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Class 9 Science: Sound Waves, Characteristics and Applications, 8 practice questions with answers

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These are 8 practice questions on Chapter 10, Sound Waves: Characteristics and Applications, of the new NCERT Class 9 Science textbook, Exploration (First Edition, April 2026). Each has a worked answer and the textbook page it comes from. They cover frequency and time period, the relation v = λν in air and in water, the delay between lightning and thunder, the distance to a wall from an echo, the audible range, and finding a depth with sonar. Try each one on paper before you open the answer.

These questions are our own, drafted with AI help. Every number was re-computed by a test script and none of the numbers is taken from the chapter, but no practising teacher has reviewed them yet. If you find a mistake, please tell us.

Where a question needs the speed of sound, it uses the textbook's own values: 340 m s⁻¹ in air and 1500 m s⁻¹ in water (Table 10.1, page 197).

What the chapter says you need

Facts from the NCERT textbook (Exploration: Textbook of Science for Grade 9, Chapter 10, printed pages 184 to 207):

Frequency, wavelength and speed

Question 1. A sound wave has a time period of 0.0016 s. What is its frequency?

Answer: 625 Hz. ν = 1/T = 1 ÷ 0.0016 = 625 Hz. (Eq. 10.1, page 194.)

Question 2. A sound wave in air has a wavelength of 0.4 m. What is its frequency? Use 340 m s⁻¹ for the speed of sound in air.

Answer: 850 Hz. From v = λν, ν = v/λ = 340 ÷ 0.4 = 850 Hz. (Eq. 10.2, page 196; speed in air, Table 10.1, page 197.)

Question 3. A tuning fork of frequency 425 Hz is sounded in air. What is the wavelength of the sound? Use 340 m s⁻¹.

Answer: 0.8 m. λ = v/ν = 340 ÷ 425 = 0.8 m. (Eq. 10.2, page 196.)

Question 4. A sound of frequency 2500 Hz travels through water. What is its wavelength there? Use 1500 m s⁻¹ for the speed of sound in water.

Answer: 0.6 m. λ = v/ν = 1500 ÷ 2500 = 0.6 m. The same frequency has a longer wavelength in water than in air, because sound is faster in water. (Eq. 10.2, page 196; speed in water, Table 10.1, page 197.)

Hearing and echoes

Question 5. During a storm you see a lightning flash and hear the thunder 3.5 s later. Roughly how far away was the lightning? Use 340 m s⁻¹, and treat the light as arriving instantly.

Answer: about 1190 m. Distance = speed × time = 340 × 3.5 = 1190 m, a little over 1 km. (Speed of sound in air, page 197; the same reasoning as Example 10.3, page 197.)

Question 6. You shout towards a cliff and hear the echo 1.3 s later. How far away is the cliff? Use 340 m s⁻¹.

Answer: 221 m. In 1.3 s the sound goes to the cliff and back: 340 × 1.3 = 442 m. The cliff is half of that away: 442 ÷ 2 = 221 m. This is more than 17 m, so the echo can be heard separately. (Echo, page 201.)

Question 7. One sound has a frequency of 26 kHz and another has 16 Hz. Can a human hear either of them? What is each one called?

Answer: neither; 26 kHz is ultrasonic and 16 Hz is infrasonic. Humans hear from 20 Hz to 20 kHz. 26 kHz is above 20 kHz, so it is an ultrasonic wave; 16 Hz is below 20 Hz, so it is an infrasonic wave. (Audible range, page 199; page 202.)

Sonar

Question 8. A ship's sonar sends an ultrasonic pulse straight down, and the echo from the sea floor returns after 2.6 s. How deep is the sea there? Use 1500 m s⁻¹ for the speed of sound in water.

Answer: 1950 m. The pulse goes down and back up: 1500 × 2.6 = 3900 m. The depth is half of that: 3900 ÷ 2 = 1950 m. (Sonar, page 203; speed in water, Table 10.1, page 197.)

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Sources

All opened and read on 21 October 2026. Page numbers are the printed page numbers of the textbook; printed page = PDF page + 183.

The definitions, equations and constants are from the textbook pages above. The questions, worked answers and checks are our own, and every number was re-computed by script. This guide is general information; follow your teacher and the textbook.