Physics previous year question Of Oscillation and Waves for NDA

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previous year question Of Oscillation and Waves for NDA

previous year question

Q 2818880709

A fork A has frequency `2% ` more than frequency of standard fork
and B has a frequency 3% less than the frequency of the same
standard fork. The forks A and B produce 6 beats/s. The frequency
of the fork A is

(A)

140 Hz

(B)

122.4 Hz

(C)

110 Hz

(D)

250 Hz

Solution:

Correct Answer is `=>` (B) 122.4 Hz

Q 2888880707

A sonometer string and a tuning fork are sounded together, they
produce 6 beats/s. When the length of string is either 95 em or
100 cm, then the frequency of the tuning fork is

(A)

120 c/s

(B)

100 c/s

(C)

190 c/s

(D)

234 c/s

Solution:

The frequency of tuning fork is `n + 6`
or `n- 6`

Frequency o f sonometer wire, `n ∝ 1/l`

`:. ( n+6)/( n- 6) = 100/95`

`95 (n+6) = 100 ( n- 6)`

`95 n + 95 xx 6 = 100 n - 600`

`600 + 570 = 5n`

`n = 1170/5 => n = 234 c/s`

Correct Answer is `=>` (D) 234 c/s

Q 2818780609

56 tuning forks are so arranged in series that each fork gives
4 beats/s with the previous one . The frequency of the last fork is
3 times that of the first. The frequency of the first fork is

(A)

(B)

(C)

110

(D)

Solution:

Given, `n_1 = 3n , n ;' = n, N =56,`

`x =4 ` beats/s,

`:. n_i = n_f + (N -1) x`

where, x is the number of beats

`3n = n + ( 56 -1) 4`

`2n = 55 xx 4 => n = ( 55 xx 4)/2 => n = 110`

Correct Answer is `=>` (C) 110

Q 2888680507

l'he periodic time of simple pendulum is 2 sin a lab at rest. If
it is suspended from the roof of a lift accelerating upwards with an
acceleration `2 m/s^2`. then its periodic time will he
(Take, `g = 10 rn/s^2` )

(A)

` sqrt ( ( 40/12)) s `

(B)

`sqrt ( (8/40 ) ) s`

(C)

`sqrt ( (12)/(40) )s`

(D)

`sqrt ( ( 40)/( 8) ) `

Solution:

Given, `n_1 = 3n , n ;' = n, N =56,`

`x =4 ` beats/s,

`:. n_i = n_f + (N -1) x`

where, x is the number of beats

`3n = n + ( 56 -1) 4`

`2n = 55 xx 4 => n = ( 55 xx 4)/2 => n = 110`

Correct Answer is `=>` (A) ` sqrt ( ( 40/12)) s `

Q 2818480309

Two sound waves passing through air have their wavelengths in the
ratio 4 : 5. 'l'heir frequencies are in the ratio

(A)

4:5

(B)

3:4

(C)

5:4

(D)

1:1

Solution:

`:. v_1/v_2 = v_1:v_2 = (n-1 λ_1)/(n_2 λ_2) => n_1 λ_1 = n_2 λ_2`

`=> n_1/n_2 = λ_2/λ_1 = 5/4 = 5: 4`

Correct Answer is `=>` (C) 5:4

Q 2808480308

If the length of second's pendulum is increased by 2% how many
seconds will it lose per day?

(A)

3600 s

(B)

3456 s

(C)

1728 s

(D)

864 s

Solution:

`T' = T sqrt ( (l+(2l)/100)/l )`

`=T (1+ 2/100)^1/2 =2 (1+1/100)`

`:. T' -T = 2/100 = 1/50 s`

Therefore, loss in Second's per day

`= (1/50)/2 xx 24 xx 60 xx 60 = 864 s`

Correct Answer is `=>` (D) 864 s

Q 2858480304

The fork A of frequency 100 is sounded with another tuning
fork B. The number of beats produced is 2. On putting some
wax on the prongs of B, the number of beats reduces to 1.
The frequency of the fork B is

(A)

102

(B)

107

(C)

(D)

Solution:

`T' = T sqrt ( (l+(2l)/100)/l )`

`=T (1+ 2/100)^1/2 =2 (1+1/100)`

`:. T' -T = 2/100 = 1/50 s`

Therefore, loss in Second's per day

`= (1/50)/2 xx 24 xx 60 xx 60 = 864 s`

Correct Answer is `=>` (A) 102

Q 2828480301

A given tube is open at both the ends. The shortest length of a
column of air that would vibrate at 331 Hz in such a tube is (the
velocity of sound is 331 m/s)

(A)

0.25 m

(B)

0.5 m

(C)

1 m

(D)

2 m

Solution:

Given,` v = 331 m/s, n =331 Hz, I = ?`

`:.` Fundamental frequency , `n = v/(2l )`

`331 = 331/(2l) => l = 0.5 m`

Correct Answer is `=>` (B) 0.5 m

Q 2868380205

The ratio of the frequencies of the fundamental notes between one
organ pipe open at both ends and another organ pipe of same length
closed at one end is

(A)

(B)

(C)

(D)

Solution:

According to question, ` (v/ (2L) )/( v/(4L)) = 2`

Correct Answer is `=>` (C) 2

Q 2838380202

Jf the frequency of a note emitted by a source changes by 20% as it
approaches an observer and then recedes away from him, the speed
of the source is nearly

(A)

22 m/s

(B)

11 m/s

(C)

33 m/s

(D)

44 m/s

Solution:

Change in apparent frequency

` n'' - n ' = ( 2 n v_s)/v`

`20/100 n = (2 n v_s )/(v)`

`v_s = (20 xx v)/(2 xx 100) = (20 xx 330 )/( 2 xx 100)`

[given, `v = 330 m/s `]

`v_s = 33 m/s`

Correct Answer is `=>` (C) 33 m/s

Q 2888280107

The earth is moving towards a stationary star with a velocity of
100 km/s. If the wavelength of the light emitted by the star is
5000 Å, then the change in wavelength apparent to an
obsever will be

(A)

167Å

(B)

zero

(C)

16.7 Å

(D)

1.67 Å

Solution:

Given, ` λ = 5000 Å , v = 100 ` km/s ,

`c = 3 xx 10^8 m/s , Delta λ = ?`

`:. Delta λ = λ v/c = 5000 xx (100 xx 1000)/( 3 xx 10^8 ) xx 10^(-10)`

`= 16.7 xx 10^(-10) m = 1.67 Å `

Correct Answer is `=>` (D) 1.67 Å

Q 2808178908

The wavelength of light received from a galaxy is 4% greater than
that received from an identical source on the earth. The velocity
of the galaxy relative to the earth is

(A)

`12 xx 10^6 m/s`

(B)

`8 xx 10^5 m/s`

(C)

`1.2 xx 10^6 m/s`

(D)

`7.5 xx 10^6 m/s`

Solution:

`:. (Delta λ )/( λ) = v/c => v = c/λ Delta λ (λ' - λ)`

`= (3 xx 10^8 xx 4)/100 = 12 xx 10^6 m/s`

Correct Answer is `=>` (A) `12 xx 10^6 m/s`

Q 2878178906

The apparent frequency noted by a moving listener away from the
stationary source is 10% less than the real frequency. If the velocity
of sound is 330 m/s, then the velocity of the listener is

(A)

8.5 m/s

(B)

40 m/s

(C)

50 m/s

(D)

16.5 m/s

Solution:

`:. ` Change in apparent frequency

`n'' - n' = (2 n v_0 )/v`

`10/100 n = (2 n v_0 )/v => (10 xx v)/(2 xx 100)= (10 xx 330 )/( 2 xx 100 )`

`[ :. ` given , ` v = 330 m/s` ]

`= 16.5 m/s`

Correct Answer is `=>` (D) 16.5 m/s

Q 2848178903

Consider the following

statements. In a stationary wave

I. all the particles perform simple harmonic motion with a
frequency which is four times that of the two component
waves.

II. particles on the opposite sides of a node vibrate with a phase
difference of `pi`

III. the amplitude of vibration of a particle at an antinode is
equal to that of either component wave.

IV. all the particles between two adjacent nodes vibrate in phase.
Of these statements

(A)

Both I and II are correct

(B)

I, Ill and IV are correct

(C)

Both II and IV are correct

(D)

I, II, Ill and IV are correct

Correct Answer is `=>` (A) Both I and II are correct

Q 2828178901

There are two strings of equal length and diameter but the
densities are in the ratio 1 : 2, they are stretched by a tension T.
The ratio of frequencies will be

(A)

2:1

(B)

1:2

(C)

`1 : sqrt 2`

(D)

`sqrt 2 : 1`

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

Q 2848078803

Two simple pendulums have the same period of oscillation. The
necessary condition for this is

(A)

t11eir lengths are equal and the suspencJed particles have the same mass

(B)

their lengths me equal but the suspended pmticles need not have ths same rna!;s

(C)

their lengths me different but the suspended p;; rticles have the same mass

(D)

the masses o· the suspended particle must Je in the inverse ratio , of the lengths of the pendulums

Correct Answer is `=>` (B) their lengths me equal but the suspended pmticles need not have ths same rna!;s

Q 2818078800

For a particle executing SHM, the equilibrium position is at `x = 0` and
the amplitude at `x = 0` and amplitude at `x = A`, the KE of the
particle will be equal to the potential energy

(A)

at x =0

(B)

at x =A

(C)

at x= A/2

(D)

when x is greater than `A/2` but less than A

Correct Answer is `=>` (D) when x is greater than `A/2` but less than A

Q 2878878706

The ratio of the fundamental frequency of an open and closed
organ pipe of same length is

(A)

2:4

(B)

2:1

(C)

3:1

(D)

1:1

Correct Answer is `=>` (A) 2:4

Q 2858878704

A body executing simple harmonic motion while passing through its
mean position will have

(A)

kinetic energy only

(B)

potential enrergy only

(C)

Both kinetic and potential energies

(D)

minimum acceleration

Correct Answer is `=>` (D) minimum acceleration

Q 2848878703

A body executing simple harmonic motion while passing through its
mean position will have

(A)

kinetic energy only

(B)

potential enrergy only

(C)

Both kinetic and potential energies

(D)

minimum acceleration

Correct Answer is `=>` (D) minimum acceleration

Q 2838878702

A body executing simple harmonic motion while passing through its
mean position will have

(A)

kinetic energy only

(B)

potential enrergy only

(C)

Both kinetic and potential energies

(D)

minimum acceleration

Correct Answer is `=>` (D) minimum acceleration

Q 2828878701

A body executing simple harmonic motion while passing through its
mean position will have

(A)

kinetic energy only

(B)

potential enrergy only

(C)

Both kinetic and potential energies

(D)

minimum acceleration

Correct Answer is `=>` (D) minimum acceleration

Q 2818778609

A source of sound is moving away from the staionary observer with
a velocity `v_s` ,. Velocity of sound is v. If n is the frequency of the source
of sound. then the apparent frequency of sound heard by the
observer is

(A)

`v/(v+ v_s ) - n`

(B)

`v/(v+ v_s ) n`

(C)

`v/(v-v_s) n`

(D)

` ( (v+v_s)/v ) n `

Correct Answer is `=>` (B) `v/(v+ v_s ) n`

Q 2808678508

(A)

`v/(v+ v_s ) - n`

(B)

`v/(v+ v_s ) n`

(C)

`v/(v-v_s) n`

(D)

` ( (v+v_s)/v ) n `

Correct Answer is `=>` (B) `v/(v+ v_s ) n`

Q 2818578409

When a wire of a sitar is plucked, then the waves produced in air
will be

(A)

stationary waves

(B)

transverse waves

(C)

longitudinal waves

(D)

a combination of transverse and stationary wave

Correct Answer is `=>` (D) a combination of transverse and stationary wave

Q 2878578406

The pitch of sound depends upon

(A)

frequency and amplitude

(B)

frequency alone

(C)

amplitude alone

(D)

the difference in frequencies from two sources

Correct Answer is `=>` (A) frequency and amplitude

Q 2848578403

To produce sound it is necessary that

(A)

the source should execute longitudinal vibrations

(B)

the source should execute transverse vibrations

(C)

the source may execute any type of vibration

(D)

the vibrations of source are not necessary

Correct Answer is `=>` (C) the source may execute any type of vibration

Q 2828578401

Sound travels in gases in the form of

(A)

longitudinal waves only

(B)

longitudinal waves only

(C)

longitudinal as well as transverse waves

(D)

stationary waves only

Correct Answer is `=>` (A) longitudinal waves only

Q 2888478307

A slow-running pendulum clock can be speeded up by

(A)

increasing the length of the rod

(B)

increasing the weight of the bob

(C)

reducing tile length of the rod

(D)

reducing the weight of the bob

Correct Answer is `=>` (C) reducing tile length of the rod

Q 2838478302

Which among the following is the necessary condition for simple
harmonic motion?

(A)

Constant period

(B)

Constant acceleration

(C)

Displacement and acceleration are proportional

(D)

Displacement and torque are proportional

Correct Answer is `=>` (C) Displacement and acceleration are proportional

Q 2818478300

The approximate intensity level of the sound which can cause
damage to the ear drum is

(A)

20dB

(B)

60 dB

(C)

100 dB

(D)

160 dB

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

Q 2888378207

The total energy of a particle executing simple harmonic motion
is proportional to the

(A)

amplitude of the motion

(B)

square of the amplitude of the motion

(C)

cube of the amplitude of the motion

(D)

square of the acceleration of the body

Correct Answer is `=>` (B) square of the amplitude of the motion

Q 2838378202

The total energy of a particle executing simple harmonic motion
is proportional to the

(A)

amplitude of the motion

(B)

square of the amplitude of the motion

(C)

cube of the amplitude of the motion

(D)

square of the acceleration of the body

Correct Answer is `=>` (B) square of the amplitude of the motion

Q 2784112057

The following figure shows displacement versus time curve for a particle executing simple-harmonic motion :

Which one of the following statements is correct?
NDA Paper 2 2017

(A)

Phase of the oscillating particles same at `t = 1 s` and `t = 3 s`

(B)

Phase of the oscillating particle is same at `t = 2 s` and `t = 8 s`

(C)

Phase of the oscillating particle is same at `t = 3 s` and `t = 7 s`

(D)

Phase of the oscillating particle is same at `t = 4 s` and `t = 10 s`

Correct Answer is `=>` (C) Phase of the oscillating particle is same at `t = 3 s` and `t = 7 s`

Q 2734801752

The time period of a simple pendulum made using a thin copper wire of length `L` is `T`. Suppose the temperature of the room in which this simple pendulum is placed increases by `30^0C`, what will be the effect on the time period of the pendulum ?
NDA Paper 2 2017

(A)

T will increase slightly

(B)

T will remain the same

(C)

T will decrease slightly

(D)

T will become more than 2 times

Correct Answer is `=>` (A) T will increase slightly

Q 2714501450

Which one of the following is the correct relation between frequency `f` and angular frequency `omega`?
NDA Paper 2 2017

(A)

`f= pi omega`

(B)

`omega=2 pi f`

(C)

`f= 2 omega //pi`

(D)

`f= 2 pi omega`

Correct Answer is `=>` (B) `omega=2 pi f`

Q 2734101052

The speed of a body that has Mach number more than 1 is
NDA Paper 2 2017

(A)

supersonic

(B)

subsonic

(C)

300 m/s

(D)

about 10 m/s

Correct Answer is `=>` (A) supersonic

Q 2346391273

How does time period (`T`) of a seconds pendulum vary with length (l)?
NDA Paper 2 2007

(A)

`T prop sqrt l`

(B)

`T prop l^2`

(C)

`T prop l`

(D)

`T` does not depend on ` l`

Solution:

A second's pendulum is a pendulum whose time
period is two seconds. Its length is `99.3 cm`.

Time period `T=2 pi sqrt(l/g)`

or `T prop sqrt l`

But length of second's pendulum is a constant equal to `99.3 crn`.

Hence, `T` does not depend on `l`.

Correct Answer is `=>` (D) `T` does not depend on ` l`

Q 2326391271

Consider the following statements Sound waves can undergo
1. reflection
2. refraction
3. interference
'Which of the statement given above is/are correct?
NDA Paper 2 2007

(A)

1 and 2

(B)

2 and 3

(C)

1 and 3

(D)

1, 2 and 3

Solution:

Sound waves can undergo reflection, refraction and
interference.

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

Q 2336291172

Consider the following statements If the same note is played on a flute and a sitar, one can still distinguish between them because they different.

I. frequency
2. intensity
3. quality
Which of the statements given above is/are correct?
NDA Paper 2 2007

(A)

1 and 2

(B)

2 and 3

(C)

Only 3

(D)

Only 2

Solution:

The difference in music from a flute and a sitar is due to
quality difference.

Correct Answer is `=>` (C) Only 3

Q 2326780671

The time period of a simple pendulum oscillating in a laboratory at north pole is `4 s.` Accounting for earth's rotation only, what will be the time period of this pendulum oscillating in a laboratory at equator?
NDA Paper 2 2007

(A)

Less than 4 s

(B)

More than 4 s

(C)

Equal to 4 s

(D)

Infinity

Solution:

The time period of a simple pendulum, `T = 2 pi sqrt(l/g)`

i.e., `T prop 1/(sqrt g)`

The value of `g` is less at equator than at poles, hence time period
of pendulum at equator is greater than that at poles.

Correct Answer is `=>` (B) More than 4 s

Q 2376778676

Some common mediums in which speed of sound waves is measured and mentioned below
1. Air
2. Steel
3. Copper
4. Water
What is the correct increasing order of the speed of sound?
NDA Paper 2 2008

(A)

`1 < 4 < 2 < 3`

(B)

`4 < 1 < 2 < 3`

(C)

`1 < 4 < 3 < 2`

(D)

`4 < 1 < 3 < 2`

Solution:

Speed of sound is maximum in solids and minimum in gases since solids are most elastic and gases are least elastic, therefore,

`V_1 < V_4 < V_3 < V_2`

Correct Answer is `=>` (C) `1 < 4 < 3 < 2`

Q 2316678570

A man standing between two parallel hills fires a gun and hears two echoes, one `2.5 s` and the other `3.5 s` after the firing. If the velocity of sound is `330 ms^(-1)` , how long will it take him to hear the third echo?
NDA Paper 2 2008

(A)

`4 s`

(B)

`5 s`

(C)

`6 s`

(D)

`8 s`

Solution:

The time taken to hear 3rd echo `= t_ 1 + t_ 2`
`=2.5+3.5=6.0 s`

Correct Answer is `=>` (B) `5 s`

Q 2316167970

If the length of second's pendulum is increased by `2%,` how many seconds will it loses per day?
NDA Paper 2 2008

(A)

3600 s

(B)

3456 s

(C)

1728 s

(D)

864 s

Solution:

A simple pendulum whose time period is `2 s` is known as second's pendulum. Length of seconds pendulum is `1 m` or `100 cm.`

The time period of second's pendulum

`T=2 pi sqrt(l/g)`

`(dT)/T=1/2 (dl)/l`

`=1/2 xx 2/100=1/100`

`dT=1/100 xx T=1/100 xx 24 xx 60xx60`

`=844 s`

Therefore, it loses `864 s` per day.

Correct Answer is `=>` (D) 864 s

Q 2306656578

A pendulum clock is set to give correct time at the sea level. The clock is moved to a hill station at an altitude `h` above sea level. In order to keep correct time on the hill station which one of the following adjustments is required?
NDA Paper 2 2008

(A)

Tile length of the pendulum has to be reduced

(B)

The length of the pendulum has to be increased

(C)

The mass of the pendulum has to be increased

(D)

The mass of the pendulum has to be reduced

Solution:

The time period of a simple pendulum `T=2 pi sqrt(l/g)` Thus, the time period of a simple pendulum depends on
(i) length of the pendulum and
(ii) the acceleration due to gravity.
As we go higher above the sea level value of g decreases and hence value of `T` increases. Therefore to keep the correct time, the length of the pendulum has to be reduced.

Correct Answer is `=>` (B) The length of the pendulum has to be increased

Q 2346656573

The simple harmonic motion of a particle is given by `y = 3sin omega t +4cos omega t`. Which one of the following is the amplitude of such motion?
NDA Paper 2 2008

(A)

(B)

(C)

(D)

Solution:

The equation of SHM of particle

`y = 3 sin omega t + 4 cos omega t`

Here, `a_1 = 3` and `b_1 = 4`

`:.` Amplitude, `r =sqrt(a_1^2+ b_1^2) = sqrt(3^2 + 4^2) = sqrt(25) = 5`

Correct Answer is `=>` (B) 5

Q 2316134979

A manometer wire having a length of 50 cm is vibrating in the fundamental mode with a frequency of 100 Hz. Which of the following is the type of propagating wave and its speed?
NDA Paper 2 2009

(A)

Longitudinal, 50 m/s

(B)

Transverse, 50 m/s

(C)

Longitudinal, 100 m/s

(D)

Transverse, 100 m/s

Solution:

For fundamental mode of vibration of sonometer

`V= sqrt(T/m)= 2l xx n`

`=2 xx 50 xx 10^(-2) xx 100`

`=100 m//s`

Correct Answer is `=>` (D) Transverse, 100 m/s

Q 2356034874

The standing wave pattern along a string of length `60 ` cm is shown in the above diagram. If the speed of the transverse waves on this string is 300 m/s, in which one of the following modes is the string vibrating?
NDA Paper 2 2009

(A)

Fundamental

(B)

First overtone

(C)

Second overtone

(D)

Third overtone

Solution:

From the diagram, we can observe that `l=(3 lambda)/2`
The vibration is in the third normal mode, i.e., second overtone or
third harmonic.

Correct Answer is `=>` (C) Second overtone

Q 2316434379

Which among the following is the necessary condition for simple harmonic motion?
NDA Paper 2 2009

(A)

Constant period

(B)

Constant acceleration

(C)

Displacement and acceleration are proportional

(D)

Displacement and torque are proportional

Solution:

Acceleration in SHM is always directed towards the mean position and so is always opposite to displacement i.e., acceleration `prop` displacement.

Correct Answer is `=>` (C) Displacement and acceleration are proportional

Q 2306612578

The audible frequency range of a human ear is
NDA Paper 2 2009

(A)

`20` Hz to `200` Hz

(B)

`2` Hz to `20` Hz

(C)

`200` Hz to `2000` Hz

(D)

`20` Hz to `20000` Hz

Solution:

Sound waves having frequency greater than `20` Hz but less than `20000` Hz are sonic or audible sound. Human ear can perceive only this range.

Correct Answer is `=>` (D) `20` Hz to `20000` Hz

Q 2325591461

When a mass `m ` is hung on a spring, the spring stretched by `6 cm`. If the loaded spring is pulled downward a little and released, then the period of vibration of the system will be
NDA Paper 2 2009

(A)

`0.27 s`

(B)

`0.35 s`

(C)

`0.49 s`

(D)

`0.64 s`

Solution:

The situation is shown alongside. When the loaded spring is pulled downwards a little and released, the system moves to and fro by distance `x cm`. Hence, time period is given by

`T=2 pi sqrt(x/g)`

Here, `x = 6 cm = 6 xx 10^(-2) g = 9.8 ms^(-2)` and

`pi= 3.14`

`:. T=2 xx 3.14 sqrt((6 xx 10^(-2))/9.8)=0.49 s`

Correct Answer is `=>` (C) `0.49 s`

Q 2355880764

When a body moves with simple harmonic motion, then the phase difference between the velocity and the acceleration is
NDA Paper 2 2010

(A)

`0^o`

(B)

`90^o`

(C)

`180^o`

(D)

`270^o`

Solution:

The velocity and acceleration for a SHM are given as

`v=A omega sin(omega t+phi+pi/2)`

and `a = A omega^2 sin (wt + phi + pi)`

The graph between velocity and acceleration is an ellipse shown as Acceleration leads the velocity by a phase angle `pi/2`

Correct Answer is `=>` (B) `90^o`

Q 2365680565

A particle oscillates in one dimension about the equilibrium position subject to a force `F_x (x)` that has an associated potential energy `U(x)`. If k is the force constant, which one of the following relations is true?
NDA Paper 2 2010

(A)

`F_x(x)=-kx^2`

(B)

`F_x(x)=-kx`

(C)

`U(x)=1/2 kx`

(D)

`U(x)=1/2 k^2 x`

Solution:

If a particle executing SHM has a displacement `x` from its equilibrium position, at an instant, the magnitude of the restoring `F_x (x)` acting on the particle at that instant is given by

`F_x(x) =- kx`

where k is known as force constant. The negative sign shows that the restoring force `F_x(x)` is always directed towards the mean
position.

Correct Answer is `=>` (B) `F_x(x)=-kx`

Q 2345078863

For a simple pendulum in simple harmonic motion, which of the following statements is/are correct'?

l. The kinetic energy is maximum at the mean position.
2. The potential energy is maximum at the mean position.
3. Acceleration is maximum at the mean position.
Select the correct answer using the codes given below
NDA Paper 2 2010

(A)

Only 1

(B)

Only 2

(C)

1 and 3

(D)

2 and 3

Solution:

For a simple pendulum in simple harmonic motion, the velocity at the mean position is maximum and minimum(= zero) at extreme positions. Due to maximum velocity the kinetic energy at mean position is maximum. At mean position, the potential energy is zero and acceleration is also zero.

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

Q 2385867767

A pendulum beats faster than a standard pendulum. In order to bring it to the standard beat, the length of the pendulum is to be
NDA Paper 2 2010

(A)

reduced

(B)

increased

(C)

reduced and the mass of the bob increased

(D)

reduced and also the mass of the bob reduced

Solution:

The time period of a simple pendulum, `T = 2 pi sqrt(l/g)`

i.e., `T prop sqrt l`

To reduce the time period of the faster pendulum, its length has
to be reduced.

Correct Answer is `=>` (A) reduced

Q 2305245168

Sound moves with higher velocity, if
NDA Paper 2 2010

(A)

pressure of the medium is decreased

(B)

temperature of the medium is increased

(C)

humidity of the medium is increased

(D)

Both '2' and '3'

Solution:

Velocity of sound in air, `v = sqrt(E/rho)`

Under isothermal conditions, `E_(iso) = p =` atmospheric pressure.

Hence, `v=sqrt(P/rho)`

But, be Laplace correction, `v = sqrt(P/rho)`

But `P/rho=(RT)/M` (By ideal gas equation)

So, velocity of sound `v=sqrt((gamma RT)/M)`

Thus, velocity of sound in a gas is
(i) directly proportional to `sqrt T` i.e., velocity of sound increases
with increase in tempmature.

(ii) inversely proportional to the square root of its density. So
with rise in humidity, density of air decreases and hence
velocity of sound ·increases.

Correct Answer is `=>` (D) Both '2' and '3'

Q 2315423369

A jet plane flies through air with a velocity of `2` Mach. While the velocity of sound is `332 m//s`, the air speed of the plane is.
NDA Paper 2 2011

(A)

`166 m//s`

(B)

`66.4 m//s`

(C)

`332 m//s`

(D)

`664 m//s`

Solution:

Mach number `= (text (Speed of body in any medium))/(text(Speed of sound in that medium))`

In the given question `2 = (text (Velocity of jet plan in medium))/332`

`=>` Velocity of jet plane in air `= 332 xx 2 = 664 m//s`

Correct Answer is `=>` (D) `664 m//s`

Q 2325123061

Which one among the following statements is not correct'?
NDA Paper 2 2011

(A)

In progressive waves, the amplitude may be constant and neighbouring points are out of phase vtith each other

(B)

In air or other gases, a progressive antinode occurs at a displacement node and a progressive node occurs at a displacement antinode

(C)

Transverse wave can be polarised while longitudinal wave can not be polarised

(D)

Longitudinal wave can be polarised while transverse wave can not be polarised

Solution:

Progressive waves are the waves which when travels in a medium, then all the particles of the medium vibrate in the same way i.e., amplitude of the different particles of the meclium is same but the phase of the particles of the medium continuously varies. In air or other gases, a progressive antinode occurs at a displacement node and a progressive node occurs at a displacement antinode. Transverse waves can be polarized while longitudinal waves cannot be polarised.

Correct Answer is `=>` (D) Longitudinal wave can be polarised while transverse wave can not be polarised

Q 2344391253

Ultrasonic waves are those sound waves having frequency.
NDA Paper 2 2011

(A)

between 20 Hz and 1000 Hz

(B)

between 1000 Hz and 20000 Hz

(C)

more than 20 kHz

(D)

less than 20 Hz

Solution:

Frequency of infrasonic waves is in between 20 Hz
and 20000 Hz.
• Frequency of ultrasonic waves is more than 20000 Hz, i.e.,
20kHz.
• Ultrasonic waves are used in sonography, cracks' detection in
metals etc.

Correct Answer is `=>` (C) more than 20 kHz

Q 2304180958

It is impossible for two oscillators, each executing simple harmonic motion, to remain in phase with each other if they have different
NDA Paper 2 2011

(A)

time periods

(B)

amplitude

(C)

spring constants

(D)

kinetic energy

Solution:

Equations of SHM is

`x=a sin omega t`

If there are two SHM's taking place, then

`x_1 = a_1 sin omega_1t_1`

and `x_ 2 = a_2 sin omega_2 t_ 2`

If the two SHM's are in phase, then

`(x_1 -x_2)=a(sin omega+1t_1 -sin omega_2 t_2 )`

`=2a sin omega' t'`

which is an another SHM.

The above equation is true only if

`a_1 = a_2`

i.e., amplitudes are equal.

Correct Answer is `=>` (B) amplitude

Q 2344580453

Which one among the following is not produced by sound waves in air'?
NDA Paper 2 2012

(A)

Polarisation

(B)

Diffraction

(C)

Reflection

(D)

Refraction

Solution:

Sound waves not produces the polarisation phenomenon.

Correct Answer is `=>` (A) Polarisation

Q 2314734659

For a simple pendulum, the graph between `T_2` and `L` (where, `T` is the time penod and `L` is the length) is
NDA Paper 2 2012

(A)

straight line passing through origin

(B)

parabolic

(C)

circle

(D)

None of the above

Solution:

`T^2 prop e`

Correct Answer is `=>` (A) straight line passing through origin

Q 2334634552

SONAR is mostly used by
NDA Paper 2 2012

(A)

doctors

(B)

engineers

(C)

astronauts

(D)

navigators

Solution:

SONAR (originally an acronym for Sound Navigation and Ranging) is a technique that uses sound propagation (usually underwater, as in submarine navigation) to navigate, communicate with or detect objects on or under the surface of the water, such as other vessels.

Correct Answer is `=>` (D) navigators

Q 2314101950

The displacement of a particle is given by `x = cos^2 omega t`. The motion is
NDA Paper 2 2013

(A)

simpie harmonic

(B)

periodic but not simple harmonic

(C)

non-periodic

(D)

None of the above

Solution:

The displacement of a particle is given by

`x=cos^2 omega t`

`x=1/2 (1+cos^2 omega t)`

By differentiating w.r.t.x

`(dx)/(dt)=1/2(0-2 sin 2 omega t)`

`(dx)/(dt)=-sin 2 omega t`

By again differentiating w.r.t. x

`(d^2x)/(dt^2)=-2 sos 2 omega t`

`=-2(2 cos^2 omega t-1)`

`=- 2 (2x -1)`

`(d^x)/(dt^2)=-4x+2`

This is not the equation of simple harmonic motion.
This is because, we know that SHM equation is

`(d^x)/(dt^2)+k/m x=0`

Our equation, `(d^x)/(dt^2)+4x=2`

It is not matched with SHM equation.
So, `x = cos^2 omega t` is periodic, but not simple harmonic.

Correct Answer is `=>` (B) periodic but not simple harmonic

Q 2344301253

If the length of a simple pendulum is being increased by 4-fold, time period of oscillation will be.
NDA Paper 2 2013

(A)

decreased by 4-fold

(B)

increased by 4-fold

(C)

decreased to half of the initial value

(D)

increased by a factor of 2 of its initial value

Solution:

`T prop sqrt l`

Correct Answer is `=>` (D) increased by a factor of 2 of its initial value

Q 2353191044

Bats can know about their prey at a distance even in the night by emitting.
NDA Paper 2 2013

(A)

infra-red lights

(B)

ultraviolet lights

(C)

chemicals from their body

(D)

ultrasonic sounds

Solution:

Bats can know about their prey at a distance even in the night by emitting ultrasonic sounds

Correct Answer is `=>` (D) ultrasonic sounds

Q 2333580442

The ceilings of a concert hall are generally curved
NDA Paper 2 2013

(A)

because they reflect the sound to the audience

(B)

because they can absorb noise

(C)

to have better aeration in the hall

(D)

as any sound from outside cannot pass through a curved ceiling

Solution:

The ceilings of a concert hall are generally curved. This
is because sound waves transmitted by the source in big halls is
absorbed by the walls, floor, seats and even by the clothes of the
crowd sitting inside the hall.

Hence, the ceiling of halls are curved to reduce this problem.
Sound gets reflected by the curved shape of the ceiling, so that
sound reaches to the every points in the hall uniformity.

Correct Answer is `=>` (A) because they reflect the sound to the audience