Waves — Page 9 | NEET Physics

Q81

The displacement

y

of a particle in a medium can be expressed as,

y = {10^{ - 6}}\,\sin

\left( {100t + 20x + {\pi \over 4}} \right)

m

where

t

is in second and

x

in meter. The speed of the wave is

A

20\,\,m/s

B

5\,m/s

C

2000\,m/s

D

5\,\pi \,m/s

Correct Answer

Option B

Solution

From equation given,

\omega = 100

and

k = 20,

v = {\omega \over k} = {{100} \over {20}} = 5m/s

Q82

If a wave gets refracted into a denser medium, then which of the following is true?

A wavelength, speed and frequency decreases.

B wavelength increases, sped decreases and frequency remains constant.

C wavelength and speed decreases but frequency remains constant.

D wavelength, speed and frequency increases.

Correct Answer

Option C

Solution

Frequency is independent of medium. For denser medium, wavelength and speed both would decrease.

Q83

A pipe of length

85

cm

is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below

1250

Hz

. The velocity of sound in air is

340

m/s

.

A

12

B

8

C

6

D

4

Correct Answer

Option C

Solution

Length of pipe

=85

cm

=0.85m

Pipe is closed from one end so it behaves as a closed organ pipe Frequency of oscillations of air column in closed organ pipe is given by,

f = {{\left( {2n - 1} \right)\upsilon } \over {4L}}

f = {{\left( {2n - 1} \right)\upsilon } \over {4L}} \le 1250

\Rightarrow {{\left( {2n - 1} \right) \times 340} \over {0.85 \times 4}} \le 1250

\Rightarrow 2n - 1 \le 12.5 \approx 6

Possible value of n = 1, 2, 3, 4, 5, 6 So, number of possible natural frequencies lie below 1250 Hz is 6.

Q84

While measuring the speed of sound by performing a resonance column experiment, a student gets the first resonance condition at a column length of

18

cm

during winter. Repeating the same experiment during summer, she measures the column length to be

x

cm

for the second resonance. Then

A

18 > x

B

x > 54

C

54 > x > 36

D

36 > x > 18

Correct Answer

Option B

Solution

For first resonant length

v = {v \over {4{\ell _1}}} = {v \over {4 \times 18}}

(in winter) For second resonant length

v' = {{3v'} \over {4{\ell _2}}} = {{3v'} \over {4x}}

(in summer) $\therefore$

{v \over {4 \times 18}} = {{3v'} \over {4 \times x}}

$\therefore$

x = 3 \times 18 \times {{v'} \over v}

$\therefore$

x = 54 \times {{v'} \over v}cm

v' > v

because velocity of light is greater in summer as compared to winter

\left( {v \propto \sqrt T } \right)

$\therefore$

x > 54\,cm

Q85

Assume that the displacement(s) of air is proportional to the pressure difference (

\Delta

p) created by a sound wave. Displacement (s) further depends on the speed of sound (v), density of air (

\rho

) and the frequency (f). If

\Delta

p ~ 10 Pa, v ~ 300 m/s,

\rho

~ 1 kg/m3 and f ~ 1000 Hz, then s will be of the order of (take the multiplicative constant to be 1) :

A 1 mm

B

{3 \over {100}}

mm

C 10 mm

D

{1 \over {10}}

mm

Correct Answer

Option B

Solution

Given, S $\propto$

\Delta

p and Proportionally constant = 1 We know,

\Delta

p = S $\beta$ k = $\rho$ v2 $\times$

{\omega \over v} \times S

=

{\rho v\omega S}

$\therefore$ S =

{{\Delta p} \over {\rho v\omega }}

=

{{\Delta p} \over {\rho v2\pi f}}

=

{{\Delta p} \over {\rho vf}}

[As Proportionally constant = 1 so assume 2 $\pi$ = 1]

= {{10} \over {1 \times 300 \times 1000}}

= {1 \over {30}}mm

\approx {3 \over {100}}mm

Q86

Two sitar strings, A and B, playing the note 'Dha' are slightly out of tune and produce beats of frequency 5 Hz. The tension of the string B s slightly increased and the beat frequency is found to decrease by 3Hz. If the frequency of A is 425 Hz, the original frequency of B is :

A 430 Hz

B 420 Hz

C 428 Hz

D 422 Hz

Correct Answer

Option B

Solution

Frequency of B, fB = 425 $\pm$ 5 = 420 or 430 Hz As tension of string B is increased So, frequency of B, fB should also increase [as f $\propto$

\sqrt T

] If initially fB = 430 Hz then when fB increases by increasing the tension then fB $-$ fA increases that means beat frequency increase.

So, fB can't be 430 Hz When fB = 420 then when fB increases fA $-$ fB decreases means beat frequency decreases.

So, correct fB = 420 Hz.

Q87

An observer moves towards a stationary source of sound with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be :

A 20%

B 10%

C 5%

D 0%

Correct Answer

Option A

Solution

f' = {f_0}\left[ {{{v - {v_0}} \over {v - {v_s}}}} \right]

\Rightarrow f' = {f_0}\left[ {{{v + {v \over 5}} \over v}} \right]

\Rightarrow f' = {{6{f_0}} \over 5}

$\Rightarrow$ % change = 20

Q88

Two engines pass each other moving in opposite directions with uniform speed of 30 m/s. One of them is blowing a whistle of frequency 540 Hz. Calculate the frequency heard by driver of second engine before they pass each other. Speed of sound is 330 m/sec :

A 450 Hz

B 540 Hz

C 648 Hz

D 270 Hz

Correct Answer

Option C

Solution

Frequency heard by the driver of second engine, F' =

\left( {{{v + {v_0}} \over {v - {v_S}}}} \right)

f given v0 = vs = 30 m/s $\therefore$ f' =

\left( {{{330 + 30} \over {330 - 30}}} \right) \times 540

= 648 Hz

Q89

A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7

\times

103 kg/m3 and its Young’s modulus is 9.27

\times

1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?

A 7.5 kHz

B 5 kHz

C 2.5 kHz

D 10 kHz

Correct Answer

Option B

Solution

As rod length = 60 cm

\therefore\,\,\,

{\lambda \over 2}

= 60 $\Rightarrow$

\,\,\,

$\lambda$ = 120 cm = 1.2 m In solid, velocity of wave, V =

\sqrt {{Y \over \rho }}

=

\sqrt {{{9.27 \times {{10}^{10}}} \over {2.7 \times {{10}^3}}}}

= 5.85 $\times$ 103 m/sec. As we know, v = f $\lambda$

\therefore\,\,\,

f =

{v \over \lambda }

=

{{5.85 \times {{10}^3}} \over {1.2}}

= 4.88 $\times$ 103 Hz

\simeq

5 kHz

Q90

A sound absorber attenuates the sound level by

20

dB

. The intensity decreases by a factor of

A

100

B

1000

C

10000

D

10

Correct Answer

Option A

Solution

We have,

{L_1} = 10\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_0}}}} \right);

{L_2} = 10\,\log \left( {{{{{\rm I}_2}} \over {{{\rm I}_0}}}} \right)

$\therefore$

\,\,{L_1} - {L_2} = 10\,\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_0}}}} \right) - 10\,\log \left( {{{{{\rm I}_2}} \over {{{\rm I}_0}}}} \right)

or,

\Delta L = 10\,\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_0}}} \times {{{{\rm I}_0}} \over {{{\rm I}_2}}}} \right)

or,

\Delta L = 10\,\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_2}}}} \right)

or,

20 = 10\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_2}}}} \right)

or,

2 = \log \left( {{{{{\rm I}_1}} \over {{{\rm I}_2}}}} \right)

or,

{{{{\rm I}_1}} \over {{{\rm I}_2}}} = {10^2}

or,

{{\rm I}_2} = {{{{\rm I}_1}} \over {100}}.

$\Rightarrow$ Intensity decreases by a factor

100.