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851.

What is the length of vocal cords in a man?

Answer»

The male vocal chords are between 17 mm and 25 mm in length.

852.

Deduce a relationship between time period and frequency of a sound wave.

Answer»

1. Let the time taken for o oscillations = 1 sec. 

2. The time taken for one oscillation = 1/υ sec. 

3. But the time taken for one oscillation is called the time period (T) and the number of oscillations per second is called the frequency (υ).

4. Hence frequency and time period are related as T =1/υ (or) υ = 1/T

the relation between v and T is 
v = 1/T

853.

What are the characteristics of a sound wave?

Answer»

The characteristics of a sound wave, which play an important role in describing the nature of a wave are

1. Wavelength (λ) 

2. Amplitude (A) 

3. Frequency (υ) 

4. Wave speed (v)

There are five main characteristics of sound waves: wavelength, amplitude, frequency, time period, and velocity. The wavelength of a sound wave indicates the distance that wave travels before it repeats itself.

Wavelength : Wavelength can be defined as the distance between two successive crests or troughs of a wave. It is measured in the direction of the wave. 

velocity :The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of an object's speed and direction of motion (e.g. 60 km/h to the north). 

Time period :The time taken by a particle of medium to complete its one vibration is called the time period of the wave.

ortime period is the time taken by a vibrating body to complete one oscillation
time period is represented by the letter T. 
second (s) is the SI unit of time period.


Amplitude : The amplitude of a periodic variable is a measure of its change in a single period. The amplitude of a non-periodic signal is its magnitude compared with a reference value. There are various definitions of amplitude, which are all functions of the magnitude of the differences between the variable's extreme values. 

frequency :Frequency is the number of occurrences of a repeating event per unit of time. It is also occasionally referred to as temporal frequency to emphasize the contrast to spatial frequency, and ordinary frequency to emphasize the contrast to angular frequency 
.
854.

Which characteristic of sound helps you to identify your friend by his voice while sitting with others in a dark room?

Answer»

The pitch of voice differs for person to person which in turn depends on frequency. So, the frequency is the characteristic of sound which helps to identify my friends voice.

855.

Show the frequencies of various musical notes in a table.

Answer»

In musical terms, the pitch of the note determines the position of the note on the musical scale which is denoted as :

NoteC
(sa)
D
(re)
E
(ga)
F
(ma)
G
(pa)
A
(dha) 
 B
(ni)
C
(sa)
Frequency (Hz)256288320341.3384426.7480512

The tuning fork set is prepared based on the above frequencies.

856.

What are the medical applications of ultrasound?

Answer»

1) Imaging of organs : Ultra sonics are useful in Electro Cardio Graphy (ECG) to form an image of heart. 

Ultrasonography is used in the formation of images of organs such as liver, gall bladder, uterus, etc. to identify the abnormalities and tumors. Ultrasonography is also used to monitor the growth of a focus inside the mother’s womb.

2) Surgical use of ultrasound : Ultrasounds are used in cataract removal, breaking up of stones in kidneys, etc. without surgery.

Ultrasound imaging uses sound waves to produce pictures of the inside of the body. It helps diagnose the causes of pain, swelling and infection in the body's internal organs and to examine an unborn child (fetus) in pregnant women. 

857.

Which produces sound of a higher pitch : a drum or a whistle?

Answer»

Drum produces higher pitch.

858.

Out of a man and a woman: (a) who has shorter vocal cords? (b) who produces sound of higher pitch?

Answer»

(a) Woman has short vocal chords. The male vocal chords are between 17 mm and 25 mm in length whereas of a woman are between 12.5 mm and 17.5 mm long.

(b) Woman produces sound of higher pitch.

859.

Composition of the nuclei of two atomic species X and Y are given as under.XYProtons =66Neutrons =68a) Give the mass number of X and Y. b) What is the relation between the two species ?

Answer»

a) Mass number of X = 6 + 6 = 12 

Mass number of Y = 6 + 8 = 14 

b) Since X and Y both have atomic numbers as 6 but mass numbers are different. 

∴ These are isotopes to each other.

860.

Mention the industrial applications of ultrasonic waves.

Answer»

1) Drilling holes and making cuts of desired shapes : 

a) Holes can also be drilled using ultrasonic vibrations. 

b) Ultrasonic cutting and drilling are very effective for fragile materials like glass, etc.

2) Ultrasonic cleaning : 

a) Ultra sonics help in cleaning the parts located in hard-to-reach places. 

b) The high frequency ultrasonic vibrations knocks off all dirt and grease particles from the objects. 

3) Ultrasonic detection of defects in metals : 

The defects in the metallic structures, which are not visible from the outside, can be detected by ultrasonic waves.

861.

“Sound travels in the form of waves”. Justify your argument.

Answer»

1. Sound is a form of energy which travels through the air and reaches our ears to give the sensation of sound. 

2. There may be two possible ways by which transfer of energy from the source of sound to our ears take place. 

a) Source of sound produces disturbances in air and they strike our ears. 

b) Some particles are shot off from the source of sound and they reach our ears.

3. If the second explanation is correct, the vibrating body would gradually lose its weight as particles are continuously shot off from it. 

4. This is impossible because it would lead to vanishing of the object. 

5. Hence the first explanation that the sound travels through disturbances in the form of waves is correct.

862.

Write the name of any two radioactive isotopes.

Answer»

1. Uranium – 235 

2. Cobalt – 60

863.

Why do we see lighting before we hear the thunder ?

Answer» The speed of light is much faster than the speed of sound (about `10^6` times).
864.

A person on a pier observes a set of incoming waves that have a sinusoidal form with a distance of `1.6 m` between the crests. If a wave laps against the pier every `4 s`, what are : (a) the frequency and (b) the speed of the waves ?

Answer» (a) `lamda = 1.6 m, T = 4 s, v = (1)/(T) = (1)/(4 s) = 0.25 Hz`
(b) `v = v lamda = (0.25 xx 1.6) m//s = 0.4 m//s`.
865.

A stone is dropped from the top of a tower 750 m high into a pond of water at the base of the tower. When is the splash heard at the top?

Answer»

(Given g = 10 ms-2 and speed of sound = 340 ms-1

Height = 750m 

h = ut + 0.5 gt2 

The initial velocity is 0 750 = 0.5 × 10 × t2 

t = 10 sec

Speed of sound is 340m/sec 

So, time taken to travel 750m upwards is,

\(\frac{750}{340}\) = 2.20 S

time taken = 10 + 2.20 = 12.2 sec.

866.

A stone is dropped from the top of a tower `500 m` high into a pond of water at the base of the tower. When is the splash heard at the top ? Given, `g = 10 m//s^2` and speed of sound `= 340 m//s`.

Answer» Height of the tower, s = 500 m
Velocity of sound, `v = 340 m s^(-1)`
Acceleration due to gravity,` g = 10 m s^(-2)`
Initial velocity of the stone, `u = 0` (since the stone is initially at rest)
Time taken by the stone to fall to the base of the tower, `t_(1)`
According to the second equation of motion:
`s=ut_(1)+(1)/(2)g t_(1)^(2)`
`500=0xxt_(1)+(1)/(2)xx10xxt_(1)^(2)`
` t_(1)^(2)=100`
`t_(1)=10 s`
Now, time taken by the sound to reach the top from the base of the tower, `t_(2)= 500 //340 = 1.47 s`
Therefore, the splash is heard at the top after time, t
Where,` t= t_(1) +t_(2) = 10 + 1.47 = 11.47 s`.
867.

A wave is represented by the equation y = Asin (10πx + 10πt +π/3),where x is in metres and t is in seconds. The expression represents(a) a wave travelling in the positive x-direction with a velocity of 1.5 m/s (b) a wave travelling in the negative x-direction with a velocity of 1.5 m/s (c) a wave travelling in the negative x-direction with a wavelength of 0.2 m (d) a wave travelling in the positive x-direction with a wavelength of 0.2 m

Answer»

Correct Answer is: (b, c)

868.

Ratan saw a flash of lightning and after sometime heard the sound of thunder. If the time gap between seeing and hearing is 3 s, then write the following steps in sequence to find the height from the ground where the lightning is produced. (a) Consider the velocity of sound (v) as `330 ms^(-1)`. (b) Note the time gap (t) between seeing and hearing. (c) From the formula, speed `= ("distance")/("time taken")`, height at which lightning was produced = speed of sound `xx` time interval. (d) Substitute the values in the equation (1) and find the value of the height of the lightning from the ground.A. b a c dB. b c a dC. c b d aD. d a b c

Answer» Correct Answer - A
From the given data, time gap = 3 s.
We know, `v = 330 ms^(-1)`
Speed `= ("distance")/("time") to (c)`
Neglecting the time taken by the light to reach the ground because velocity of light is very large, that is, `C = 3 xx 10^(8) ms^(-1)`. On substituting the value, the height of the lightning from the ground is equal to `330 xx 3 = 990 m.`
869.

Lightning and thunder take place in the sky at the same time and at the same distance from us. Lightning is seen earlier and thunder is heard later. Can you explain?

Answer»

The speed of sound is less than the speed of light. Due to this, light reaches to us faster than sound. Hence, during lightning we see the streak of light earlier than hearing the sound of thunder.

870.

A thunder clap was heard 6 seconds after a lightening flash was seen. If the speed of sound in air is 340 m/s at the time of observation, the distance of the listener from the thunder clap is ………(A) 56.6 m (B) 346 m (C) 1020 m (D) 2040 m

Answer»

Correct option is: (D) 2040 m

871.

State the applications of acoustics observed in nature.

Answer»

Application of acoustics in nature:

i. Bats apply the principle of acoustics to locate objects. They emit short ultrasonic pulses of frequency 30 kHz to 150 kHz. The resulting echoes give them information about location of the obstacle. This helps the bats to fly in even in total darkness of caves.

ii. Dolphins navigate underwater with the help of an analogous system. They emit subsonic frequencies which can be about 100 Hz. They can sense an object about 1.4 m or larger.

872.

If 125 oscillations are produced in 5 seconds, what is the frequency in hertz?

Answer»

Frequency is f = \(\frac{No\, of\, Vibrations}{time} = \frac{125}{5} = 25 Hz\)

873.

A simple pendulum makes 10 oscillations in 20 seconds. What is the time period and frequency of its oscillation?

Answer»

Time period : 2 s

frequency : 0.5 oscillations/sec

874.

The time taken for one oscillation is called A) frequency B) time period C) wavelengthD) wave speed

Answer»

B) time period

875.

If the depth of the sea is 1.125 km, the time taken for the reflected sound to reach the sonar is _________ s. (velocity of sound in water is `1500 ms^(-1)`)A. `1`B. `1.5`C. `2`D. `2.5`

Answer» Correct Answer - B
Speed of sound in water
`= ("Distance covered by sound")/("Time")`
`"Time " = ("distance covered by sound")/("speed of sound in water") = (2 xx 1125)/(1500) = 1.5 s`
876.

Three persons `P_(1), P_(2)` and `P_(3)` are at different points A, B and C, as shown in the figure. `P_(1)` and `P_(2)` clap at the same time. For `P_(3)` to hear two distinct claps, the minimum distance betweenl `P_(1)` and `P_(2)` should be ______ m. (The velocity of sound in air is `330 ms^(-1)`) A. 33B. 330C. 363D. 660

Answer» Correct Answer - A
Human ear can hear two sounds separately only if they reach the ear after an interval of 1/10 of a second. So, the time taken by the sound from `P_(1)` to `P_(2)` should be greater than `1//10 s = 0.1 s.`
The distance between `P_(1)` and `P_(2)`
`= v xx t = 330 xx 0.1 = 33 m.`
877.

If the same time taken for the reflected sound to reach the sonar os 2 s, the depth of the sea is _______ m. (velocity of sound in water is `1500 ms^(-1)`)A. 1500B. 3000C. 750D. 500

Answer» Correct Answer - A
Speed of sound in water `= ("distance covered by sound")/("time")`
distance covered by sound `= v xx t`
`= 1500 xx 1 = 1500 m.`
878.

The time period of a simple pendulum is given by the formula, `T = 2pi sqrt(l//g)`, where T = time period, l = length of pendulum and g = acceleration due to gravity. If the length of the pendulum is decreased to 1/4 of its initial value, then what happens to its frequency of oscillations ?

Answer» The time period of a simple pendulum is
`T = 2pi sqrt(l//g)`.
When length is decreased to `1//4^(th)` of its initial value then `l^(1) = l//4`.
`therefore` New time period, `(T^(1))`
`= 2pi sqrt((l//4)/(g)) = (2pi)/(2) sqrt(l/g) = 1/2 (T)`
`therefore` Time taken to perform one oscillation is reduced to half of its initial value.
`therefore` Frequency of oscillation becomes double that of the initial value.
879.

A man is standing not exactly at the exactly at the centre, in between two parallel walls separated by a distance 990 m. He fires a gun and hears the first echo after 2 s, the time taken to hear 3rd echo after firing is ______ s. (take velocity of sound in air is `330 ms^(-1)`)A. 5B. 7C. 6D. 8

Answer» Correct Answer - B
He hears the first echo after 2 s. The distance between the 1st wall and the man is
`d = v xx t = 330 xx 1 = 330 m.`
So, the distance between the man and 2nd wall is
`990 - 330 = 660 m.`
So, time taken for 2nd echo time `= ("distance")/("velocity") = (2 xx 660)/(330) = 4 s`
The 3rd and 4th echo reach at the same time.
Time taken for 3rd echo = time taken to hear the first echo + time taken to hear the 3rd echo
`= 2 + (2 xx 660)/(330) = 2 + 2 xx 2 = 6 s.`
880.

Why the sound is made to undergo multiple reflection in a mega phone ?

Answer» Mega phone, loudspeaker, hearing aid, etc., are the devices which work on the reflection of sound.
The main part of the mega phone or the loudspeaker is a horn shaped tube.
This tube prevents the spreading of sound waves in all directions. The sound entering the tube undergoes multiple reflections and comes out of the tube with a high directionality and it can propagate through longer distances.
881.

A police targets a thief, who is at rest at point Q. He fires a bullet from a point P which is at a distance of 990 m from Q and it reaches the point Q in 3 s. When the bullet is fired at P it produced a sound, which was heard by the theif. What is the difference in time taken by the sound and the bullet to reach point Q. Will the thief be able to escape from the bullet fired by police ? Take the velocity of sound in air as `330 ms^(-1)`.

Answer» Distance between points P and Q, (d) = 990 m.
Time, (t) = 3 s.
Speed with which the bullet moves is
`= (990 m)/(3s) = 330 ms^(-1)`.
As the speed of body matches the speed of sound, the time interval is zero. The thief cannot escape the bullet.
882.

Decibel is the unit for measuring intensity of sound. It is expressed in Decibels (dB). Some common sounds and their decibel ratings. Near total silence – 0 dB, A whisper -15 dB, Normal conversation – 60 dB, A lawnmower – 90 dB, A car horn -110 dB, A jet engine -120 dB. A gunshot or fire-cracker – 140 dB, If a person is being exposed to the sound of 80 dB continuously it may lead to hearing problems.Which of the following can be said based on the above?A) Listening to whispering sounds continuously can lead to hearing problems. B) Hearing to continuous explosive sounds of guns can lead to hearing problems. C) Total silence can harm the ears and some sound is required for maintaining the health of ear drums. D) Car horns and other vehicle horn sounds do not cause any damage to the ears.

Answer»

B) Hearing to continuous explosive sounds of guns can lead to hearing problems.

883.

A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. (i) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10 m s-1 (b) recedes from the platform with a speed of 10 m s-1? (ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 m s-1.

Answer»

Given: vs = 10 m/s, v = 340 m/s, n0 = 400 Hz

Apparent frequency (n), velocity of sound (vs) in each case

Formulae:

i. n = n0 \((\frac{v}{v-v_s})\)

ii. n = n0 \(\frac{v}{v+v_s}\)

Calculation:

a. As the train approaches the platform, using formula (i),

n = 400 \((\frac{340}{340-10})\) = 421.12 Hz

b. As the train recedes from the platform, using formula (ii),

n = 400 \((\frac{340}{340+10})\) = 388.57 Hz

ii. The relative motion of source and observer results in the apparent change in the frequency but has no effect on the speed of sound. 

Hence, the speed of sound remains unchanged in both the cases.

884.

Name one musical instrument each in which the sound is produced: (a) by vibrating a stretched string. (b) by vibrating air enclosed in a tube. (c) by vibrating a stretched membrane (d) by vibrating metal plates.

Answer»

(a) Veena 

(b) Flute 

(c) Table 

(d) Cymbals

885.

Given figure represents about A) Musical instrument B) Cell phone C) Tumbler closed with cloth D) Sound has energy

Answer»

Correct option is D) Sound has energy

886.

Intensity of sound depends on A) pitch B) amplitude C) frequencyD) none of these

Answer»

Correct option is B) amplitude

887.

……….. produce more shrillness sound. A) Man B) Woman C) Child D) None of these

Answer»

Correct option is C) Child

888.

Sound can’t travel in A) solid B) liquid C) gas D) vacuum

Answer»

Correct option is D) vacuum

889.

………… can’t travel in vacuum.A) Solid B) Liquid C) Gas D) Sound

Answer»

Correct option is D) Sound

890.

i. What type of wave is a sound wave? ii. Can sound travel in vacuum? iii. What are reverberation and echo? iv. What is meant by pitch of a sound?

Answer»

i. Sound wave is a longitudinal wave. 

ii. Sound cannot travel in vacuum. 

iii. 

a. Reverberation is the phenomenon in which sound waves are reflected multiple times causing a single sound to be heard more than once.

b. An echo is the repetition of the original sound because of reflection by some surface.

iv. The characteristic of sound which is determined by the value of frequency is called as the pitch of the sound.

891.

Expand SONAR.

Answer»

SOUND NAVIGATION AND RANGING.

892.

Write some harmful effects of noise pollution.

Answer»

Some harmful effects of noise pollution are:

  1. Lack of sleep 
  2. Hypertension 
  3. Anxiety 
  4. Temporary or even permanent loss of hearing.
893.

Select the correct alternativeSound can not travel in1. solid2. liquid3. gas4. vacuum

Answer»

Sound can not travel in vacuum.

894.

a) An electric bell is suspended by thin wires in a glass vessel and set ringing. Describe and explain what happens if the air is gradually pumped out of the glass vessel. b) Why cannot a sound be heard on the moon? How do astronauts talk to one another on the surface of moon?

Answer»

a) When the air is gradually pumped out of the glass vessel, the ringing of the electric bell cannot be heard as there is a creation of vacuum inside the glass vessel. As there are no air molecules inside the glass jar, sound waves cannot travel. 

b) Sound cannot be heard on the moon because there is no air on the moon which acts as a carrier of sound waves. Astronauts talk to one another with the help of radio waves as these waves can travel through the vacuum.

895.

Explain why a ringing bell suspended in a vacuum chamber cannot be heard outside.

Answer»

When the ringing bell is suspended in a vacuum chamber it cannot be heard outside because there are no air molecules that vibrate to carry the sound waves.

896.

Name one solid, one liquid and one gas through which sound can travel.

Answer»

Solid: Metal- Iron, Aluminium etc. 

Liquid: Water 

Gas: Air

897.

Which object is vibrating when the following sounds are produced? a) the sound of a sitar b) the sound of a table c) the sound of a tuning fork d) the buzzing of a bee or mosquito e) the sound of a flute

Answer»

a) Vibrations in the stretched strings of sitar 

b) Vibrations from the stretched membrane 

c) Vibrations from the prongs 

d) Vibrations from the wings 

e) Vibrations from the air columns

898.

If a ringing bicycle bell is held tightly by hand, it stops producing sound. Why?

Answer»

The ringing bicycle bell does not produce sound when it is held tightly with hand because the vibrations are stopped and so the sound production stops.

899.

Which of the following cannot transmit sound?Water, Vacuum, Aluminium, Oxygen gas

Answer»

Vacuum cannot transmit sound because sound needs a material medium to travel.

900.

Write a short note on quality (timbre) of sound note.

Answer»

i. Timbre of a sound refers to the quality of the sound which depends upon the mixture of tones and overtones in the sound. Same sound played on different musical instruments feels significantly different and the musical instrument from which the sound generated can be easily identified.