CBSE Class 7 Science Advanced Physics Sound MCQs Set 03

Download CBSE MCQs for Class 7 Physics: Sound

Access targeted multiple-choice questions for Sound designed to align with the latest CBSE academic syllabus for Class 7 Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Chapter-wise Objective Questions: Sound

View or download the dedicated Sound MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Physics evaluations.

Multiple Choice Questions

Question 1: Match the statements of Column A with those of Column B.
Column AColumn B
(A) Carpets(a) Birds
(B) Syrinx(b) Ultrasonic sounds
(C) Dolphins(c) Reduces reverberation
(D) Internal organs(d) Echoes

(a) A → c; B → a; C → d; D → b
(b) A → d; B → c; C → a; D → b
(c) A → c; B → d; C → a; D → b
(d) A → b; B → d; C → a; D → c
Show Answer & Explanation

Answer: (a) A → c; B → a; C → d; D → b

Explanation:
1. Carpets absorb sound and so reduce reverberation: A → c.
2. The syrinx is the sound-producing organ of birds: B → a.
3. Dolphins find their way with echoes: C → d.
4. Ultrasonic sounds are used to scan internal organs: D → b.
5. The matching A → c; B → a; C → d; D → b is option (a).

Teacher's Note:
1. Soft materials absorb sound and cut down reverberation.
2. Ultrasound scanning is a medical use of high-frequency sound.

Question 2: Rearrange the following steps of an experiment in sequential order to determine the speed of sound in air.
(A) The person P1 will fire the cracker and person P2 will start the stopwatch as soon as he sees the light.
(B) The speed of sound in air is the ratio of the distance between the two places to the time interval.
(C) Note the time taken by the sound to travel from P1 to P2.
(D) The person P2 will stop the stopwatch as he hears the sound of the cracker.
(E) Let two persons P1 and P2 are separated by a distance of 1 km in an open place.
(F) Measure the distance between the two places by using a survey chain.
(a) ABCDEF
(b) ABDCFE
(c) ADBFCE
(d) EADCFB
Show Answer & Explanation

Answer: (d) EADCFB

Explanation:
1. First the two persons are placed 1 km apart in an open place: (E).
2. P1 fires the cracker and P2 starts the stopwatch on seeing the flash: (A).
3. P2 stops the stopwatch on hearing the bang: (D).
4. The time taken is noted: (C).
5. The distance between the places is measured with a survey chain: (F).
6. Finally speed = distance ÷ time: (B).
7. The order is E, A, D, C, F, B: option (d).

Teacher's Note:
1. Set up first, observe next, calculate last.
2. Light reaches P2 almost instantly, so the flash is used as the 'start' signal.

Question 3: Ultrasonic sound waves are used in ______.
(a) Dish washers
(b) Medical diagnostics
(c) Detection of flaws in materials
(d) All of these
Show Answer & Explanation

Answer: (d) All of these

Explanation:
1. Ultrasonic waves are used to clean in dishwashers.
2. They are used in medical diagnostics, such as scanning.
3. They are used to detect flaws (cracks) in metal parts.
4. All three uses are correct, so the answer is (d) All of these.

Teacher's Note:
1. Ultrasonic means above the audible range, above 20,000 Hz.
2. Its uses include cleaning, scanning and testing.

Question 4: Pitch of the sound is determined by the _____ of vibration.
(a) Frequency
(b) Amplitude
(c) Velocity
(d) Wavelength
Show Answer & Explanation

Answer: (a) Frequency

Explanation:
1. Pitch tells us how shrill or how flat a sound is.
2. It depends on the frequency of vibration: higher frequency gives a higher pitch.
3. The answer is frequency.

Teacher's Note:
1. Amplitude decides loudness, not pitch.
2. Frequency decides pitch.

Question 5: s⁻¹ is the unit of ______.
(a) Amplitude
(b) Intensity
(c) Frequency
(d) Time period
Show Answer & Explanation

Answer: (c) Frequency

Explanation:
1. Frequency is the number of vibrations per second, so its unit is per second, s⁻¹ (hertz).
2. Amplitude is a length, intensity is in W m⁻² and time period is in s.
3. The answer is frequency.

Teacher's Note:
1. 1 Hz = 1 s⁻¹.
2. Time period has the unit s, the inverse unit.

Question 6: Which of the following materials are sound absorbers?
(a) Thermocol
(b) Carpets
(c) Gunny bags
(d) All of these
Show Answer & Explanation

Answer: (d) All of these

Explanation:
1. Soft, porous materials absorb sound.
2. Thermocol, carpets and gunny bags are all such materials.
3. The answer is (d) All of these.

Teacher's Note:
1. Hard smooth surfaces reflect sound; soft rough ones absorb it.
2. Auditoriums use absorbers to cut down echo.

Question 7: A thunder is heard 2 s after the flash of lightning is seen. Lightning has occurred at a distance of _____ from you. (Velocity of sound in air is 330 m s⁻¹)
(a) 220 m
(b) 330 m
(c) 660 m
(d) 600 m
Show Answer & Explanation

Answer: (c) 660 m

Explanation:
1. Light reaches us almost instantly, so the 2 s is the time taken by sound.
2. Distance = speed × time = 330 × 2 = 660 m.
3. The answer is 660 m.

Teacher's Note:
1. Use distance = speed × time.
2. The flash of light takes negligible time.

Question 8: What is the order of magnitude of velocity of sound in solids, liquids and gases? (vₛ = velocity in solids, vℓ = velocity in liquids, vᵍ = velocity in gases)
(a) vᵍ > vₛ > vℓ
(b) vℓ > vₛ > vᵍ
(c) vₛ > vℓ > vᵍ
(d) vₛ = vᵍ = vℓ
Show Answer & Explanation

Answer: (c) vₛ > vℓ > vᵍ

Explanation:
1. Particles are closest in solids and farthest in gases, so sound passes fastest through solids.
2. Liquids come next and gases are the slowest.
3. So vₛ > vℓ > vᵍ: option (c).

Teacher's Note:
1. Closer particles pass the disturbance on faster.
2. Remember: solid > liquid > gas.

Question 9: A simple pendulum is made of a string of length 25 cm and a metal bob of diameter 2 cm. What is the length of the pendulum?
(a) 25 cm
(b) 27 cm
(c) 26 cm
(d) 23 cm
Show Answer & Explanation

Answer: (c) 26 cm

Explanation:
1. The length of a pendulum is measured from the point of suspension to the centre of the bob.
2. Radius of the bob = 2 ÷ 2 = 1 cm.
3. Length = 25 + 1 = 26 cm: option (c).

Teacher's Note:
1. Add the radius of the bob, not the diameter.
2. The centre of the bob is where its weight acts.

Question 10: The time period in seconds for 20 oscillations noted down in an experiment are 38.4, 40.5, 39.6, 42.8 and 41.5. Then the time period of the given simple pendulum is ______ s.
(a) 38.4
(b) 41.5
(c) 40.56
(d) None of these
Show Answer & Explanation

Answer: (d) None of these

Explanation:
1. Average time for 20 oscillations = (38.4 + 40.5 + 39.6 + 42.8 + 41.5) ÷ 5 = 202.8 ÷ 5 = 40.56 s.
2. Time period = 40.56 ÷ 20 = 2.028 s.
3. The value 40.56 s is for 20 oscillations, not one, so none of the options (a), (b), (c) is the time period: (d).

Teacher's Note:
1. Divide the total time by the number of oscillations (20).
2. Average first, then divide.

Question 11: The time period of a pendulum depends on ______.
(a) Its length
(b) Amplitude
(c) Mass of the bob
(d) frequency
Show Answer & Explanation

Answer: (a) Its length

Explanation:
1. For small swings, T = 2π√(ℓ/g), so the time period depends on length (and g).
2. It does not depend on the mass of the bob or the amplitude.
3. The answer is its length.

Teacher's Note:
1. Longer pendulum, longer time period.
2. Frequency is the inverse of the time period, not a cause of it.

Question 12: To find the depth of ocean bed ______ are used.
(a) Ultrasonic sounds
(b) Infrasonic sounds
(c) Audible sounds
(d) Subsonic sounds
Show Answer & Explanation

Answer: (a) Ultrasonic sounds

Explanation:
1. SONAR sends ultrasonic waves to the sea bed and times the echo.
2. Ultrasonic waves travel in a narrow beam and are not heard by us.
3. The answer is ultrasonic sounds.

Teacher's Note:
1. SONAR means sound navigation and ranging.
2. Depth = speed × time ÷ 2.

Question 13: A ship at rest is sending ultrasonic sound waves to detect a shoal of fish. The time gap between the signal sent from the ship and received (reflected) signal is 0.5 s (velocity of sound in water is 1400 m s⁻¹). Arrange the following steps in sequential order to find the distance of the shoal of fish from the ship.
(A) Use the formula, speed = distance travelled by sound ÷ total time taken = 2d ÷ t
(B) The distance ‘d’ travelled by the sound is twice the depth of the ocean bed.
(C) Note the total time ‘t’ taken by the reflected sound to reach the ship from the shoal of fish from the given data.
(D) Substitute the values in the given formula and find the value of ‘d’.
(a) ABCD
(b) BACD
(c) DACB
(d) CBAD
Show Answer & Explanation

Answer: (d) CBAD

Explanation:
1. First note the total time t = 0.5 s: (C).
2. The sound travels to the shoal and back, so the path is 2d: (B).
3. Use speed = 2d ÷ t: (A).
4. Substitute and solve: d = 1400 × 0.5 ÷ 2 = 350 m: (D).
5. The order is C, B, A, D: option (d).

Teacher's Note:
1. An echo path is there and back, so divide by 2.
2. Read the data, build the formula, then substitute.

Question 14: Match the statements of Column A with those of Column B.
Column AColumn B
(A) Depth of ocean bed(a) Decibel
(B) Sound(b) SONAR
(C) Loudness of sound(c) 0.1 s
(D) Persistence of hearing(d) Vibrating body

(a) A → d; B → c; C → a; D → b
(b) A → b; B → a; C → d; D → c
(c) A → b; B → d; C → a; D → c
(d) A → c; B → a; C → d; D → b
Show Answer & Explanation

Answer: (c) A → b; B → d; C → a; D → c

Explanation:
1. The depth of the ocean bed is measured with SONAR: A → b.
2. Sound is produced by a vibrating body: B → d.
3. Loudness is measured in decibels: C → a.
4. Persistence of hearing is about 0.1 s: D → c.
5. This is option (c).

Teacher's Note:
1. Persistence of hearing is the time for which a sound seems to stay in the ear.
2. Two sounds less than 0.1 s apart cannot be heard separately.

Question 15: Write the following steps in sequential order to prove that the angle of incident sound is equal to the angle of reflected sound.
(A) Measure the angle made by the two pipes with the normal.
(B) Place a ticking clock ‘s’ closer to pipe ‘p’.
(C) Repeat the experiment by changing the inclination of P and correspondingly, the inclination of Q.
(D) Measure the angles in each case and you find the sound which is incident and reflected from the wooden plank makes equal angles with the normal.
(E) Now keep your ear on the open end of Q and adjust the position of Q to listen to the ticking sound.
(F) Mount a smooth and hard wooden plank vertically on the surface of a table and clamp two pipes P and Q inclined at certain angles on the table.
(a) FBEACD
(b) BFEDCA
(c) ABCDEF
(d) CDEABF
Show Answer & Explanation

Answer: (a) FBEACD

Explanation:
1. Fix the plank and the two pipes first: (F).
2. Place the ticking clock near pipe P: (B).
3. Listen at the open end of Q and adjust it until the ticking is heard: (E).
4. Measure the angles with the normal: (A).
5. Repeat with other inclinations: (C).
6. Compare the angles in all cases: (D).
7. The order is F, B, E, A, C, D: option (a).

Teacher's Note:
1. Set up, produce sound, find the reflected sound, measure, repeat, conclude.
2. The conclusion about equal angles is always last.

Question 16: Bats make use of _____ waves.
(a) Ultrasonic sound
(b) Infrasonic sound
(c) Audible range of sound
(d) Subsonic sound
Show Answer & Explanation

Answer: (a) Ultrasonic sound

Explanation:
1. Bats send out ultrasonic squeaks and listen to the echoes.
2. The echoes tell them where obstacles and prey are.
3. The answer is ultrasonic sound.

Teacher's Note:
1. Ultrasonic sound is above 20,000 Hz.
2. This is called echolocation.

Question 17: When two tuning forks A and B having frequencies 312 Hz and 112 Hz, respectively, are vibrated simultaneously, the pitch of sound produced _____.
(a) By A will be more
(b) By B will be more
(c) By A and B will be equal
(d) By A and B will be more than 312 Hz.
Show Answer & Explanation

Answer: (a) By A will be more

Explanation:
1. Pitch increases with frequency.
2. A has the higher frequency (312 Hz > 112 Hz), so its pitch is higher.
3. The answer is (a).

Teacher's Note:
1. Higher frequency means higher (shriller) pitch.
2. Compare only the two frequencies.

Question 18: If a body vibrates 10 times in a second, then its frequency is _____.
(a) 10 Hz
(b) 0.1 Hz
(c) 1 Hz
(d) 0.01 Hz
Show Answer & Explanation

Answer: (a) 10 Hz

Explanation:
1. Frequency is the number of vibrations per second.
2. The body makes 10 vibrations in 1 s, so f = 10 Hz.
3. The answer is 10 Hz. (0.1 s would be its time period.)

Teacher's Note:
1. Frequency = number of vibrations ÷ time.
2. Do not confuse frequency with time period.

Question 19: Choose the false statement from the following:
(a) Compression is the region of high pressure and rarefaction is the region of low pressure.
(b) Compression is the region of low pressure and rarefaction is the region of high pressure.
(c) Compression is the region of high density and rarefaction is the region of low density.
(d) Both (b) and (c)
Show Answer & Explanation

Answer: (b) Compression is the region of low pressure and rarefaction is the region of high pressure.

Explanation:
1. In a compression the particles are crowded together, so pressure and density are high.
2. In a rarefaction they are spread apart, so pressure and density are low.
3. Statements (a) and (c) are true. Only statement (b) is false.
4. The answer is (b).

Teacher's Note:
1. Compression = crowded = high pressure and density.
2. Option (d) would need (c) to be false too, but (c) is correct.

Question 20: The sound produced by a siren given by a company reaches a person on the road at a distance of 220 m in _____ s (velocity of sound in air is 330 m s⁻¹).
(a) 2/3
(b) 1/3
(c) 4
(d) 0.5
Show Answer & Explanation

Answer: (a) 2/3

Explanation:
1. Time = distance ÷ velocity = 220 ÷ 330.
2. = 2/3 s.
3. The answer is 2/3.

Teacher's Note:
1. Use t = d ÷ v.
2. Cancel by 110.

Question 21: Violin is an example of a _____.
(a) Wind instrument
(b) String instrument
(c) Reed instrument
(d) Percussion instrument
Show Answer & Explanation

Answer: (b) String instrument

Explanation:
1. A violin makes sound by vibrating strings, played with a bow.
2. So it is a string instrument: (b).

Teacher's Note:
1. Strings: violin, sitar, guitar.
2. Drums are percussion; flutes are wind instruments.

Question 22: The bob of an oscillating simple pendulum arrives at one of the extreme positions 100 times in 300 s. Then the time period of the pendulum is _____ s.
(a) 3.0
(b) 1.5
(c) 1.0
(d) 2.5
Show Answer & Explanation

Answer: (a) 3.0

Explanation:
1. The bob reaches a given extreme position once in every oscillation.
2. So 100 oscillations take 300 s.
3. Time period = 300 ÷ 100 = 3.0 s: option (a).

Teacher's Note:
1. Reaching the same extreme again means one full oscillation.
2. Time period = total time ÷ number of oscillations.

Question 23: What is the loudness of sound in the range 50 dB – 60 dB?
(a) Not audible
(b) Just audible
(c) Moderate
(d) Painful to ear
Show Answer & Explanation

Answer: (c) Moderate

Explanation:
1. Very low levels, near 0 dB, are just audible.
2. Sound of 50 to 60 dB is like normal conversation, a moderate (normal) level.
3. Very high levels, above about 120 dB, are painful.
4. The answer is moderate.

Teacher's Note:
1. Normal conversation is about 60 dB.
2. The book's hint calls this range normal.

Question 24: The speed of sound in vacuum is _____ m s⁻¹.
(a) 330
(b) 240
(c) 420
(d) 0
Show Answer & Explanation

Answer: (d) 0

Explanation:
1. Sound needs a material medium to travel.
2. A vacuum has no particles, so sound cannot travel.
3. The speed is 0.

Teacher's Note:
1. The bell-jar experiment shows this.
2. Light can travel in vacuum; sound cannot.

Question 25: The time period of a simple pendulum is given by the formula, T = 2π√(ℓ/g), where T = time period, ℓ = length of pendulum, g = acceleration due to gravity. If the length of the pendulum is decreased to 1/4th of its initial value, rearrange the following steps in sequential order to find the frequency of oscillations at the new length.
(A) From the formula, time period of a simple pendulum is directly proportional to the square root of its length.
(B) When length becomes 1/4th of its initial value, time period will be 1/2 of its initial value, i.e., T/2.
(C) Frequency = 1 ÷ Time period.
(D) So, frequency of oscillation will be double.
(a) ABCD
(b) ADBC
(c) ACBD
(d) ACDB
Show Answer & Explanation

Answer: (a) ABCD

Explanation:
1. State the relation: T is proportional to √ℓ: (A).
2. With ℓ reduced to 1/4, T becomes √(1/4) = 1/2 of the old value: (B).
3. Frequency = 1/T: (C).
4. Halving T doubles the frequency: (D).
5. The order is A, B, C, D: option (a).

Teacher's Note:
1. Law first, new value next, then frequency.
2. Frequency and time period are inverses.

Question 26: What is the time taken to displace a vibrating body from one extreme position to the other?
(a) Time period
(b) Time period ÷ 2
(c) 2 (Time period)
(d) Time period ÷ 4
Show Answer & Explanation

Answer: (b) Time period ÷ 2

Explanation:
1. One full oscillation goes from one extreme to the other and back again.
2. Going from one extreme to the other is half of that.
3. So the time is half the time period: option (b).

Teacher's Note:
1. Draw the swing: left extreme to right extreme is half a cycle.
2. Back to the start is a full time period.

Question 27: __________ of sound is determined by its amplitude.
(a) Pitch
(b) Timbre
(c) Loudness
(d) Quality
Show Answer & Explanation

Answer: (c) Loudness

Explanation:
1. A larger amplitude of vibration gives a louder sound.
2. Pitch depends on frequency, and timbre (quality) on the waveform.
3. The answer is loudness.

Teacher's Note:
1. Amplitude controls loudness.
2. Frequency controls pitch.

Question 28: The velocity of sound is minimum in ________.
(a) Iron
(b) CO₂
(c) Water
(d) Aluminium
Show Answer & Explanation

Answer: (b) CO₂

Explanation:
1. Sound travels slowest in gases and faster in liquids and solids.
2. Of the four, only CO₂ is a gas (about 260 m s⁻¹).
3. The answer is CO₂.

Teacher's Note:
1. Gas < liquid < solid for speed of sound.
2. Only one option is a gas.

Question 29: What is the ratio of frequency to time period of a seconds’ pendulum?
(a) 1 : 2
(b) 1 : 1
(c) 1 : 4
(d) 2 : 1
Show Answer & Explanation

Answer: (c) 1 : 4

Explanation:
1. A seconds' pendulum has a time period T = 2 s.
2. Its frequency f = 1/T = 1/2 Hz.
3. Ratio f : T = 1/2 : 2 = 1 : 4.
4. The answer is 1 : 4.

Teacher's Note:
1. A seconds' pendulum takes 1 s for each swing, so 2 s for one oscillation.
2. Use f = 1/T.

Question 30: If a body vibrates 40 times in 4 seconds, then calculate its frequency.
(a) 10 Hz
(b) 20 s⁻¹
(c) 50 Hz
(d) 40 s⁻¹
Show Answer & Explanation

Answer: (a) 10 Hz

Explanation:
1. Frequency = number of vibrations ÷ time = 40 ÷ 4.
2. = 10 Hz.
3. The answer is 10 Hz.

Teacher's Note:
1. Frequency is vibrations per second.
2. 40 vibrations in 4 s means 10 in 1 s.

Question 31: _________ principle is used to locate shoals of fish under the seabed.
(a) Refraction of sound
(b) Reflection of light
(c) Reflection of sound
(d) Reverberation of sound
Show Answer & Explanation

Answer: (c) Reflection of sound

Explanation:
1. SONAR sends ultrasonic pulses and detects the echoes from the fish.
2. An echo is reflected sound.
3. The answer is reflection of sound.

Teacher's Note:
1. Echo = reflection of sound.
2. Light does not travel far in sea water.

Question 32: Match the statements of Column A with those of Column B.
Column AColumn B
(A) Bell jar experiment(p) 0.1 s
(B) Bats(q) 10 Hz
(C) Persistence of hearing(r) Sound can not travel in vacuum
(D) Infrasonic(s) 50 Hz to 80,000 Hz

(a) A → r; B → s; C → q; D → p
(b) A → r; B → s; C → p; D → q
(c) A → r; B → q; C → s; D → p
(d) A → r; B → q; C → p; D → s
Show Answer & Explanation

Answer: (b) A → r; B → s; C → p; D → q

Explanation:
1. The bell-jar experiment shows that sound cannot travel in vacuum: A → r.
2. Bats hear a wide range of frequencies up to ultrasonic: B → s.
3. Persistence of hearing is 0.1 s: C → p.
4. Infrasonic sound has a frequency below 20 Hz, such as 10 Hz: D → q.
5. This is option (b).

Teacher's Note:
1. Infrasonic means below the audible range.
2. Bats work in the ultrasonic region.

Question 33: ________ property distinguishes two sounds of the same frequency and loudness.
(a) Pitch
(b) Quality of sound
(c) Loudness of sound
(d) Speed of sound
Show Answer & Explanation

Answer: (b) Quality of sound

Explanation:
1. Pitch depends on frequency and loudness on amplitude, which are already the same.
2. Quality (timbre) depends on the waveform and tells the sounds apart.
3. The answer is quality of sound.

Teacher's Note:
1. This is how we know who is speaking on the phone.
2. Two instruments playing the same note differ in timbre.

Question 34: Arrange the following steps of an experiment in a sequential order to determine the time period of a simple pendulum.
(A) Pull the bob of the simple pendulum from its initial position of rest and release.
(B) Determine the average of t₁ and t₂ which is denoted as ‘t’.
(C) Start the stopwatch when the bob reaches the extreme position and count the number of oscillations.
(D) Suspend a single pendulum from a rigid support.
(E) Stop the stop watch at the end of 20 oscillations and determine the time taken to complete 20 oscillations (t₁).
(F) Repeat the experiment and determine the time taken for 20 oscillations (t₂).
(G) The time period of the oscillation is found using, T = t ÷ 20.
(a) DACEBGF
(b) DACEFGB
(c) DACFBGE
(d) DACEFBG
Show Answer & Explanation

Answer: (d) DACEFBG

Explanation:
1. Suspend the pendulum: (D). Pull and release: (A).
2. Start the stopwatch and count oscillations: (C). Stop after 20 and record t₁: (E).
3. Repeat to get t₂: (F).
4. Average the two readings: (B). Then T = t ÷ 20: (G).
5. The order is D, A, C, E, F, B, G: option (d).

Teacher's Note:
1. Averaging needs both t₁ and t₂ first.
2. Counting 20 oscillations reduces the timing error.

Question 35: When two drums A and B, having frequencies of 500 Hz and 216 Hz, respectively are vibrated simultaneously, then the shrillness of sound produced by ____________.
(a) Drum A will be more
(b) Drum B will be more
(c) A and B will be equal
(d) A and B will be more than 500 Hz
Show Answer & Explanation

Answer: (a) Drum A will be more

Explanation:
1. Shrillness means pitch, which increases with frequency.
2. Drum A (500 Hz) has the higher frequency than B (216 Hz).
3. The answer is drum A will be more.

Teacher's Note:
1. Shrill = high pitch = high frequency.
2. Compare 500 Hz with 216 Hz.

Question 36: The sound produced by a company horn reaches a person on the road at a distance of 990 m in the time of ________ s. Consider the velocity of sound in air as 330 m s⁻¹.
(a) 3
(b) 9
(c) 11
(d) 90
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. Time = distance ÷ velocity = 990 ÷ 330.
2. = 3 s: option (a).

Teacher's Note:
1. t = d ÷ v.
2. 990 is exactly 3 times 330.

Question 37: An observer stands at a distance of 6600 m from a cliff and fires a gun. If the speed of sound in air is 330 m s⁻¹, then the time taken by him to hear an echo is ____________.
(a) 20 s
(b) 400 s
(c) 40 s
(d) 60 s
Show Answer & Explanation

Answer: (c) 40 s

Explanation:
1. The sound travels to the cliff and back, a total of 2 × 6600 = 13200 m.
2. Time = 13200 ÷ 330 = 40 s.
3. The answer is 40 s.

Teacher's Note:
1. An echo path is twice the distance.
2. Use v = 2d ÷ t.

Question 38: A person stands in between two parallel cliffs which are 132 m apart. If he blows a whistle and hears two successive echoes after 0.4 s and 0.8 s, then the distance of the person from the nearer cliff is ________.
(a) 44 m
(b) 88 m
(c) 132 m
(d) 66 m
Show Answer & Explanation

Answer: (a) 44 m

Explanation:
1. Sound goes at the same speed to both cliffs, so each echo time is proportional to the distance of that cliff: 2d₁ ÷ 0.4 = 2d₂ ÷ 0.8.
2. This gives d₂ = 2d₁.
3. The cliffs are 132 m apart, so d₁ + d₂ = 132, i.e. 3d₁ = 132.
4. d₁ = 44 m: option (a).

Teacher's Note:
1. The speed is not given, so use the ratio of the echo times.
2. The longer echo time means the farther cliff.

Question 39: The time period of a simple pendulum of length ‘ℓ’ on the Earth is ‘T’. If it is taken to the moon, then the time period of the pendulum on the moon is __________. (take gₑ = 6 gₘ)
(a) 36 T
(b) T√6
(c) 6 T
(d) T ÷ 6
Show Answer & Explanation

Answer: (b) T√6

Explanation:
1. T = 2π√(ℓ/g), so T is proportional to 1/√g.
2. Tₘ ÷ T = √(gₑ ÷ gₘ) = √6.
3. So Tₘ = T√6: option (b).

Teacher's Note:
1. Weaker gravity on the moon makes the pendulum slower.
2. Take the square root of 6, not 6.

Question 40: A thunder is heard by a girl in 6 seconds after the lightning is seen and the speed of sound in air is 330 m s⁻¹. What is the distance from the girl to where lightning occurred?
(a) 2000 m
(b) 1900 m
(c) 1880 m
(d) 1980 m
Show Answer & Explanation

Answer: (d) 1980 m

Explanation:
1. Distance = speed × time = 330 × 6.
2. = 1980 m: option (d).

Teacher's Note:
1. Light reaches the girl almost at once.
2. Use d = v × t.

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