Here we have a collection of 24 carefully selected numerical for class 9 chapter 11 Sound. Take your time to solve these as some of them might seem challenging.
Formulas
- Speed of sound: v = distance ÷ time
- Wave relation: v = λ × ν
- Frequency: ν = 1 ÷ T
- Time period: T = 1 ÷ ν
- Echo distance: d = (v × t) ÷ 2
- Distinct echo condition: time gap ≥ 0.1 s
- At 22 ºC in air: v = 344 m/s
- At 0 ºC in air: v = 331 m/s
Class 9 Science (NCERT) – Chapter 11: Sound (Numericals Only)
Basic
Q1) A sound travels 680 m in 2 s. Find the speed of sound.
Solution:
Given: distance = 680 m, time = 2 s
Use v = distance ÷ time
v = 680 ÷ 2 = 340 m/s
Answer: 340 m/s
Q2) A sound wave has frequency 200 Hz and wavelength 1.7 m. Find its speed.
Solution:
Given: ν = 200 Hz, λ = 1.7 m
Use v = λ × ν
v = 1.7 × 200 = 340 m/s
Answer: 340 m/s
Q3) The speed of sound in air is 344 m/s at 22 ºC. Find the distance covered in 0.5 s.
Solution:
Given: v = 344 m/s, time = 0.5 s
Use distance = v × time
Distance = 344 × 0.5 = 172 m
Answer: 172 m
Q4) A sound wave has time period 0.01 s. Find its frequency.
Solution:
Given: T = 0.01 s
Use ν = 1 ÷ T
ν = 1 ÷ 0.01 = 100 Hz
Answer: 100 Hz
Q5) A sound wave has frequency 50 Hz. Find its time period.
Solution:
Given: ν = 50 Hz
Use T = 1 ÷ ν
T = 1 ÷ 50 = 0.02 s
Answer: 0.02 s
Q6) At 22 ºC, speed of sound in air is 344 m/s. Find the wavelength of a wave of frequency 172 Hz.
Solution:
Given: v = 344 m/s, ν = 172 Hz
Use v = λ × ν ⇒ λ = v ÷ ν
λ = 344 ÷ 172 = 2 m
Answer: 2 m
Standard
Q7) The speed of sound in air at 0 ºC is 331 m/s. How far does sound travel in 3 s?
Solution:
Given: v = 331 m/s, time = 3 s
Distance = v × time
Distance = 331 × 3 = 993 m
Answer: 993 m
Q8) A sound wave in air has speed 346 m/s (at 25 ºC). If its wavelength is 0.5 m, find its frequency.
Solution:
Given: v = 346 m/s, λ = 0.5 m
Use v = λ × ν ⇒ ν = v ÷ λ
ν = 346 ÷ 0.5 = 692 Hz
Answer: 692 Hz
Q9) An echo is heard after 0.6 s. Take speed of sound as 344 m/s. Find the distance of the reflecting surface.
Solution:
Given: t = 0.6 s, v = 344 m/s
Echo time includes to-and-fro travel, so use d = (v × t) ÷ 2
d = (344 × 0.6) ÷ 2
d = 206.4 ÷ 2 = 103.2 m
Answer: 103.2 m
Q10) A person is 20 m away from a wall. Using v = 344 m/s, find the echo time gap. Will it be a distinct echo? (Use 0.1 s rule)
Solution:
Given: d = 20 m, v = 344 m/s
Total distance travelled by sound = 2d = 2 × 20 = 40 m
Time gap t = total distance ÷ v = 40 ÷ 344 = 0.116 s
Compare with 0.1 s: 0.116 s ≥ 0.1 s
Answer: Time gap = 0.116 s, so a distinct echo will be heard.
Q11) Using v = 344 m/s at 22 ºC, NCERT shows minimum distance for distinct echo as 17.2 m. Verify it using the 0.1 s rule.
Solution:
Given: v = 344 m/s, minimum time gap = 0.1 s
Total distance needed = v × t = 344 × 0.1 = 34.4 m
Obstacle distance = total distance ÷ 2 = 34.4 ÷ 2 = 17.2 m
Answer: 17.2 m
Q12) A sound wave has wavelength 1.2 m and frequency 250 Hz. Find its speed and time period.
Solution:
Given: λ = 1.2 m, ν = 250 Hz
Speed: v = λ × ν = 1.2 × 250 = 300 m/s
Time period: T = 1 ÷ ν = 1 ÷ 250 = 0.004 s
Answer: Speed = 300 m/s, Time period = 0.004 s
Advance
Q13) The speed of sound in air is 346 m/s and in oxygen is 316 m/s. Find how much more distance sound travels in air than in oxygen in the same time of 2 s.
Solution:
Given: v(air) = 346 m/s, v(oxygen) = 316 m/s, time = 2 s
Distance in air = 346 × 2 = 692 m
Distance in oxygen = 316 × 2 = 632 m
Difference = 692 − 632 = 60 m
Answer: Sound travels 60 m more in air than in oxygen (in 2 s).
Q14) The speed of sound in sea water is 1531 m/s. Find the time taken for sound to travel 3.062 km in sea water.
Solution:
Given: v = 1531 m/s, distance = 3.062 km = 3062 m
Use time = distance ÷ speed
Time = 3062 ÷ 1531 = 2 s
Answer: 2 s
Q15) The speed of sound in steel is 5960 m/s. A sound travels through a steel rod of length 29.8 m. Find the time taken.
Solution:
Given: v = 5960 m/s, distance = 29.8 m
Use time = distance ÷ speed
Time = 29.8 ÷ 5960 = 0.005 s
Answer: 0.005 s
Q16) Two sounds are produced: A = 256 Hz, B = 512 Hz. If speed of sound is same, find the ratio of their wavelengths (λA : λB).
Solution:
For same medium, v is constant and v = λ × ν
So wavelength is inversely proportional to frequency: λ ∝ 1 ÷ ν
λA : λB = (1 ÷ 256) : (1 ÷ 512)
λA : λB = 512 : 256 = 2 : 1
Answer: 2 : 1
Q17) A sound wave has frequency 400 Hz. In air at 22 ºC, take v = 344 m/s. Find wavelength, then find how many wavelengths fit in 17.2 m.
Solution:
Given: ν = 400 Hz, v = 344 m/s
Wavelength λ = v ÷ ν = 344 ÷ 400 = 0.86 m
Number of wavelengths in 17.2 m = 17.2 ÷ 0.86
17.2 ÷ 0.86 = 20
Answer: λ = 0.86 m, number of wavelengths = 20
Q18) A ship sends a sound pulse downward in sea water. The speed of sound in sea water is 1531 m/s. The echo returns in 1.2 s. Find the depth of the sea.
Solution:
Given: v = 1531 m/s, t = 1.2 s
Use depth d = (v × t) ÷ 2
d = (1531 × 1.2) ÷ 2
d = 1837.2 ÷ 2 = 918.6 m
Answer: 918.6 m
HOTS
Q19) At 0 ºC, speed of sound in air = 331 m/s. At 22 ºC, speed = 344 m/s. For a fixed frequency of 500 Hz, find the change in wavelength (λ22ºC − λ0ºC).
Solution:
Given: ν = 500 Hz
At 0 ºC: λ0 = v ÷ ν = 331 ÷ 500 = 0.662 m
At 22 ºC: λ22 = v ÷ ν = 344 ÷ 500 = 0.688 m
Change = 0.688 − 0.662 = 0.026 m
Answer: Wavelength increases by 0.026 m
Q20) In a hall, a person is 15 m from a wall. Using v = 344 m/s, calculate the echo time gap and decide if a distinct echo is heard.
Solution:
Given: d = 15 m, v = 344 m/s
Total distance = 2d = 2 × 15 = 30 m
Time gap t = 30 ÷ 344 = 0.087 s
Compare with 0.1 s: 0.087 s < 0.1 s
Answer: Time gap = 0.087 s, so distinct echo will NOT be heard.
Q21) A person stands between two parallel walls. Distance to wall A is 20 m and to wall B is 30 m. Using v = 344 m/s, find the echo time gaps from each wall and the difference between them.
Solution:
Given: dA = 20 m, dB = 30 m, v = 344 m/s
Echo time from A: tA = (2 × 20) ÷ 344 = 40 ÷ 344 = 0.116 s
Echo time from B: tB = (2 × 30) ÷ 344 = 60 ÷ 344 = 0.174 s
Difference = 0.174 − 0.116 = 0.058 s
Answer: tA = 0.116 s, tB = 0.174 s, difference = 0.058 s
Q22) The speed of sound in helium is 965 m/s and in hydrogen is 1284 m/s. For a distance of 642 m, find the time taken in each gas and the time difference.
Solution:
Given: distance = 642 m
Time in helium: tHe = 642 ÷ 965 = 0.665 s
Time in hydrogen: tH2 = 642 ÷ 1284 = 0.5 s
Time difference = 0.665 − 0.5 = 0.165 s
Answer: Helium = 0.665 s, Hydrogen = 0.5 s, difference = 0.165 s
Q23) A sound wave in air at 25 ºC has v = 346 m/s. If its time period is 0.002 s, find frequency and wavelength.
Solution:
Given: v = 346 m/s, T = 0.002 s
Frequency ν = 1 ÷ T = 1 ÷ 0.002 = 500 Hz
Wavelength λ = v ÷ ν = 346 ÷ 500 = 0.692 m
Answer: Frequency = 500 Hz, Wavelength = 0.692 m
Q24) The speed of sound in aluminium is 6420 m/s and in glass (flint) is 3980 m/s. If a sound takes 0.01 s to pass through a material, find the distance travelled in each and the distance difference.
Solution:
Given: time = 0.01 s
In aluminium: distance = 6420 × 0.01 = 64.2 m
In glass (flint): distance = 3980 × 0.01 = 39.8 m
Difference = 64.2 − 39.8 = 24.4 m
Answer: Aluminium = 64.2 m, Glass = 39.8 m, difference = 24.4 m
