Practice Start Test

Waves

JEE Physics · 96 questions · Page 2 of 10 · Click an option or "Show Solution" to reveal answer

Q11
A tuning fork of frequency 480 Hz is used in an experiment for measuring speed of sound (v) in air by resonance tube method. Resonance is observed to occur at two successive lengths of the air column, l1 = 30 cm and l2 = 70 cm. Then, v is equal to -
A 338 ms–1
B 384 ms–1
C 379 ms–1
D 332 ms–1
Correct Answer
Option B
Solution

We know, λ\lambda = 2(

l2−l1{l_2} - {l_1}

) given

l2{l_2}

= 70 cm and

l1{l_1}

= 30 cm ∴\therefore λ\lambda = 2(70 - 30) = 80 cm Also we know, v = λ\lambda

ff

= 0.8 ×\times 480 = 384 m/s

Q12
A toy-car, blowing its horn, is moving with a steady speed of 5 m/s, away from a wall. An observer, towards whom the toy car is moving, is able to hear 5 beats per second. If the velocity of sound in air is 340 m/s, the frequency of the horn of the toy car is close to :
A 680 Hz
B 510 Hz
C 340 Hz
D 170 Hz
Correct Answer
Option D
Solution

Let the frequency of the horn = f Apparent frequency heared by the observer directly,

fdir=(VV−Vs)f=(340340−5)f=340335f{f_{dir}} = \left( {{V \over {V - {V_s}}}} \right)f = \left( {{{340} \over {340 - 5}}} \right)f = {{340} \over {335}}f

Apparent frequency heared by the observer on reflection from the wall,

find=(VV+Vs)f=(340340+5)f=340345f{f_{ind}} = \left( {{V \over {V + {V_s}}}} \right)f = \left( {{{340} \over {340 + 5}}} \right)f = {{340} \over {345}}f

Also, given that, find −- fdir == 5 ⇒\Rightarrow

340f345{{340f} \over {345}}

−-

340f335{{340f} \over {335}}

== 5 ⇒\Rightarrow f ==

5340×335×34510{5 \over {340}} \times {{335 \times 345} \over {10}}

= 169.9

≃\simeq

170 Hz

Q13
An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 3 ×108 ms–1)
A 15.3 GHz
B 10.1 GHz
C 12.1 GHz
D 17.3 GHz
Correct Answer
Option D
Solution

This question is from Doppler's effect of light.

When observer is moving towards the source then the frequency of wave measured by the observer will be fobserved = factual

c+vc−v\sqrt {{{c + v} \over {c - v}}}

where c = speed of light and v = speed of observer According to the question, v =

c2{c \over 2}

∴\therefore fobserved = factual

c+c2c−c2\sqrt {{{c + {c \over 2}} \over {c - {c \over 2}}}}

= 10×\times

3c2c2\sqrt {{{{{3c} \over 2}} \over {{c \over 2}}}}

= 10

3\sqrt 3

= 17.3 GHz

Q14
An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency ?
A 0.5%0.5\%
B zero
C 20%20\%
D 5%5\%
Correct Answer
Option C
Solution
n′=n[v+v0v]=n[v+v5v]=n[65]n' = n\left[ {{{v + {v_0}} \over v}} \right] = n\left[ {{{v + {v \over 5}} \over v}} \right] = n\left[ {{6 \over 5}} \right]
n′n=65;n′−nn{{n'} \over n} = {6 \over 5};{{n' - n} \over n}
=6−55×100=20%= {{6 - 5} \over 5} \times 100 = 20\%
Q15
A travelling wave is described by the equation y(x,t)=[0.05sin⁡(8x−4t)]y(x,t) = [0.05\sin (8x - 4t)] m The velocity of the wave is : [all the quantities are in SI unit]
A 4 ms−1\mathrm{4~ms^{-1}}
B 2 ms−1\mathrm{2~ms^{-1}}
C 8 ms−1\mathrm{8~ms^{-1}}
D 0.5 ms−1\mathrm{0.5~ms^{-1}}
Correct Answer
Option D
Solution

∵y(x,t)=[0.05sin⁡(8x−4t)]m\because y(x, t)=[0.05 \sin (8 x-4 t)] \mathrm{m}

 Speed of wave =∣ Coefficient of t Coefficient of x∣=48=0.5 ms−1\begin{aligned} \text{ Speed of wave } & =\left|\frac{\text{ Coefficient of } t}{\text{ Coefficient of } x}\right| \\\\ & =\frac{4}{8}=0.5 \mathrm{~ms}^{-1} \end{aligned}
Q16
A student is performing the experiment of resonance column. The diameter of the column tube is 6 cm. The frequency of the tuning fork is 504 Hz. Speed of the sound at the given temperature is 336 m/s. The zero of the metre scale coincides with the top end of the resonance column tube. The reading of the water level in the column when the first resonance occurs is :
A 13 cm
B 18.4 cm
C 16.6 cm
D 14.8 cm
Correct Answer
Option D
Solution

∴\therefore

l+1.8=λ4l + 1.8 = {\lambda \over 4}

Also

λ=vf=336504\lambda = {v \over f} = {{336} \over {504}}
⇒l+1.8=3364×504\Rightarrow l + 1.8 = {{336} \over {4 \times 504}}
⇒l=14.86\Rightarrow l = 14.86

cm

Q17
When two tuning forks (fork 11 and fork 22) are sounded simultaneously, 44 beats per second are heated. Now, some tape is attached on the prong of the fork 2.2. When the tuning forks are sounded again, 66 beats per second are heard. If the frequency of fork 11 is 200200 HzHz, then what was the original frequency of fork 22 ?
A 202202 HzHz
B 200200 HzHz
C 204204 HzHz
D 196196 HzHz
Correct Answer
Option D
Solution

No. of beats heard when fork

22

is sounded with fork

11
=Δn=4= \Delta n = 4

Now we know that if on loading (attaching tape) an unknown fork, the beat frequency increases (from

44

to

66

in this case) then the frequency of the unknown fork

22

is given by,

n=n0−Δn=200−4=196Hzn = {n_0} - \Delta n = 200 - 4 = 196Hz
Q18
A tuning fork of known frequency 256256 HzHz makes 55 beats per second with the vibrating string of a piano. The beat frequency decreases to 22 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was
A 256+2Hz256 + 2Hz
B 256−2Hz256 - 2Hz
C 256−5Hz256 - 5Hz
D 256+5Hz256 + 5Hz
Correct Answer
Option C
Solution

A tuning fork of frequency

256256
HzHz

makes

55

beats/ second with the vibrating string of a piano. Therefore the frequency of the vibrating string of piano is

(256±5)\left( {256 \pm 5} \right)
HzHz

ie either

261261
HzHz

or

251251
Hz.Hz.

When the tension in the piano string increases, its frequency will increases.

Now since the beat frequency decreases, we can conclude that the frequency of piano string is

251251
HzHz
Q19
A tuning fork arrangement (pair) produces 44 beats/sec with one fork of frequency 288288 cps.cps. A little wax is placed on the unknown fork and it then produces 22 beats/sec. The frequency of the unknown fork is
A 286286 cpscps
B 292292 cpscps
C 294294 cpscps
D 288288 cpscps
Correct Answer
Option B
Solution

A tuning fork produces

44

beats/sec with another tuning fork of frequency

288288

cps. From this information we can conclude that the frequency of unknown fork is

288+4288+4
cpscps

or

288−4288-4
cpscps

i.e.

292292
cpscps

or

284284
cps.cps.

Here when a little wax is placed on the unknown fork, it decreases the frequency of unknown fork.

Here also beats per second decreases to 2 from 4.

So the difference between frequency decreases.

This is possible only when before placing the wax, the frequency of unknown fork is greater than the frequency of the given tuning fork.

So the frequency of the unknown tuninh fork is = 292 Hz

Q20
A metal wire of linear mass density of 9.89.8 g/mg/m is stretched with a tension of 1010 kgkg-wtwt between two rigid supports 11 metre apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency n.n. The frequency nn of the alternating source is
A 5050 HzHz
B 100100 HzHz
C 200200 HzHz
D 2525 HzHz
Correct Answer
Option A
Solution

KEY CONCEPT : For a string vibrating between two rigid support, the fundamental frequency is given by

n=12ℓTμ=12×10×9.89.8×10−3=50Hzn = {1 \over {2\ell }}\sqrt {{T \over \mu }} = {1 \over {2 \times }}\sqrt {{{10 \times 9.8} \over {9.8 \times {{10}^{ - 3}}}}} = 50Hz

As the string is vibrating in resonance to a.c of frequency

n,n,

therefore both the frequencies are same.

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