Waves — Page 8 | JEE Physics

Q71

A uniform string of length

20

m

is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the supports is : (take

{\,\,g = 10m{s^{ - 2}}}

)

A

2\sqrt 2 s

B

2\pi \sqrt 2 s

C

2\pi \sqrt 2 s

D

2

s

Correct Answer

Option A

Solution

We know that velocity in string is given by

v = \sqrt {{T \over \mu }} \,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)

where

\mu = {m \over {\rm I}} = {{mass\,\,\,of\,\,\,string} \over {length\,\,\,of\,\,\,string}}

The tension

T = {m \over \ell } \times x \times g\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

From

(a)

and

(b)

{{dx} \over {dt}} = \sqrt {gx}

{x^{ - 1/2}}\,dx = \sqrt g \,dt

$\therefore$

\int\limits_0^\ell {{x^{ - 1/2}}} dx - \sqrt g \int\limits_0^\ell {dt}

2\sqrt \ell = \sqrt g \times t

$\therefore$

t = 2\sqrt {{\ell \over g}} = 2\sqrt {{{20} \over {10}}} = 2\sqrt 2

Q72

A wave

y=a

\sin \left( {\omega t - kx} \right)

on a string meets with another wave producing a node at

x=0.

Then the equation of the unknown wave is

A

y = a\,\sin \,\left( {\omega t + kx} \right)

B

y = - a\,\sin \,\left( {\omega t + kx} \right)

C

y = a\,\sin \,\left( {\omega t - kx} \right)

D

y = - a\,\sin \,\left( {\omega t - kx} \right)

Correct Answer

Option B

Solution

To form a node there should be superposition of this wave with the reflected wave.

The reflected wave should travel in opposite direction with a phase change of $\pi$ .

The equation of the reflected wave will be

y = a\sin \left( {\omega t + kx + \pi } \right)

\Rightarrow y = - a\sin \left( {\omega t + kx} \right)

Q73

A uniform thin rope of length 12 m and mass 6 kg hangs vertically from a rigid support and a block of mass 2 kg is attached to its free end. A transverse short wavetrain of wavelength 6 cm is produced at the lower end of the rope. What is the wavelength of the wavetrain (in cm) when it reaches the top of the rope ?

A 12

B 3

C 9

D 6

Correct Answer

Option A

Solution

T1 = 2g T2 = 8g V =

\sqrt {{T \over \mu }}

$\therefore$ V $\propto$

\sqrt T

Also V = f $\lambda$ $\therefore$ V1 = f1 $\lambda$ 1 and V2 = f2 $\lambda$ 2 We know frequency of sources are same. $\therefore$ f1 = f2 So

\sqrt {{{{T_1}} \over {{T_2}}}} = {{{\lambda _1}} \over {{\lambda _2}}}

$\Rightarrow$

\sqrt {{{2g} \over {8g}}} = {6 \over {{\lambda _2}}}

$\Rightarrow$ $\lambda$ 2 = 12 cm

Q74

A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound ? [Given reference intensity of sound as 10–12 W/m2 ]

A 20 cm

B 10 cm

C 40 cm

D 30 cm

Correct Answer

Option C

Solution

Sound level = 10

{\log _{10}}\left( {{I \over {{I_0}}}} \right)

$\Rightarrow$ 120 = 10

{\log _{10}}\left( {{I \over {{{10}^{ - 12}}}}} \right)

$\Rightarrow$ 12 =

{\log _{10}}\left( {{I \over {{{10}^{ - 12}}}}} \right)

$\Rightarrow$ 1012 =

{{I \over {{{10}^{ - 12}}}}}

$\Rightarrow$ I = 1 W/m2 Also we know, I =

{P \over {4\pi {r^2}}}

$\Rightarrow$ 1 =

{2 \over {4\pi {r^2}}}

$\Rightarrow$ r = 40 cm

Q75

When temperature increases, the frequency of a tuning fork

A increases

B decreases

C remains same

D increases or decreases depending on the material

Correct Answer

Option B

Solution

KEY CONCEPT : The frequency of a tuning fork is given by the expression

f = {{{m^2}k} \over {4\sqrt 3 \pi {\ell ^2}}}\sqrt {{Y \over \rho }}

As temperature increases,

\ell

increases and therefore

f

decreases.

Q76

A cylindrical tube, open at both ends, has a fundamental frequency,

f,

in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now :

A

f

B

f/2

C

3/4

D

2f

Correct Answer

Option A

Solution

The fundamental frequency of open tube

{v_0} = {v \over {2{l_0}}}\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

That of closed pipe

{v_c} = {\upsilon \over {4{l_c}}}\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

According to the problem

{l_c} = {{{l_0}} \over 2}

Thus

{v_c} = {\upsilon \over {{l_0}/2}} \Rightarrow {v_c}{\upsilon \over {2l}}\,\,\,\,...\left( {iii} \right)

From equations

(i)

and

(iii)

{v_0} = {v_c}

Thus,

{v_c} = f

\,\,\,\left( {\,\,} \right.

as

{v_0} = f

is given

\left. {\,\,} \right)

Q77

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

Q78

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.

Q79

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.

Q80

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