# Propagation of Sound Waves Exe-8A Numericals Concise Physics ICSE Class 9

Propagation of Sound Waves Exe-8A Numericals Concise Physics ICSE Class 9 Selina Publishers. There is the solutions of Numerical Questions of your latest textbook which is applicable in 2024-25 academic sessionVisit official Website CISCE for detail information about ICSE Board Class-9 Physics.

## Propagation of Sound Waves Exe-8A Numericals Concise Physics ICSE Class 9 Selina Publishers

 Board ICSE Class 9 Subject Physics Writer / Publication Concise Selina Publishers Chapter-8 Propagation of Sound Waves Exe-8A Production and Propagation of Sound Waves Topics Solution of Exe-8A Numericals Academic Session 2024-2025

### Production and Propagation of Sound Waves

Propagation of Sound Waves Exe-8A Concise Physics ICSE Class 9 Selina Publishers

#### NUMERICALS:

##### Question-1:  The heart of a man beats 75 times a minute. What is its (a) frequency and (b) time period ?

Answer- (a) Frequency = No. of times the heart beats in 1 s

=75/60 =1.25 Hz

(b) Time period, T = 1/f = 1/1.25 = 0.8 s

##### Question-2: The time period of a simple pendulum is 2 s. Find its frequency.

Answer- The Time period of a simple pendulum is given as 2 seconds.

The frequency is given as the inverse of time period, that is, f=1/T

f = 1/2 = 0.5Hz

##### Question-3:  The separation between two consecutive crests in a transverse wave is 100 m. If wave velocity is 20 m/s, find the frequency of wave.

Answer- The distance between any two consecutive crests or trough of a wave is known as the wavelength of the wave.

Hence, here λ=100 m

Also, given , v=100 m/s

The wavelength, wave velocity and frequency relationship is given as follows.

λ=v/f.

From this equation, frequency is derived as f=v/λ.

f = 20/100 = 0.2 Hz.

##### Question-4: A longitudinal wave travels at a speed of 0.3 m/s and the frequency of wave is 20 Hz. Find the separation between the two consecutive compressions.

The wavelength, wave velocity and frequency relationship are given as follows.

λ = v/f =0.3/20 = 0.015m.

That is 1.5×10^−2m

Hence, the separation between two consecutive compressions, i.e., the wavelength of the wave is 0.015 m or 1.5 cm.

##### Question-5: A source of wave produces 40 crests and 40 troughs in 0.4 s. What is the frequency of the wave?

Answer-  One crest and one trough constitute one cycle.

40 crests and 40 troughs involve 40 cycles.

Time taken for one cycle is called time period.

So, Time period = 0.4 s / (40) = 0.01 s

Frequency = 1 / Time period

Frequency = 1 / (0.01 s) = 100 Hz.

##### Question-6: An observer A fires a gun and another observer B at a distance 1650 m away from A hears its sound. If the speed of sound is 330. m/s, find the time when B will hear the sound after firing by A.

Answer- The speed of the sound, distance traveled and the time taken relationship is given as follows.

Speed of sound = Distance traveled / Time taken

From this equation the time taken is derived as

Time taken=Distance traveled / Speed of sound

That is, Time taken = 1650/330 = 5 seconds.

##### Question-7: The time interval between a lightning flash and the first sound of thunder is 5 s. If the speed of sound in air is 330 m s¹, find the distance of flash from the observer.

Answer-  Since light travels at a much faster speed as compared to sound, time taken for light to reach the earth’s surface is negligible.

Speed of sound in air (V) = 330 m/s

Time in which thunder is heard after lighting is seen (t) = 5 s

Thus, distance between flash and observer = V × t = 330 × 5 = 1650 m.

##### Question-8:  A boy fires a gun and another boy at a distance hears the sound of fire 2-5 s after seeing the flash. If speed of sound in air is 340 m/s, find the distance between the boys.

Answer-  Speed of sound in air (V) = 340 m/s

Time in which sound of fire is heard after flash is seen (t) = 2.5s

Thus, distance between flash and observer = V x t = (340 x 2.5) = 850 m.

##### Question-9:  An observer sitting in line of two tanks, watches the flashes of two tanks firing at each other at the same time, but he hears the sounds of two shots 2 s and 3-5 s after seeing the flashes. If distance between the two tanks is 510 m, find the speed of sound.

Answer-9  The time difference between the two tanks is 3.5 – 2 s = 1.5 s (or the sound takes this much time to travel between the tanks)

the distance between the two tanks is = 510 m

So, the speed of sound would be given by v = distance / time

= 510 / 1.5 m/s

= 340 m/s

##### Question-10: How long will sound take to travel in (a) an iron rail and (b) air, both 3.3 km in length ? Take speed of sound in air to be 330 m s and in iron to be 5280 m/s.

time =distance/speed = (3.3×1000)/5280 = 0.625s.

(b)=> time =distance/speed = (3.3×1000)/330 = 10s

##### Question-11:  Assuming the speed of sound in air equal to 340 m s and in water equal to 1360 m s-¹, find the time taken to travel a distance 1700 m by sound in

(i) air and (ii) water.

Answer- (i)=> Distance travelled (D) = 1700

Speed of sound in air (V) = 340 ms

Time taken (t) = D/V = 1700/340s = 5s

(ii)=> Distance travelled (D) = 1700

Speed of sound in water (V’) = 1360 ms

Time taken (t) = DV = 1700/1360s = 1.25s

—  : end of Propagation of Sound Waves Exe-8A Numericals Concise Physics ICSE Class 9 Selina Publishers  : —

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