Physics previous year question Of Measurement, Motion and Force for NDA

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previous year question Of Measurement, Motion and Force for NDA

previous year question

Q 3112545439

Which one of the following statements is not. correct'?
NDA Paper 2 2015

(A)

It the velocity and acceleration have opposite sign, then the object is slowing down

(B)

It the velocity is zero at an instant. then the acceleration should also be zero at that instant

(C)

II the velocit•t is zero for a tirne interval, then the acceleration is zero at any instant within the time interval

(D)

If the position and velocity have opposite sign, then the object is moving towards the origin

Solution:

Correct Answer is `=>` (B) It the velocity is zero at an instant. then the acceleration should also be zero at that instant

Q 2863056845

A 7 kg object is subjected to two forces `vec F_1 = (20hat i + 30 hat j)` and `vecF_2 = (8 hat i + 5 hat j)` N. The magnitude of resulting acceleration in `m//s^2` will

(A)

(B)

(C)

(D)

5.3

Solution:

Resultant: Force `F = F_1 + F_2`

`= ( 20 hat i + 30 hat j ) + ( 8 hat i - 5 hat j)`

`= ( 28 hat i + 25hat j ) N`

`:. ` Acceleration ` a = F/m = ( 28 hat i + 25 hat j ) /7`

`| a | = sqrt ( 1409)/7 = 5.3 m//s^2`

Correct Answer is `=>` (D) 5.3

Q 2833056842

The displacement y (in metres) of a body varies with time t (in seconds) as

`y = (-2)/3 t^2 + 16 t -12`

Then, the body will come to rest in

(A)

16 s

(B)

12 s

(C)

8 s

(D)

None of these

Solution:

Given `y = -2/3t^2 + 16 t -12`

`:. v = (dy)/(dt) = d/(dt) (-2/3 t^2 + 16t -12)`

`= -2/3 (2t) + 16 = -4/3 t + 16`

Since, the body comes to rest, v = 0

`:. - (4t)/3 + 16 = 0`

`=> t = 12s`

Correct Answer is `=>` (B) 12 s

Q 2823056841

A machine is delivering constant power to drive a body along a straight line. What is the relation between the distance travelled by the body against time?

(A)

`s^2 prop t^3`

(B)

`s^2 prop t^-3`

(C)

`s^3 prop t^2`

(D)

`s^3 prop t^(1//2)`

Solution:

We know that,

power `= text(work)/text(time)`

`=[ML^2L^-3] = ` constant

`:. ML^2T^-3 =` constant

or `L^2 prop T^3 => s^2 prop t^3`

Correct Answer is `=>` (A) `s^2 prop t^3`

Q 2813856749

A man travels along a straight road for the first half length with a velocity u and the second half length with a velocity v. Then, the mean velocity is given by

(A)

`(u +v)/2`

(B)

`(2uv)/( u +v)`

(C)

`sqrt(uv)`

(D)

zero

Solution:

Let the total distance travelled be = s + s = 2s

Total time taken, `t = t_1 + t_2`

`= s/u + s/v`

`:.` Average velocity `= (2s)/(t_1 + t_2)`

`= (2s)/( s/u + s/v) = (2uv)/(u +v)`

Correct Answer is `=>` (B) `(2uv)/( u +v)`

Q 2803856748

A car travels the first one-third distance at a speed of 10 km/h, the next one-third distance at 20 km/h and the last one-third distance at 60 km/h. Then, the average speed of the car is

(A)

30 km/h

(B)

24 krn/h

(C)

18 krn/h

(D)

None of these

Solution:

Average speed `= text(Total distance )/text(Total time)`

`= (3s)/( s/10 + s/20 + s/60)`

`= (60 xx 3 )/10 =18` km/h

Correct Answer is `=>` (C) 18 krn/h

Q 2873856746

A body of mass 6 kg is rotated in cirde of radius 3 m with a uniform speed of 10 m/s, the force which must act on the body to maintain the motion is

(A)

100 N

(B)

200 N

(C)

300 N

(D)

20 N

Solution:

Given, m = 6 kg, v = 10 m/s,

r = 3 m, F =?

`:. F = (mv^2)/r = (6xx (10)^2)/3 = 200 N`

Correct Answer is `=>` (B) 200 N

Q 2823856741

When a moving bus suddenly applies brakes, then the passengers sitting in it fall in the forward direction This can be explained by

(A)

the theory of relativity

(B)

Newton's first law

(C)

Newton's second law

(D)

Newton's third law

Solution:

Given, m = 6 kg, v = 10 m/s,

r = 3 m, F =?

`:. F = (mv^2)/r = (6xx (10)^2)/3 = 200 N`

Correct Answer is `=>` (B) Newton's first law

Q 2813756649

The area under acceleration-time graph represents

(A)

velocity

(B)

displacement travelled

(C)

distance travelled

(D)

change in velocity

Solution:

Area under acceleration-time graph

= acceleration x time

= change in velocity

Correct Answer is `=>` (D) change in velocity

Q 2883756647

A body goes from P to Q with a velocity of 40 `ms^-1` and comes back from Q to P vrith a velocity of `60 ms^- 1` • Then, the average velocity of the body during the whole journey is

(A)

` 50ms^-1`

(B)

`48 ms^-1`

(C)

`45 ms^-1`

(D)

zero

Solution:

Average velocity `= text(Total displacement covered)/text(Total time taken ) = 0`

When the body returns to the starting point, then total displacement covered is zero.

Correct Answer is `=>` (D) zero

Q 2833756642

If the energy E of hv photon is equal to hv, where v is the frequency and h is Planck's constant then the dimensions of Planck's constant ts

(A)

`[ML^2T^-3]`

(B)

`[M^0L^2T^-1]`

(C)

`[ML^2T^-1]`

(D)

`[ML^2T^-2]`

Solution:

Given, E = hv

where, E = energy of a photon,

h = Planck's constant,

v = frequency

`:. h = E/v= ([ML^2T^-2])/([M^0L^0T^-1]) = [ML^2T^-1]`

Correct Answer is `=>` (C) `[ML^2T^-1]`

Q 2823756641

Newton's laws of motion do not hold good for objects

(A)

at rest

(B)

rnoving slowly

(C)

rnoving with high velocity

(D)

rnoving with velocity comparable to velocity of light

Solution:

Given, E = hv

where, E = energy of a photon,

h = Planck's constant,

v = frequency

`:. h = E/v= ([ML^2T^-2])/([M^0L^0T^-1]) = [ML^2T^-1]`

Correct Answer is `=>` (D) rnoving with velocity comparable to velocity of light

Q 2813656549

Frictional force

1 . is self-adjusting force.
2. is a non-conservative force.
3. is a necessary evil.

(A)

1 and 2

(B)

1 and 3

(C)

only 1

(D)

All of the above

Solution:

Given, E = hv

where, E = energy of a photon,

h = Planck's constant,

v = frequency

`:. h = E/v= ([ML^2T^-2])/([M^0L^0T^-1]) = [ML^2T^-1]`

Correct Answer is `=>` (D) All of the above

Q 2863656545

Momentum has the same unit as that of

(A)

couple

(B)

torque

(C)

impulse

(D)

moment of momentum

Solution:

Given, E = hv

where, E = energy of a photon,

h = Planck's constant,

v = frequency

`:. h = E/v= ([ML^2T^-2])/([M^0L^0T^-1]) = [ML^2T^-1]`

Correct Answer is `=>` (C) impulse

Q 2833656542

Newton's second law of motion connects

(A)

momentum and acceleration

(B)

change of momentum and velocity

(C)

rate of change of momentum and external force

(D)

rate of change of force and momentum

Solution:

Given, E = hv

where, E = energy of a photon,

h = Planck's constant,

v = frequency

`:. h = E/v= ([ML^2T^-2])/([M^0L^0T^-1]) = [ML^2T^-1]`

Correct Answer is `=>` (C) rate of change of momentum and external force

Q 2813656540

If the displacement of an object is proportional to square of time, then the objeet moves with

(A)

uniform velocity

(B)

uniform acceleration

(C)

increasing acceleration

(D)

decreasing acceleration

Solution:

Given, E = hv

where, E = energy of a photon,

h = Planck's constant,

v = frequency

`:. h = E/v= ([ML^2T^-2])/([M^0L^0T^-1]) = [ML^2T^-1]`

Correct Answer is `=>` (B) uniform acceleration

Q 2883556447

When the distanee, an object travels is direetly proportional to the length of time, it is said to travel with

(A)

zero velocity

(B)

uniform velocity

(C)

constant velocity

(D)

constant acceleration

Solution:

Given, E = hv

where, E = energy of a photon,

h = Planck's constant,

v = frequency

`:. h = E/v= ([ML^2T^-2])/([M^0L^0T^-1]) = [ML^2T^-1]`

Correct Answer is `=>` (B) uniform velocity

Q 2863556445

Average velocity of an object is equal to the mean of its initial and final velocities, if the acceleration is

(A)

uniform

(B)

variable

(C)

Both (a) and (b)

(D)

None of these

Solution:

Given, E = hv

where, E = energy of a photon,

h = Planck's constant,

v = frequency

`:. h = E/v= ([ML^2T^-2])/([M^0L^0T^-1]) = [ML^2T^-1]`

Correct Answer is `=>` (A) uniform

Q 2813456349

The dimensions of coefficient of viscosity is

(A)

`[MLT^-1 ]`

(B)

`[ML^-1T]`

(C)

`[ML^-1T^-1]`

(D)

`[ML^-1]`

Solution:

Given, E = hv

where, E = energy of a photon,

h = Planck's constant,

v = frequency

`:. h = E/v= ([ML^2T^-2])/([M^0L^0T^-1]) = [ML^2T^-1]`

Correct Answer is `=>` (C) `[ML^-1T^-1]`

Q 2833145042

Dimension of impulse is

(A)

`[ML^-2T^-3 ]`

(B)

`[ML^-2]`

(C)

`[MLT^-1]`

(D)

`[MLT^-2]`

Solution:

Given, E = hv

where, E = energy of a photon,

h = Planck's constant,

v = frequency

`:. h = E/v= ([ML^2T^-2])/([M^0L^0T^-1]) = [ML^2T^-1]`

Correct Answer is `=>` (C) `[MLT^-1]`

Q 2813034849

The dimension of impulse is equal to that of

(A)

force

(B)

linear momentum

(C)

pressure

(D)

angular momentum

Solution:

Given, E = hv

where, E = energy of a photon,

h = Planck's constant,

v = frequency

`:. h = E/v= ([ML^2T^-2])/([M^0L^0T^-1]) = [ML^2T^-1]`

Correct Answer is `=>` (B) linear momentum

Q 2803034848

`Wb//m^2` is equal to

(A)

dyne

(B)

tesla

(C)

watt

(D)

henry

Solution:

Given, E = hv

where, E = energy of a photon,

h = Planck's constant,

v = frequency

`:. h = E/v= ([ML^2T^-2])/([M^0L^0T^-1]) = [ML^2T^-1]`

Correct Answer is `=>` (B) tesla

Q 2863734645

What is the dimension of momentum?

(A)

Impulse

(B)

Power

(C)

Stress

(D)

Pressure

Solution:

Given, E = hv

where, E = energy of a photon,

h = Planck's constant,

v = frequency

`:. h = E/v= ([ML^2T^-2])/([M^0L^0T^-1]) = [ML^2T^-1]`

Correct Answer is `=>` (A) Impulse

Q 2843734643

1 Fermi is equivalent to

(A)

`10^- 15m`

(B)

`10^-12` m

(C)

`10^-13` rn

(D)

`10^- 19` m

Solution:

Given, E = hv

where, E = energy of a photon,

h = Planck's constant,

v = frequency

`:. h = E/v= ([ML^2T^-2])/([M^0L^0T^-1]) = [ML^2T^-1]`

Correct Answer is `=>` (A) `10^- 15m`

Q 2764801755

Which one of the following physical quantity has the same unit as that of pressure?
NDA Paper 2 2017

(A)

Angular momentum

(B)

stress

(C)

Strain

(D)

Work

Solution:

Given, E = hv

where, E = energy of a photon,

h = Planck's constant,

v = frequency

`:. h = E/v= ([ML^2T^-2])/([M^0L^0T^-1]) = [ML^2T^-1]`

Correct Answer is `=>` (B) stress

Q 2714101050

The speed of a car travelling on a straight road is listed-below: at successive intervals of s :

`text ( Time sec 0 1 2 3 4 )`
`text ( Speed m/s 0 2 4 6 8 )`

Which of the following is/are correct?
The car travels
(1) with a uniform acceleration of 2 `m//s^2`
(2) `16m` in `4 s`.
(3) with an average speed of `4 m//s`.
Select the correct answer using the code given below :
NDA Paper 2 2017

(A)

1 , 2 and 3

(B)

2 and 3 only

(C)

1 and 2 only

(D)

1 only

Solution:

Given, E = hv

where, E = energy of a photon,

h = Planck's constant,

v = frequency

`:. h = E/v= ([ML^2T^-2])/([M^0L^0T^-1]) = [ML^2T^-1]`

Correct Answer is `=>` (A) 1 , 2 and 3

Q 2326591471

How long does it take light from the sun to reach to the Earth (approximately)?
NDA Paper 2 2007

(A)

2 min

(B)

4 min

(C)

8 min

(D)

16 min

Solution:

The light rays from the sun reach the Earth in 8 min.

Correct Answer is `=>` (C) 8 min

Q 2356391274

Two teams are pulling a rope with equal and opposite forces each of 5 kN in a tug of war so that a condition of equilibrium exists. What will be the tensile force in the rope?
NDA Paper 2 2007

(A)

Zero

(B)

`2.5` kN

(C)

`5` kN

(D)

`10` kN

Solution:

In tug of war between two teams when equilibrium exists, then tensile force is equal to either of two equal forces.

Correct Answer is `=>` (C) `5` kN

Q 2316180979

A bullet travelling horizontally hits a block kept at rest on a horizontal surface and gets embedded into it, the two together then move with a uniform velocity. Which one of the following conservation laws holds?
NDA Paper 2 2007

(A)

Conservation of angular momentum

(B)

Conservation of kinetic energy

(C)

Conservation of linear momentum

(D)

Conservation of velocity

Solution:

When a bullet travelling horizontally hits a block and gets embedded into it and then the two together move with uniform velocity, such a collision is an example of perfectly inelastic collision. For a perfectly inelastic collision, the linear momentum is conserved by kinetic energy changes.

Correct Answer is `=>` (C) Conservation of linear momentum

Q 2326880771

A bus moving at a speed of `24 m//s` begins to slow at a rate of `3 m//s` each second. How far does it go before stopping?
NDA Paper 2 2007

(A)

`96 m`

(B)

`72 m`

(C)

`60 m`

(D)

`48 m`

Solution:

Here, initial velocity of bus, `u = 24m // s`

Retardation `= 3m //s^2`

Final velocity, `v = 0`

Using the relation,

`v^2- u^2 = 2as`

`0- (24)^2 = 2 xx 3 xx S`

`=>(a = -3m // s^2` , since the bus is retarded)

`=> s = (24 xx 24)/(2 xx3) = 96m`

Correct Answer is `=>` (A) `96 m`

Q 2336580472

A bus moving at a speed of `24 m//s` begins to slow at a rate of `3 m//s^2` each second. How far does it go before stopping?
NDA Paper 2 2007

(A)

`96 m`

(B)

`72 m`

(C)

`60 m`

(D)

`48 m`

Solution:

Here, initial velocity of bus, `u = 24m //s`

Retardation `= 3m // s^2`

Final velocity, `v = 0`

`:.` Using the relation

`v^2-u^2=2as`

`=>0-(24)^2=-2 xx 3 xx s`

`(a = -3m // s^2 , text(since the bus is retarded))`

`=> s=(24 xx24)/(2 xx 3)=96 m`

Correct Answer is `=>` (A) `96 m`

Q 2386478377

A car accelerates from rest with acceleration `1.2 m//s^2` A bus moves with constant speed of `12 m//s` in a parallel lane. How long does the car take from its start to meet the bus?
NDA Paper 2 2008

(A)

`17 s`

(B)

`8 s`

(C)

`20 s`

(D)

`12 s`

Solution:

Let the car take t seconds from its start to meet the bus. So, the distance travelled by bus = Distance travelled by car

`12t=1/2 xx 1.2 xx t^2`

`t=(12 xx 2)/(1.2)=20 s`

Correct Answer is `=>` (C) `20 s`

Q 2316178079

A rubber ball dropped from `24 m` height loses its energy by `25%`. What is the height to which it rebounds?
NDA Paper 2 2008

(A)

`6 m`

(B)

`12 m`

(C)

`18 m`

(D)

`24 m`

Solution:

`mgh' =75/100 (mgh)`

`h'=3/4 h=3/4 xx 24=18 m`

Correct Answer is `=>` (C) `18 m`

Q 2346067873

A body starting from the rest moves along a straight line with constant acceleration. Which one of the following graphs represents the variation of speed (`v`) and distance (`s`)'?
NDA Paper 2 2008

Solution:

To draw a graph between speed (`v`) and distance (`s`), we have to find out the relation between `v` and `s`. By using equation of motion,

`v^2-u^2=2as`

Here, `u=0` (as body is initially at rest)

`:. v^2 prop 2as`

or `v prop sqrt s` (as a is constant)

Therefore, the graph between `v` and `s` will be a rectangular hyperbola.

Correct Answer is `=>` (B)

Q 2306167078

Consider the following statements If the net external torque acting on an object is switched off, then
1. linear momentum will remain unchanged.
2. angular momentum will remain unchanged.
Which of the statements given above is/are correct?
NDA Paper 2 2008

(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

By the relation between torque and angular momentum

We have, `tau_(ext)=(dL)/(dt)`

if `tau_(ext) = 0` then `tau = 0` or `L` must be constant.

Therefore, in the absence of any external torque, total angular momentum of a system must remain conserved.

Correct Answer is `=>` (B) Only 2

Q 2346756673

Which one of the following remains constant while throwing a ball upward?
NDA Paper 2 2008

(A)

Displacement

(B)

Kinetic energy

(C)

Acceleration

(D)

Velocity

Solution:

If a body is projected vertically upwards, then acceleration remains constant and is given as `a = - g` (as acceleration is downwards while motion is upwards).

Correct Answer is `=>` (C) Acceleration

Q 2306456378

What is the correct sequence in which the lengths of the following units increase?
1. Angstrom
2. Micron
3. Nanometer
Select the correct answer using the codes given below
NDA Paper 2 2008

(A)

1,2,3

(B)

3,1,2

(C)

1,3,2

(D)

2,3,1

Solution:

The given units are used to measure small distances

1 Angstrom (`Å`) `= 10^(-10)` m

1 nano meter (`nm`) `= 10^(-9)` m

1 micron (`mu m`) `= 10^(-6)` m

So, the correct sequence of increasing order of these units
1,3,2.

Correct Answer is `=>` (C) 1,3,2

Q 2356456374

If a body travels half the distance with velocity `v_1` and next half with velocity `v_2` , then which one of the following will be the average velocity of the body?
NDA Paper 2 2008

(A)

`sqrt(v_1v_2)`

(B)

`(v_1+v_2)/2`

(C)

`v_2/v_1`

(D)

`(2v_1v_2)/(v_1+v_2)`

Solution:

The average velocity of the body

`=(text(total displacement))/(text (Total time))`

The body covers equal distances with different velocities.

`:.` Average velocity `= (S/2 +S/2)/(S/(2v_1)+S/(2v_2))`

`= S/(S/2 (1/v_1+1/v_2))`

`=(2v_1v_2)/(v_1+v_2)`

Correct Answer is `=>` (D) `(2v_1v_2)/(v_1+v_2)`

Q 2316334279

A long jumper runs before jumping because he
NDA Paper 2 2009

(A)

covers a greater distance

(B)

maintains momentum conservation

(C)

gains energy by running

(D)

gains momentum

Solution:

A long jumper always runs for some distance before
taking a jump so that inertia of motion may help him in his
muscular efforts to take a longer jump. By this, he maintains
momentum conservation.

Correct Answer is `=>` (B) maintains momentum conservation

Q 2356112974

One light year is equal to
NDA Paper 2 2009

(A)

`9.46 xx 10^(-15) m`

(B)

`9.46 xx 10^15 m`

(C)

`9. 46 xx 10^(-13) m`

(D)

`9.46 xx 10^13 m`

Solution:

A light year (`1 ly`) is distance covered by the light in
vacuum in 1 year. We know that light travels `3 xx 10^8 m` in vacuum
in `1 s`. Therefore

`1 ly = (3 xx 10^8 ms^(-1)) xx (365 xx 24xx 60 xx 60) s`

or `1 ly=9.46 xx 10^15 m`

Correct Answer is `=>` (B) `9.46 xx 10^15 m`

Q 2306012878

Which of the following is the best example of the law of conservation of mass?
NDA Paper 2 2009

(A)

When `12 g` of carbon is heated in vacuum, there is no change in mass

(B)

Weight of platinum wire is the same before and after heating

(C)

A sample of air increases in volume when heated at constant pressure but mass remains unchanged

(D)

`12 g` of carbon combines with 32 g of oxygen to give 44 g of carbon dioxide

Solution:

`12 g` of carbon combines with `32 g` of oxygen to give `44 g` of carbon dioxide. This is the best example of the law of conservation of mass.

Correct Answer is `=>` (D) `12 g` of carbon combines with 32 g of oxygen to give 44 g of carbon dioxide

Q 2346012873

Water flows out of the hole of a bucket and follows a parabolic path. If the bucket falls freely under gravity, the water flow (ignoring air resistance)
NDA Paper 2 2009

(A)

·follows a straight-line path relative to the falling bucket

(B)

follows a parabolic path relative to the falling bucket

(C)

stops

(D)

decreases but continues to flow

Solution:

Velocity of efflux of a liquid is equal to the velocity which a body acquires in falling freely from the free liquid surface to the orifice. Hence, if the bucket falls freely undergravity, the water flow follows a parabolic path relative to the falling bucket

Correct Answer is `=>` (B) follows a parabolic path relative to the falling bucket

Q 2305091868

In the relation `alpha = beta t + lambda, alpha` and `lambda` are measured in metre (`m`) and `t` is measured in second (`s`). The SI unit of `beta` must be
NDA Paper 2 2010

(A)

`m`

(B)

`ms`

(C)

`s`

(D)

`ms^(-1)`

Solution:

In the relation,

`alpha=beta t+alpha`

`alpha` and `lambda`. are measured in metre (m) and t is second (s).
By Homogeneity principle of dimensions, units

Unit of `alpha` = units of `beta t`

`=>` metre = unit of `beta xx` second

`= ms^(-1)`

Correct Answer is `=>` (D) `ms^(-1)`

Q 2375591466

A book is kept on the surface of a table. If the gravitational pull of the Earth on the book is the force of action, then the force of reaction is exerted by
NDA Paper 2 2009

(A)

the book on the table

(B)

the book on the Earth

(C)

the table on the book

(D)

the table on the Earth

Solution:

By Newton's third law, if a body exerts a force on another body, then the other body exerts a reaction force (same in magnitude). Therefore, is present situation, the book exerts the force of reaction on the Earth.

Correct Answer is `=>` (B) the book on the Earth

Q 2335491362

An object of mass `5 kg` is attached to the end of a rope. If the rope is pulled upward with an acceleration `0.30 ms^( -2)`, what is the tension in the rope?
NDA Paper 2 2010

(A)

`30.5 N`

(B)

`40.5 N`

(C)

`50.5 N`

(D)

`60.5 N`

Solution:

The situation is shown along side. By the Newton's law,

`T-mg=ma`

`T-5g=5 xx 0.30`

`=> T=1.5+5 xx 9.8`

`=1.5+49=50.5 N`

Correct Answer is `=>` (C) `50.5 N`

Q 2375180966

The weight of a body is `9.8 N` at the place where `g = 9.8 ms^(-2)` . Its mass is
NDA Paper 2 2009

(A)

zero

(B)

9.8 kg

(C)

10 kg

(D)

1 kg

Solution:

Weight of the body `= 9.8 N`

i.e., `W=mg`

`=>` Mass of the body, `m= W/g=(9.8 N)/(9.8 ms^(-2))=1 kg`

Correct Answer is `=>` (D) 1 kg

Q 2325180961

Which one of the following graphs represents uniform motion?
NDA Paper 2 2010

Solution:

For uniform motion, displacement-time graph is a straight line as in option (4). Also, the velocity-time graph is a straight line but parallel to time axis.

Correct Answer is `=>` (D)

Q 2335867762

Which one of the following characteristics of the particle does the shaded area of the velocity-time graph shown above represent?
NDA Paper 2 2010

(A)

Momentum

(B)

Acceleration

(C)

Distance covered

(D)

Speed

Solution:

If we measure the velocity of a moving object at different times and draw a graph between time (t)and velocity (v), then it is called a 'velocity-time' graph. The distance covered by an object in a time interval is equal to the area enclosed between time-velocity graph and time axis for that time interval, whatever be the shape of the graph.

Correct Answer is `=>` (C) Distance covered

Q 2315367260

A body is at rest on the surface of the earth. Which one among the following statements is correct regarding this?
NDA Paper 2 2010

(A)

No force is acting on the body

(B)

Only weight of the body acts on it

(C)

Net downward force is equal to the net upward force

(D)

None of the above

Solution:

For a body at rest on the surface of earth, the net downward force is equal to the net upward force. i.e., `R=mg`

Correct Answer is `=>` (C) Net downward force is equal to the net upward force

Q 2315556469

A circus performer of mass `M` is walking along a wire as shown in the figure given above. The tension `T` in the wire is (`g =` acceleration due to gravity)
NDA Paper 2 2010

(A)

approximately `Mg`

(B)

less than `Mg`

(C)

more than `Mg`

(D)

depends on whether the performer stands on one or two feet

Solution:

According to the given condition, three concurrent forces are in equilibrium when the resultant of any two of them is equal and opposite to the third.

i.e, `2T cos theta=Mg`

`=> T=(Mg)/(2 cos theta)`

i.e., the tension in the string in less than `Mg`

Correct Answer is `=>` (B) less than `Mg`

Q 2375256166

For a particle revolving in a circular path, the acceleration of the particle is
NDA Paper 2 2010

(A)

along the tangent

(B)

along the radius

(C)

zero

(D)

along the circumference of the circle

Solution:

For a particle revolving in a circular path, acceleration `a` is directed opposite to r (radius of circular path), i.e., its direction is radially inward. Due to this reason acceleration of uniform circular motion is known as 'radial acceleration' or centripetal acceleration' given as
`a=omega^2 r=v^2/r`

Correct Answer is `=>` (B) along the radius

Q 2305134068

A jet engine works on the principle of conservation of
NDA Paper 2 2011

(A)

linear momentum

(B)

angular momentum

(C)

energy

(D)

mass

Solution:

A jet engine works on the principle of conservation of linear momentum assuming that in outer space no external forces are acting on it.

Correct Answer is `=>` (A) linear momentum

Q 2305523468

·when a moving bus suddenly applies brakes, the passengers sitting in it fall in the forward direction. This can be explained by.
NDA Paper 2 2011

(A)

the theory of relativity

(B)

Newton's first law

(C)

Newton's second law

(D)

Newton's third law

Solution:

The passengers fall forward when a fast moving bus stops suddenly because the lower part of the bodies of the passengers come into rest along with the bus while the upper part of the bodies due to inertia of motion, continue to move forward. This is in accordance with Newton's, first law of motion which is also known as law of inertia.

Correct Answer is `=>` (B) Newton's first law

Q 2365523465

A man is at rest in the middle of a horizontal plane of perfectly smooth surface of ice. He can move himself to the shore by making use of Newton's.
NDA Paper 2 2011

(A)

first law of motion

(B)

second law of motion

(C)

third law of motion

(D)

first, second and third laws of motion

Solution:

According to Newton's third law of motion, to every action, there is an equal and opposite reaction. By the use of Newton's third law of motion a man at rest in the middle of a smooth surface of ice can move himself to the shore.

Correct Answer is `=>` (C) third law of motion

Q 2314391259

Which one among the following is correct
resultant of balanced forces?
NDA Paper 2 2011

(A)

It is zero

(B)

It is non-zero

(C)

It varies continuously

(D)

None of these

Solution:

When two or more forces are in balance, then the
resultant of them will be zero.

Correct Answer is `=>` (A) It is zero

Q 2374391256

The position-time (`x-t`) graph for motion of a body is given below

Which one among the following is depicted fry the above graph?
NDA Paper 2 2011

(A)

Positive acceleration

(B)

Negative acceleration

(C)

Zero acceleration

(D)

None of these

Solution:

According to the shown graph, equal changes are
observed in distance and time. Hence, change in velocity is zero.

`:.` Acceleration, `a=(Delta v)/(Delta t)=0`

Correct Answer is `=>` (C) Zero acceleration

Q 2334391252

A body is thrown upward against the gravity `g` with initial velocity `u.` Which one among the following is the correct expression for its final velocity when it attains the maximum height?
NDA Paper 2 2011

(A)

`u^2/(2g)`

(B)

`(2g)/u^2`

(C)

`(u^2 g)/2`

(D)

None of these

Solution:

The final velocity will become zero, when the body thrown vertically upward attains its maximum height.

Correct Answer is `=>` (D) None of these

Q 2324391251

If the ratio of the weight of a man in stationary lift and when it is moving downwards with uniform acceleration is `3 : 2`, then the value of `a` is.
NDA Paper 2 2011

(A)

`(3g)/2`

(B)

`g/3`

(C)

`g`

(D)

`(2g)/3`

Solution:

Weight of a man in stationary lift `(w_1) =mg`

Weight of a man in lift moving downward `(w_2 ) = m(g-a)`

According to the question,

`(mg)/(m(g-a))=3/2`

`=> 2g=3g-3a`

`=> a=g/3`

Correct Answer is `=>` (B) `g/3`

Q 2304291158

Momentum of a body is

1. a vector quantity.
2. a conserved quantity in an isolated system.
3. same as force in linear motion.
Select the correct answer using the codes given below

NDA Paper 2 2011

(A)

1 and 3

(B)

2 and 3

(C)

1 and 2

(D)

All of these

Solution:

Momentum (`p = mv`) is a vector quantity and is a
conserved quantity is an isolated system, according to the law of
conservation of linear momentum and angular momentum.
However, it is not same as force, because we have

`F= ma`

and `p =mv`

`=> F =(Delta p)/(Delta t)`

Correct Answer is `=>` (C) 1 and 2

Q 2314056850

An object is in uniform circular motion on a
plane. Suppose that you measure its
displacement from the centre along one
direction, say, along the X-axis. Which one
among the following graphs could represent this
displacement (`x`)?
NDA Paper 2 2012

Solution:

Graph (1) shows uniform circular motion.

Correct Answer is `=>` (A)

Q 2374656556

A staircase has five steps each `10 cm` high and `10 cm` wide. What is the minimum horizontal velocity to be given to the ball, so that it hits directly the lowest plane from the top of the staircase?
NDA Paper 2 2012

(A)

`2 ms^(-1)`

(B)

`1 ms^(-1)`

(C)

`sqrt 2 ms^(-1)`

(D)

`1/2 ms^(-1)`

Solution:

Graph (1) shows uniform circular motion.

Correct Answer is `=>` (C) `sqrt 2 ms^(-1)`

Q 2304545458

A racing car accelerates on a straight line from rest to a speed of 50 m/s in 25 s. Assuming uniform acceleration of the throughout, the distance covered in this will be
NDA Paper 2 2016

(A)

625 m

(B)

1250 m

(C)

2500 m

(D)

50 m

Solution:

Acceleration of the car is uniform. Therefore,
according to first equation of motion
`v=u+at`
where, `v =` final velocity of the car `=50 m//s`
`a =` acceleration of the car
`t =` time interval `= 25 s`
`u =` initial velocity `= 0`

`50 = 0 + a xx 25 => a = 2` ` m//s^2`
Now, applying second equation of motion we get

`s = ut + 1/2 at^2 = 0 + 1/2 at^2`

`=1/2 xx 2 xx 25 xx 25 = 625` m

Correct Answer is `=>` (A) 625 m

Q 2364334255

The displacement of a particle at time t is given
by `x= a hat i+bt hat j+c/2 t^2 hat k`

where `a, b` and `c` are positive constants. Then, the particle is
NDA Paper 2 2013

(A)

accelerated along `hat k` direction

(B)

decelerated along `hat k`-direction

(C)

decelerated along `hat j`-direction

(D)

accelerated along `hat j`-direction

Solution:

The displacement of a particle is given as

`x= a hat i+bt hat j+c/2 t^2 hat k`

By differentiating w.r.t. `x, (dx)/(dt)=0+b hat j+ c that k`

Again differentiating w.r.t. `x, (d^2 x)/(dt^2)= c hat k`

Therefore, the particle is accelerated along `hat k`-direction.

Correct Answer is `=>` (A) accelerated along `hat k` direction

Q 2364523455

A force `F` is applied on a body (which moves on a
straight line) for a duration of `3 s`. The
momentum of the body changes from `10 g cm//s`
to `40 g cm//s`. The magnitude of the force `F` is
NDA Paper 2 2013

(A)

`10` dynes

(B)

`10 N`

(C)

`120` dynes

(D)

`12` dynes

Solution:

It is given that, `p_1 = 10 g cm//s => p_2 = 40 g cm//s`

Time, `t => 3s`

We know that, `F=(p_2-p_1)/t`

`F=(40-10)/3=30/3=10` dynes

`F=10 ` dynes

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

Q 2384423357

If an object undergoes a uniform circular motion, then its
NDA Paper 2 2013

(A)

acceleration remains uniform

(B)

velocity changes

(C)

speed changes

(D)

velocity remains uniform

Solution:

If an object is moving in a circle with a constant speed, then it is accelerating, since, the direction of its velocity changes. This is why you might notice that you are going faster when you are rounding a curve even though you did not step on the accelerator.

Correct Answer is `=>` (B) velocity changes

Q 2364323255

If d denotes the distance covered by a car in time `t` and `s` denotes the displacement by the car during the same time, then
NDA Paper 2 2013

(A)

`d le |s|`

(B)

`d=|s|`

(C)

`d ge |s|`

(D)

`d < |s|`

Solution:

Displacement (s) by the car is never greater than the distance `(d)` covered by a car. This is because displacement is distance between initial and final position. `d ge |s|`

Correct Answer is `=>` (C) `d ge |s|`

Q 2304223158

An ant is moving on thin (negligible thickness) circular wire. How many coordinates do you require to completely describe the motion of the ant?
NDA Paper 2 2013

(A)

one

(B)

Two

(C)

Three

(D)

zero

Solution:

Displacement (s) by the car is never greater than the distance `(d)` covered by a car. This is because displacement is distance between initial and final position. `d ge |s|`

Correct Answer is `=>` (B) Two

Q 2314223150

A car is moving with a uniform speed. However, its momentum is changing. Then the car.
NDA Paper 2 2013

(A)

may be on an elliptical path

(B)

is moving on a straight path without acceleration

(C)

is moving on a straight path with acceleration

(D)

is moving without any acceleration

Solution:

A car is moving with a uniform speed but its momentum changes, it means the velocity changes. Velocity changes with constant speed is possible in a car, when it is moving in a circular path.

Correct Answer is `=>` (A) may be on an elliptical path

Q 2314112959

The motion of a particle is given by a straight line in the graph given above drawn with displacement `(x)` and time `(t)`. Which one among the following statements is correct?
NDA Paper 2 2013

(A)

The velocity of the particle is uniform

(B)

The velocity of the particle is non-uniform

(C)

The speed is uniform and the particle is moving on a circular path

(D)

The speed is non-uniform and the particle is moving on a straight line path

Correct Answer is `=>` (A) The velocity of the particle is uniform

Q 2334412352

An object is undergoing a non-accelerated motion. Its rate of change in momentum is
NDA Paper 2 2013

(A)

a non-zero constant

(B)

zero

(C)

not a constant

(D)

None of these

Solution:

We know that,

Momentum, `p=mv=>(dp)/(dt)=m(dv)/(dt)`

`=> (dp)/(dt)=0`

Correct Answer is `=>` (B) zero

Q 2384101957

A motor vehicle is moving on a circle with a uniform speed. The net acceleration of the vehicle is.
NDA Paper 2 2013

(A)

zero

(B)

towards the centre of the circle

(C)

away from the centre along the radius of the circle

(D)

perpendicular to the radius ancl along the velocity

Solution:

If a body is moving around a circle, even if it is moving
at a constant speed it is accelerating. This is because it is
changing direction (it is not moving in a straight line). The
direction of this acceleration is towards the centre of the circle
and the magnitude is given by `a = v^2 // r, v` is the speed and `r` is
the radius of the circle.

Correct Answer is `=>` (B) towards the centre of the circle

Q 2303691548

Which one among the following situations is best represented by the velocity time plot shown above?
NDA Paper 2 2013

(A)

Uniform motion of a particle on a circle

(B)

Accelerated motion of a particle which has a non-zero initial velocity

(C)

Decelerated motion of a particle which has an initial non-zero velocity

(D)

Decelerated motion of a particle which has no initial velocity

Correct Answer is `=>` (D) Decelerated motion of a particle which has no initial velocity

Q 2303280148

The plot given above represents the velocity of a
particle (in m/s)with time (in seconds) assuming
that the plot represents a semi-circle, distance
traversed by the particle at the end of `7 s` is
approximately
NDA Paper 2 2013

(A)

`19.25 m`

(B)

`7 m`

(C)

`3.2 m`

(D)

`4.75 m`

Solution:

Distance =Area under the graph

`=1/2 xx 22/7 xx (7/2)^2=19.25 m`

Correct Answer is `=>` (A) `19.25 m`

Q 2313178940

A bullet of mass `20 g` is fired in the horizontal direction with a velocity `150 m//s ` from a pistol of mass `1 kg`. Recoil velocity of the pistol is
NDA Paper 2 2013

(A)

`3 m//s`

(B)

`3 km//s`

(C)

`300 m//s`

(D)

`1/3 m//s`

Solution:

`20/1000 xx 150=1 xx V`

`V=3 m//s`

Correct Answer is `=>` (A) `3 m//s`

Q 2373078846

The plot given above represents displacement `x` of a particle with time `t`. The particle is.
NDA Paper 2 2013

(A)

moving with uniform velocity

(B)

moving with acceleration

(C)

moving with deceleration

(D)

executing a periodic motion

Solution:

`20/1000 xx 150=1 xx V`

`V=3 m//s`

Correct Answer is `=>` (B) moving with acceleration

Q 2333078842

Motion of a particle can be described in x-direction by `x = a sin omega t` and y-direction by `y = b cos omega t`. The particle is moving on.
NDA Paper 2 2013

(A)

a circular path of radius `a`

(B)

a circular path of radius `b`

(C)

an elliptical path

(D)

a straight line

Solution:

`sin^2 wt+cos^2wt=x^2/a^2+y^2/b^2=1`

Correct Answer is `=>` (C) an elliptical path

Q 2109134918

The motion of a car along a straight path is shown by the following figure The car starts from `O` and reaches at `A, B` and `C` at different instants of time. During its motion from `O` to `C` and back to `B`, the distance covered and the magnitude of the displacement are, respectively
NDA Paper 2 2016

(A)

`25` km and `60` km

(B)

`95` km and `35` km

(C)

`60` km and `25` km

(D)

`85` km and `35` km

Solution:

During motion from `O` to `C` and back to `B`, distance
covered

`d =` actual path taken `= O C + BC`

`= 60 km+ (60 - 35) km = 85 km`

Similarly displacement

`x =` least distance between `O` and `B`

`= OB = 35 km`

Correct Answer is `=>` (D) `85` km and `35` km

Q 2139734612

The impulse on a particle due to a force acting on it during a given time interval is equal to the change in its
NDA Paper 2 2016

(A)

force

(B)

momentum

(C)

work done

(D)

energy

Solution:

According to Newton's second law of motion

` vec F = (d vec p )/(dt) = m vec a`

where, ` vec F =` applied external force

`m =` mass of the particle

` vec a =` acceleration of the particle

`:.` Impulse `= vec J = vec F*dt = d vec p =` change in momentum

Correct Answer is `=>` (B) momentum

Q 2159401314

The `SI` unit of acceleration is.
NDA Paper 2 2016

(A)

`ms^(-1)`

(B)

`ms^(-2)`

(C)

`cms^(-2)`

(D)

`kms^(-2)`

Solution:

Rate of change in velocity is called acceleration.

`:.` Acceleration, `a = text ( change in velocity) (m // s)// text (time interval) (s)`

Therefore, unit of acceleration is `m//s^(2)`

Correct Answer is `=>` (B) `ms^(-2)`

Q 1819367219

A force F is applied on a body (which
moves on a straight line) for a duration
of 3 s. The momentum of the body
changes from 10 g cm/s to 40 g cm/s.
The magnitude of the force F is
NDA Paper 2 2013

(A)

10 dynes

(B)

10 newtons

(C)

120 dynes

(D)

12 dynes

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

Q 1809267118

The impulse on a particle due to a force
acting on it during a given time interval
is equal to the change in its

(A)

force

(B)

momentum

(C)

work done

(D)

energy

Correct Answer is `=>` (B) momentum

Q 1879167016

A brass ball is tied to a thin wire and
swung so as to move uniformly in a
horizontal circle. Which of the following
statements in this regard is/are true?

1. The ball moves with constant velocity.
2. The ball moves with constant speed.
3. The ball moves with constant
acceleration.
4. The magnitude of the acceleration of
the ball is constant.
Select the correct
code given below

(A)

Only 1

(B)

1 and 3

(C)

1, 2 and 4

(D)

2 and 4

Q 1809856718

A racing car accelerates on a straight
line from rest to a speed of 50 m/s in
25 m. Assuming uniform acceleration
of the throughout, the distance covered
in this time will be

(A)

625 m

(B)

1250 m

(C)

2500 m

(D)

50 m

Correct Answer is `=>` (A) 625 m

Q 1869856715

The motion of a car along a straight
path is shown by the following figure

The car starts from 0 and reaches at A,
8 and C at different instants of time.
During its motion from 0 to C and back
to 8, the distance covered and the
magnitude of the displacement are,
respectively

(A)

25 km and 60 km

(B)

95 km and 35 km

(C)

60 km and 25 km

(D)

85 km and 35 km

Correct Answer is `=>` (D) 85 km and 35 km

Q 1859856714

(A)

25 km and 60 km

(B)

95 km and 35 km

(C)

60 km and 25 km

(D)

85 km and 35 km

Correct Answer is `=>` (D) 85 km and 35 km

Q 1889756617

If the distance S covered by a moving
car in rectilinear motion with a speed v
in time t is given by S = vt, then the car
undergoes

(A)

a uniform acceleration

(B)

a non-uniform acceleration

(C)

a uniform velocity

(D)

a non-uniform velocity

Correct Answer is `=>` (C) a uniform velocity

Q 1879556416

In SI unit of force 'Newton' (N) is given
by (where 'm' stands for 'metre' and 's'
stands for second)

(A)

`1 N 1 `kg/`ms^2`

(B)

`1 N 1 `kgm/`s^2`

(C)

`1 N 1 kgs^2`/m

(D)

`1 N 1` kgm`s^2`

Correct Answer is `=>` (B) `1 N 1 `kgm/`s^2`

Q 1869556415

In SI unit of force 'Newton' (N) is given
by (where 'm' stands for 'metre' and 's'
stands for second)

(A)

`1 N 1 `kg/`ms^2`

(B)

`1 N 1 `kgm/`s^2`

(C)

`1 N 1 kgs^2`/m

(D)

`1 N 1` kgm`s^2`

Correct Answer is `=>` (B) `1 N 1 `kgm/`s^2`

Q 1723412341

If the distance `S` covered by a moving car in rectilinear motion with a speed `v` in time `t` is given by `S = vt`, then the car undergoes
NDA Paper 2 2014

(A)

a uniform acceleration

(B)

a non-uniform acceleration

(C)

a uniform velocity

(D)

a non-uniform velocity

Solution:

If the distance `S` covered by a moving car in rectilinear motion with a speed `v` in time `t` is given by `S = vt`, then the car undergoes a uniform velocity.

Correct Answer is `=>` (C) a uniform velocity

Q 1733101942

A passenger in a moving train tosses a coin upward which falls behind him. It implies that the motion of the train is
NDA Paper 2 2014

(A)

accelerated

(B)

uniform

(C)

retarded

(D)

along the circular tracks

Solution:

If the distance `S` covered by a moving car in rectilinear motion with a speed `v` in time `t` is given by `S = vt`, then the car undergoes a uniform velocity.

Correct Answer is `=>` (A) accelerated

Q 1761167925

The dimension of 'impulse' is the same as that of
NDA Paper 2 2014

(A)

pressure

(B)

angular momentum

(C)

work

(D)

linear momentum

Solution:

Impulse = Change in linear momentum

`F = (Delta P)/(Delta t) => F Delta t = Delta P`

Impulse `= Delta P`

Hence, the dimension of impulse is same as linear

momentum.

Correct Answer is `=>` (D) linear momentum

Q 1751867724

A bullet is fired vertically up from a `400 m` tall tower with a speed `80 m//s`. If g is taken as `10 m//s^2`, the time taken by the bullet to reach the ground will be.
NDA Paper 2 2014

(A)

`8 s`

(B)

`16 s`

(C)

`20 s`

(D)

`24 s`

Solution:

Total time `T = t_1 + t_2 + t_3`

`u = 80 m//s`

Velocity of the bullet at point `A`,

`V_A = 0` m/s,

`a = -10` m/s

`v = u + at`

` 0 = 80 - 10 t_1`

`t_1 = 8s`

` t_2 = t_1` [as there is no air n:Jsistance]

Calculation for `t_3`

Velocity of the bullet at point `B`,

`V_B = u = 80` m/s

` h = 400 m`

Using `s = ut + 1/2 at^ 2`

`=> 400 = 80 xx t_3 + 1/2 xx 10 xx t_3^ 2`

`=> 5 t_3^2 + 80 t_ 3 - 400 = 0`

`=> t_3^2 + 16 t_3 - 80 = 0`

`=> t_3(t_3 + 20)- 4(t_3 + 20) = 0`

`=> t_ 3 = -20 s quad t_3 = 4 s`

Time cannot be negative hence neglecting `t_ 3 = 4 s`

Total time `T = 8 + 8+ 4 = 20 s`

Correct Answer is `=>` (C) `20 s`

Q 1781667527

Two cars `A` and `B` have masses `m_A` and `m_B` respectively, with `m_A > m _B`. Both the cars are moving in the same direction with equal kinetic energy. If equal braking force is applied on both, then before coming to rest
NDA Paper 2 2014

(A)

A will cover a greater distance

(B)

B will cover a greater distance

(C)

both will cover the same distance

(D)

distance covered by them will depend on their respective velocities

Solution:

` => (P_A^2)/(2m_A) = (P_B^2)/(2m_B)`

is linear momentum

When `P => P_A < P_B` given `[m_A > m_B ]`

Now from Newton's second law

` F_(ext) = (Delta P) /(Delta t) => Delta P = F_(ext) Delta t`

`=> Delta P_A < Delta P_B => F_(ext) Deltat_A < F_(ext) Deltat_B`

` => Deltat_A < Deltat_B` [equal braking force is hence on comparing

time `Delta t_A` and `Delta t_B` applied on both the laws] we can say

that car `B` will cover a greater distance.

Correct Answer is `=>` (B) B will cover a greater distance

Q 1771267126

If the motion of an object is represented by a straight line parallel to the time axis in a distance-time graph, then the object undergoes
NDA Paper 2 2014

(A)

an accelerated motion

(B)

a decelerated motion

(C)

a uniform non-zero velocity motion

(D)

a zero velocity motion

Solution:

It is clear from the graph that the position of the object is not changing with the change in time, hence the object is at rest or in other words we can say that it is a zero velocity motion.

Correct Answer is `=>` (D) a zero velocity motion

Q 1780823717

A force applied on a body is represented as

` overset rightarrow(F) = 6 hat i-8 hat j+ 10 hat k`

and accelerates it at `1 m//s^2` . The mass of the body is

NDA Paper 2 2014

(A)

`10 kg`

(B)

`10 sqrt(2) kg`

(C)

`2 sqrt(10) kg`

(D)

`8 kg`

Solution:

` overset rightarrow(F) = 6 hat i-8 hat j+ 10 hat k`

` overset rightarrow |F| = sqrt ( (6)^2 +(-8)^2+ (10)^2)`

` = sqrt (36 + 64 + 100) = 10sqrt(2) N`

` overset rightarrow |a| = 1m//s^2`

using `F = ma`

`10 sqrt(2) = m xx 1`

`=> m = 10 sqrt(2) kg`

Correct Answer is `=>` (B) `10 sqrt(2) kg`

Q 1730612512

A particle is moving in a circular path of radius `r` at `a` constant speed `v`. Which one of the following graphs correctly represents its acceleration `a`?
NDA Paper 2 2014

Solution:

The given situation is uniform circular motion.

In this case, the centripetal acceleration `a_c = v^2/r` as `| overset rightarrow (v) |`

is constant `=> a prop1/r`. Hence, the graph between 'a' and 'r' is

Correct Answer is `=>` (D)

Q 1658245104

The displacement-time graph of a particle acted upon by a constant force is
NDA Paper 2 2015

(A)

a straight line

(B)

a circle

(C)

a parabola

(D)

any curve depending upon initial conditions

Correct Answer is `=>` (D) any curve depending upon initial conditions

Q 1607191088

A man is sitting in a train which is moving with a

velocity of `60` Km/h. His speed with respect to the train

is
NDA Paper 2 2015

(A)

` (10)/3 `m/s

(B)

`60` m/s

(C)

infinite

(D)

zero

Solution:

According to the concept of relative velocity, both the

train and man moves with the velocity of `60` km/h.

So, velocity of man w.r.t. train = velocity of man

velocity of train =` 60 - 60 ` = zero `(0)`

Correct Answer is `=>` (D) zero

Q 1617380289

The following figure represents

the velocity-time graph of a

moving car on a road:

Which segment of the graph

represents the retardation?
NDA Paper 2 2015

(A)

`AB`

(B)

`BC`

(C)

`CD`

(D)

None

Solution:

The segment `BC` represents retardation as the slope of

the `BC` curve is negative.

Correct Answer is `=>` (B) `BC`

Q 1686056877

In `SI` unit of force, 'Newton' `(N)` is given by (where `'m'` stands for 'metre' and `'s'` stands for 'second')
NDA Paper 2 2015

(A)

`1 N = 1 kg//ms^2`

(B)

`1 N = 1 kg-m//s^2`

(C)

`1 N = 1 kg -s^ 2//m`

(D)

`1 N = 1 kg -m s^2`

Solution:

As we know, `F = ma` When mass `(m)` is taken as `1` kg and acceleration `(a)` is taken as `1 m//s^2` the force will be `1 N`. So, `1 N = 1 kg - m//s^2`

Correct Answer is `=>` (B) `1 N = 1 kg-m//s^2`

Q 1676756676

Two forces, one of 3 newton and another of 4 newton are applied on a standard 1 kg body, placed on a horizontal and frictionless surface, simultaneously along the x-axis and the y-axis, respectively, as shown below: The magnitude of the resultant acceleration is
NDA Paper 2 2015

(A)

`7 m//s ^2`

(B)

`1 m//s ^2`

(C)

`5 m//s ^2`

(D)

`sqrt(7) m//s ^2`

Solution:

As two forces are perpendicular to each other, so resultant force is given by `F_text(net) = sqrt(F_1^2 + F_2^2) = sqrt((3)^2 + (4)^2) = sqrt(25) = 5N` Now, from second law of Newton, `F = ma => a = F/m = 5/1 = 5 m// s^2`

Correct Answer is `=>` (C) `5 m//s ^2`

Q 1632712632

A brass ball is tied to a thin wire and swung so as to
move uniformly in a horizontal circle. Which of the
following statements in this regard is/are true?

1. The ball moves with constant velocity.
2. The ball moves with constant speed.
3. The ball moves with constant acceleration.
4. The magnitude of the acceleration of the ball is
constant.

Select the correct answer using the code given below :
NDA Paper 2 2015

(A)

Only 1

(B)

1 and 3

(C)

1, 2 and 4

(D)

2 and 4

Solution:

Since, the direction of the ball changes continuously

during motion in horizontal circle, so the magnitude of

velocity (i.e. speed) and magnitude of acceleration

remains constant. But their direction changes

continuously.

Correct Answer is `=>` (D) 2 and 4

Q 1660280115

The motion of a car along a straight path is shown by the
following figure

The car starts from O and reaches at A, B and C at
different instants of time. During its motion from O to C
and back to B, the distance covered and the magnitude of
the displacement are, respectively
NDA 2016

(A)

25 km and 60 km

(B)

95 km and 35 km

(C)

60 km and 25 km

(D)

85 km and 35 km

Correct Answer is `=>` (D) 85 km and 35 km

Q 1670178916

The impulse on a particle due to a force acting on it

during a given time interval is equal to the change in its
NDA 2016

(A)

force

(B)

momentum

(C)

work done

(D)

energy

Solution:

According to Newton's second law of motion

` underset(F) rightarrow = (d underset(p) rightarrow)/dt = (m underset(a) rightarrow)`

where, ` underset(F) rightarrow` = applied external force

m = mass of the particle

`underset(a) rightarrow` = acceleration of the particle

:. Impulse `= underset(J) rightarrow = underset(F) rightarrow dt = d underset(p) rightarrow =` change in momentum

Correct Answer is `=>` (B) momentum

Q 1680167917

The SI unit of acceleration is
NDA 2016

(A)

`ms^(-1)`

(B)

`ms^(-2)`

(C)

`cms^(-2)`

(D)

`kms^(-2)`

Solution:

Rate of change in velocity is called acceleration.

:. Acceleration, `a = text ( change in velocity) (m // s)// text (time interval) (s)`

Therefore, unit of acceleration is `m//s^(2)`

Correct Answer is `=>` (B) `ms^(-2)`