

Q 2853678544

The power output of a machine that lifts a `600 kg` weight through a height of `20 m` in `1` min is

(A)

`1.96 kW`

(B)

`0.98 kW`

(C)

`12 kW`

(D)

`3.92 kW`

Solution:

Given, `m = 600 kg, h = 20 m`,

`t = 1` min `= 60 s , P = ? `

We know that, `text(power) = text(work done)/text(time) = ( m g h)/t \ \ \ \ [ because W = mgh]`

` = ( 600 xx 9.8 xx 20)/60`

` = 1960 W = 1.96 kW `

Correct Answer is `=>` (A) `1.96 kW`

Q 2813678540

A `4 kg` mass and a `1 kg` mass are moving with equal kinetic energies. The ratio of their momenta is

(A)

`3 :1`

(B)

`1 : 1`

(C)

`4 : 1`

(D)

`2 : 1`

Solution:

Given `m_1 = 4 kg , m_2 = 1 kg`

According to question,

`K_1 = K_2`

The ratio of their momentum is `p_1/p_2`

` = ( K_1 m_1)/(K_2 m_2) \ \ \ \ \ \ [ because K = p^2/(2m) ]`

` => p_1/p_2 = sqrt(4/1) = 2 : 1 `

Correct Answer is `=>` (D) `2 : 1`

Q 2823578441

A `80 kg` man runs up a staircase of `4m` in `8 s`. If the value of acceleration due to gravity is `10 m//s^2`?, then his average power output is

(A)

`400 W`

(B)

`800 W`

(C)

`1600 W`

(D)

`3200 W`

Solution:

Given, `m = 80 kg , g = 10 m//s^2`

`h = 4m , t = 8 s , P = ?`

`therefore ` Average power = ` text(Work done)/text(Time) = ( m gh)/t \ \ \ \ [ because W = m gh]`

` = (80xx10xx4)/8 = 400 W`

Correct Answer is `=>` (A) `400 W`

Q 2813478340

Mass of `B` is four times that of `A . B` moves with a velocity half that of `A`. Then. `B` has

(A)

kinetic energy equal to that of `A`

(B)

half the kinetic energy of `A`

(C)

twice the kinetic energy of `A`

(D)

kinetic energy one fourth of `A`

Solution:

Given `m_B = 4 m_A => v_B = v_A/2`

`therefore (KE)_B = 1/2 m_B v_B^2 = 1/2 ( 4 m_A) ( v_A/2)^2`

` = 1/2 xx 4 m_A xx v_A^2/4`

`=> (KE)_n = 1/2xx1/2 m_A v_A^2 = 1/2 (KE)_A`

`=> (KE)_B = 1/2 (KE)_A`

Correct Answer is `=>` (B) half the kinetic energy of `A`

Q 2823378241

Which one of the following diagrams illustrates the relation between kinetic energy `(K)` and the velocity `(v)` of a body ?

Correct Answer is `=>` (A)

Q 2843278143

Which one of the following statements correctly defines power ?

(A)

Energy supplied per unit time

(B)

Energy of an object due to its motion

(C)

Energy of an objects due to its position

(D)

none of the above

Correct Answer is `=>` (A) Energy supplied per unit time

Q 2813178049

For a perfectly elastic collision the coefficient of rescitution `(e)` is

(A)

`1`

(B)

`-1`

(C)

`0`

(D)

infinity

Correct Answer is `=>` (A) `1`

Q 2873178046

A force of `5 N` acts through a distance `20` min the direction of motion The work done is

(A)

`50 J`

(B)

`30 J`

(C)

`100 J`

(D)

zero

Correct Answer is `=>` (C) `100 J`

Q 2843178043

Which of the following are the characteristies of an inelastic collision' ?

`I` Momentum is conserved.
`II` Total energy is Conserved.
`III` Kinetic energy is conserved.
`IV` All the forces must be of conservative nature
Choose the correct answer from the codes given below

(A)

Both `III` and `IV`

(B)

Both `I` and `II`

(C)

`I , II ` and `IV`

(D)

Both `II` and `IV`

Correct Answer is `=>` (B) Both `I` and `II`

Q 2883856747

The work done on an object does not depends upon the

(A)

displacement

(B)

force applied

(C)

final velocity

(D)

initial velocity

Correct Answer is `=>` (D) initial velocity

Q 2863856745

A body is falling freely under the action of gravity alone in vacuum. Which one of the following remains constant during the fall?

(A)

Potential energy

(B)

Kinetic energy

(C)

Total linear momentum

(D)

Total mechanical energy

Correct Answer is `=>` (D) Total mechanical energy

Q 2813856740

The SI unit of mechanical energy is

(A)

Joule

(B)

watt

(C)

Newton-second

(D)

Joule-second

Correct Answer is `=>` (A) Joule

Q 2803756648

When a body falls freely .

(A)

its PE is converted into KE

(B)

its mechanical in to PE

(C)

its mechanical energy is converted in to heat energy

(D)

None of these

Correct Answer is `=>` (A) its PE is converted into KE

Q 2813756640

A body is thrown vertically upward and then falls back on the ground. Its potential energy is maximum

(A)

on the ground

(B)

at the maximum height

(C)

during the return journey

(D)

Both on the ground and at the maximum height

Correct Answer is `=>` (B) at the maximum height

Q 2853656544

The energy associated with state of compression or expansion of an elastic spring is called its

(A)

rational kinetic energy

(B)

elastic potential energy

(C)

Total energy

(D)

magnetic energy

Correct Answer is `=>` (B) elastic potential energy

Q 2803556448

The linear momentum `(p)` and kinetic energy `(E)` for a body of mass `m` are related as

(A)

`p = sqrt(2mE)`

(B)

`p = sqrt((2m)/E)`

(C)

`p = sqrt(E/(2m))`

(D)

`p = 2 mE`

Correct Answer is `=>` (A) `p = sqrt(2mE)`

Q 2833556442

An object of mass `2 kg` is lifted vertically through a distance of `1.5 m`. The work done in process is

(A)

`29.4 J`

(B)

`19.4 J`

(C)

`17.4 J`

(D)

`20.4 J`

Correct Answer is `=>` (A) `29.4 J`

Q 2883456347

Choose the wrong statement

(A)

Work done is a scalar quantity

(B)

Work done by a body does not depend on the time taken to complete the work

(C)

Work done can never be zero

(D)

SI unit of work is joule

Correct Answer is `=>` (C) Work done can never be zero

Q 2843456343

When work is done on the body

(A)

it gains energy

(B)

it loses energy

(C)

its energy remains constant

(D)

None of the above

Correct Answer is `=>` (A) it gains energy

Q 2356791674

If the linear momentum of a body is increased by `5%`, by what percentage shall the kinetic energy increase (approximately)?
NDA Paper 2 2007

(A)

`10 %`

(B)

`20 %`

(C)

`25 %`

(D)

`50 %`

Solution:

Linear momentum `= sqrt(2m xx text(Energy))`

or `p=sqrt(2 mE)`

`:. (Delta p)/p xx 100=1/2 (Delta E)/E xx 100`

or `5=1/2 ((Delta E)/E xx 100)`

or `(Delta E)/E xx 100 =10 %`

Hence, kinetic energy increases by `1 0%`.

Correct Answer is `=>` (A) `10 %`

Q 2316791670

What would be the power of an engine which supplies 18 kJ of energy per minute'?
NDA Paper 2 2007

(A)

200 W

(B)

250 W

(C)

300 W

(D)

1080 W

Solution:

Power of an engine is

`P=W/t=18/60`

`=(18 xx 10^3)/60`

`=300 W`

Correct Answer is `=>` (C) 300 W

Q 2376391276

Consider the following statements. Work is not done, when
1. a man is walking on a horizontal road with a load on his head.
2. a man is climbing up a hill with a load on his head.
3. moon is revolving round the earth.
Which of the statement(s) given above is/are correct?
NDA Paper 2 2007

(A)

1 and 3

(B)

1 and 2

(C)

Only 2

(D)

Only 1

Solution:

1. In order to balance the load on his head, the man applies a force on it in the upward direction equal to its weight, his displacement is along the horizontal direction. Thus, the angle between force F and displacements is `90^@`. Therefore, work done

`W = Fs cos theta = Fs cos 90^o = 0`

2. The man lifts the load vertically up a hill. Force has to be applied on the load against the force of gravitation. Hence, the man does work.

3. The moon is revolving round the earth in a circular orbit is acted upon by gravitational force. The work done by the gravitational force over every complete orbit of the earth is zero.

Correct Answer is `=>` (A) 1 and 3

Q 2336178972

The neutron, proton, electron and `alpha` particles are moving with equal kinetic energies. How can the particles be arranged in the increasing order of their velocities?
NDA Paper 2 2008

(A)

Alpha particle-neutron-proton-electron

(B)

Proton-electron-neutron-alpha particle

(C)

Electron-proton-neutron-alpha particle

(D)

Neutron-proton-electron-alpha particle

Solution:

Given that kinetic energies of neutron, proton, electron and `alpha` particles are equal, i.e,

`1/2m_nv_n^2=1/2m_pv_p^2=1/2m_ev_e^2=1/2m_alpha v_alpha^2`

But `m_alpha > m_n > m_p > m_e`

`therefore v_alpha < v_n < v_p < v_e`

Correct Answer is `=>` (C) Electron-proton-neutron-alpha particle

Q 2356123074

Which one of the following statements correctly defines power?
NDA Paper 2 2009

(A)

Energy supplied per unit time

(B)

Energy of an object due to its motion

(C)

Energy of an object due to its position

(D)

None of the above

Solution:

The power of a body or a system or a machine is
defined as the rate at which it can do work.

`P=W/t=(text(Energy))/(text(Time))`

Correct Answer is `=>` (A) Energy supplied per unit time

Q 2315191069

If a light body and a heavy body have equal momentum, then
NDA Paper 2 2009

(A)

the lighter body has greater kinetic energy than the heavier body

(B)

the lighter body has lesser kinetic energy than the heavier .body

(C)

the kinetic energy of the lighter body is equal to the kinetic energy of the heavier body

(D)

the kinetic energy of both the bodies are independent of momentum

Solution:

Let subscripts 1 and 2 refers to light body and heavy body respectively in the following relations

`E_1=1/2 m_1v_1^2=(p_1^2)/(2m_1)`
`E_1=1/2 m_2v_2^2=(p_2^2)/(2m_2)`

But `p_1=p_2`

`:. E_1/E_2=m_2/m_1`

Here, `m_2 > m_1 => m_2/m_1 > 1`
`:. E_1/E_2 > 1=> E_1 > E_2`

Hence, the lighter body has greater kinetic energy than the heavier body.

Correct Answer is `=>` (A) the lighter body has greater kinetic energy than the heavier body

Q 2385880767

A body is thrown vertically upwards and then
falls back on the ground. Its potential energy is
maximum
NDA Paper 2 2010

(A)

on the ground

(B)

at the maximum height

(C)

during the return journey

(D)

both on the ground and at the maximum height

Solution:

Potential energy is the energy possessed by a body by
virtue of its position. When a body is at its maximum height, then
its potential energy will be maximum.

`PE=mgh_(max)`

Correct Answer is `=>` (B) at the maximum height

Q 2345323263

A body initially at rest is acted upon by a constant force. The rate of change of its kinetic energy varies.
NDA Paper 2 2011

(A)

linearly with square root of time

(B)

linearly with time

(C)

linearly with square of time

(D)

inversely with time

Solution:

When a constant force is acted on a body which is initially at rest, then it starts moving and its kinetic energy changes.
The rate of change of kinetic energy(P) `=(text(kinetic energy))/(text(Time)(t))` So, `Pprop 1/t`

Therefore, the rate of change of kinetic energy of the body varies inversely with time.

Correct Answer is `=>` (D) inversely with time

Q 2334334252

A particle is moving freely. Then its
NDA Paper 2 2013

(A)

kinetic energy is always greater than zero

(B)

potential energy is greater than zero and kinetic energy is less than zero

(C)

potential energy is less than zero and kinetic energy is greater than zero

(D)

potential energy is zero and kinetic energy is less than zero

Solution:

A particle is moving freely, then its kinetic energy is always greater than zero. This is because the kinetic energy of an object is the energy, which it possesses due to its motion.

Correct Answer is `=>` (A) kinetic energy is always greater than zero

Q 2333278142

Power required by a boy of mass `30` `kg` to run up a staircase of `40` steps in `10 s` is

(Height of each step is `15 cm`)
(Take, `g =10 m//s^2`)
NDA Paper 2 2013

(A)

`1800 W`

(B)

`180 W`

(C)

`18000 W`

(D)

`18 W`

Solution:

`P=W/t=(mgh)/t`

`=(30 xx 10 xx 40 xx 0.15)/10=180 W`

Correct Answer is `=>` (B) `180 W`

Q 1678645506

Which one among the following happens when a swing rises to a certain height from its rest position?
NDA Paper 2 2015

(A)

Its potential energy decreases while kinetic energy increases

(B)

Its kinetic energy decreases while potential energy increases

(C)

Both potential and kinetic energy decrease

(D)

Both potential and kinetic energy increase

Solution:

When a swing rises to a certain height from its rest position, its kinetic energy decreases while potential energy increases.

Correct Answer is `=>` (B) Its kinetic energy decreases while potential energy increases

Q 1622234131

Conservation of momentum in a collision between

particles can be understood on the basis of
NDA Paper 2 2015

(A)

Newton's first law of motion

(B)

Newton's second law of motion

(C)

Both Newton's second law motion and Newton's third law of motion

(D)

Conservation of energy

Solution:

Newton's second law states that rate of change of

momentum is proportional to the force applied and

Newton's third law also states about the action and

reaction forces. Thus, we can say that principle of

conservation can be understood by both of these laws.

Correct Answer is `=>` (C) Both Newton's second law motion and Newton's third law of motion