CBSE Class 7 Science Advanced Physics Sound MCQs Set 02

Answer: (b) The pitch of sound increases with a decrease in the thickness of a vibrating string.

Explanation:
1. A shorter air column or string vibrates faster, so its pitch rises; options (a) and (c) are wrong.
2. A thinner string also vibrates faster, so the pitch rises, which makes (b) correct.
3. Pitch does not depend on amplitude, so (d) is wrong. The answer is (b).

Teacher's Note:
1. Thinner and shorter strings give higher pitch.
2. Amplitude affects loudness, not pitch.

Answer: (b) 1.5

Explanation:
1. Depth = 1.125 km = 1125 m. The sound travels down and up, so the total path is 2 × 1125 = 2250 m.
2. t = 2250 ÷ 1500 = 1.5 s.

Teacher's Note:
1. Convert kilometres to metres first.
2. Double the depth because the sound goes down and comes back.

Answer: (d) 495 m

Explanation:
1. Nearer wall: 2d₁ = 330 × 1 = 330, so d₁ = 165 m.
2. Farther wall: 2d₂ = 330 × 2 = 660, so d₂ = 330 m.
3. Distance between the walls = 165 + 330 = 495 m.

Teacher's Note:
1. Each echo needs the sound to go to the wall and come back, so use half the distance.
2. The walls are on opposite sides of the man, so add the two distances.

Answer: (c) 363 m

Explanation:
1. The clap of P₁ travels 330 m to P₂ in 330 ÷ 330 = 1 s.
2. To be heard separately, the clap of P₃ must arrive at least 0.1 s later, so it must take 1 + 0.1 = 1.1 s.
3. Distance P₂P₃ = 330 × 1.1 = 363 m.

Teacher's Note:
1. The ear separates two sounds only if they are at least 0.1 s apart.
2. Add 0.1 s to the time taken by the first clap.

Answer: (a) 0.5

Explanation:
1. T = 1 ÷ f = 1 ÷ 10 = 0.1 s for one vibration.
2. Five vibrations take 5 × 0.1 = 0.5 s.

Teacher's Note:
1. Find the time for one vibration first.
2. Then multiply by the number of vibrations.

Answer: (c) 1:11

Explanation:
1. Distances from the bomb: 330 × 5 = 1650 m and 330 × 6 = 1980 m.
2. Minimum separation (both on the same side) = 1980 − 1650 = 330 m.
3. Maximum separation (opposite sides) = 1980 + 1650 = 3630 m.
4. Ratio = 330 : 3630 = 1 : 11.

Teacher's Note:
1. The persons may be on the same side or on opposite sides of the bomb.
2. Difference of distances gives the minimum, sum gives the maximum.

Answer: (b) 1.65

Explanation:
1. Distance = speed × time = 330 × 5 = 1650 m.
2. 1650 m = 1.65 km.

Teacher's Note:
1. Convert metres to kilometres by dividing by 1000.
2. Do not divide by 2; this is not an echo.

Answer: (b) 2

Explanation:
1. Frequency = number of vibrations ÷ time = 20 ÷ 10 = 2 Hz.

Teacher's Note:
1. Frequency means vibrations per second.
2. Divide the vibrations by the time in seconds.

Answer: (a) 33

Explanation:
1. The two claps start together, so the gap in their arrival at P₃ equals the extra distance of P₁ divided by the speed.
2. For two distinct claps this gap must be at least 0.1 s.
3. Distance = 330 × 0.1 = 33 m.

Teacher's Note:
1. The extra distance travelled by the farther clap is the distance P₁P₂.
2. The least gap the ear can separate is 0.1 s.

Answer: (b) 6

Explanation:
1. First echo after 2 s: nearer wall at 330 × 2 ÷ 2 = 330 m; the other wall is 990 − 330 = 660 m away.
2. Second echo (farther wall) comes after 2 × 660 ÷ 330 = 4 s.
3. The third echo is the sound that has gone to one wall, then to the other, and back to the man: total path 330 + 990 + 660 = 1980 m, taking 1980 ÷ 330 = 6 s.

Teacher's Note:
1. Work out each wall's distance first.
2. The 3rd echo is a double reflection and needs the sound to cover the gap between the walls once.

Answer: (a) 1500

Explanation:
1. Total distance = 1500 × 2 = 3000 m, which is down and back.
2. Depth = 3000 ÷ 2 = 1500 m.

Teacher's Note:
1. The depth is half of the distance travelled by the sound.
2. Round trip time is given, not one-way time.

Answer: (b) The frequency of vibration determines the pitch of any sound.

Explanation:
1. Pitch depends on frequency, not amplitude, so (a) is wrong and (b) is right.
2. A woman's voice has a higher pitch than a man's, so (c) is wrong. A tabla is not shriller than a flute, so (d) is wrong.
3. The correct statement is (b).

Teacher's Note:
1. Pitch equals frequency; loudness equals amplitude.
2. Women and children have higher-pitched voices.

Answer: (b) 30°

Explanation:
1. By the law of reflection, angle of reflection = angle of incidence.
2. Both are measured from the normal, so the reflected wave makes 30° with the normal.

Teacher's Note:
1. Both angles are measured from the normal.
2. Do not confuse this with the angle made with the surface, which would be 60°.

Answer: (c) remains the same

Explanation:
1. For small swings the time period of a simple pendulum depends only on its length and on g.
2. A change in amplitude does not alter it, so the time period remains the same.

Teacher's Note:
1. Amplitude changes how far the bob swings, not how long one swing takes.
2. The time period changes only if the length changes.

Answer: (b) 60°

Explanation:
1. Angle with the normal = 90° − 30° = 60°.
2. The angle of reflection equals the angle of incidence, so it is 60°.

Teacher's Note:
1. Always measure incidence and reflection angles from the normal.
2. The 30° given here is measured from the surface.

Answer: (d) c b a d

Explanation:
1. Note the total time (c), recognise that the path is twice the depth (b), write the speed formula (a), then substitute (d).
2. The order is c b a d.
3. Check: d = 1500 × 4 ÷ 2 = 3000 m = 3 km.

Teacher's Note:
1. Collect the data, set up the relation, then substitute.
2. The echo path is twice the depth.

Answer: (a) A → e, B → d, C → c, D → a, E → b

Explanation:
1. Frequency is the number of oscillations divided by the time taken: A → e.
2. Loudness depends on amplitude: B → d. Angle of incidence equals angle of reflection: C → c.
3. Ultrasound detects flaws in metals: D → a. The ghatam is a ghana vadya: E → b.
4. This is option (a).

Teacher's Note:
1. Match the sure pairs first and eliminate options.
2. Option (c) even uses d twice, so it cannot be right.

Answer: (d) c a b

Explanation:
1. First do the activity (c), then observe the two sounds (a), and then draw the conclusion (b).
2. The order is c a b.

Teacher's Note:
1. Procedure, observation, conclusion: that is the natural order of any experiment.
2. The conclusion is always the last step.

Answer: (d) painful to the ear.

Explanation:
1. 15 B to 16 B means 150 dB to 160 dB.
2. Sound above about 90 dB is painful, so it is painful to the ear.

Teacher's Note:
1. The printed question has dB; it is changed to B (bel) because the book's hint and answer treat it as 15 B (150 dB).
2. 1 bel = 10 decibels.

Answer: (d) zero

Explanation:
1. Sound needs a medium and cannot travel in a vacuum.
2. So its velocity there is zero.

Teacher's Note:
1. 3 × 10⁸ m s⁻¹ is for light.
2. No medium means no propagation.

Answer: (a) the louder is the sound produced.

Explanation:
1. More energy gives a larger amplitude of vibration.
2. A larger amplitude makes the sound louder.

Teacher's Note:
1. Strike a bell gently and then hard: more energy gives a louder note.
2. Loudness and energy go up together.

Answer: (a) All vibrations produce sound.

Explanation:
1. Every vibrating body produces sound, but only vibrations between 20 Hz and 20 000 Hz are audible to us.
2. So the statement that all vibrations produce sound is correct as per the book.

Teacher's Note:
1. This question repeats Test Your Concepts Question 23.
2. Producing and hearing are different things.

Answer: (a) its amplitude.

Explanation:
1. The largest displacement from the mean position is the definition of amplitude.

Teacher's Note:
1. Amplitude is measured from the mean position, not from extreme to extreme.
2. Frequency is a different quantity.

Answer: (a) Oboes

Explanation:
1. Instruments in which an air column vibrates are wind instruments.
2. The oboe is a wind instrument; the veena and violin are stringed and the mridangam is a percussion instrument.
3. The answer is oboes.

Teacher's Note:
1. Match the group with the vibrating part: air column means wind instrument.
2. Strings and membranes vibrate in the other groups.

Answer: (a) 33

Explanation:
1. Two claps are heard separately if they reach P₃ at least 0.1 s apart.
2. Extra distance = 330 × 0.1 = 33 m.

Teacher's Note:
1. The ear needs a gap of 0.1 s.
2. The extra path of the farther clap is the distance between P₁ and P₂.

Answer: (b) The pitch of sound increases with a decrease in the thickness of a vibrating string.

Explanation:
1. Shorter air columns or strings raise the pitch, so (a) and (c) are wrong.
2. Thinner strings vibrate faster and give a higher pitch, so (b) is right; amplitude does not decide pitch.
3. The answer is (b).

Teacher's Note:
1. This repeats Concept Application Question 45.
2. Thin and short means shrill.

Answer: (a) increases

Explanation:
1. Time period depends on the length of the pendulum; a longer pendulum has a longer period.
2. So the time period increases.

Teacher's Note:
1. T = 2π√(ℓ/g).
2. Length up means period up.

Answer: (b) 30°

Explanation:
1. Angle of reflection = angle of incidence = 30°.

Teacher's Note:
1. Both angles are measured from the normal.
2. The angle with the surface would be 60°.

Answer: (a) b a c d

Explanation:
1. Note the time gap (b), take the speed of sound (a), write height = speed × time (c), substitute (d).
2. The order is b a c d.
3. Height = 330 × 3 = 990 m.

Teacher's Note:
1. Collect data first, then the formula, then substitute.
2. Light takes negligible time, so only the sound travel time counts.

Answer: (d) A → d, B → c, C → a, D → b, E → e

Explanation:
1. Cloud height = speed of sound × time interval: A → d.
2. T = 1/10 = 0.1 s: B → c. Gunny bags absorb sound: C → a.
3. Above 90 dB is painful: D → b. The syrinx is a ring of cartilage in birds: E → e.
4. This is option (d).

Teacher's Note:
1. Do the easy matches first and eliminate wrong options.
2. Only one option has A → d and C → a.

Answer: (c) b c a d

Explanation:
1. Fill the container (b), hit the coins under water (c), put the ear against the container and repeat (a), and conclude (d).
2. The order is b c a d.

Teacher's Note:
1. Setup, first trial, improved trial, conclusion.
2. The conclusion cannot come before the observation.

Answer: (a) not audible

Explanation:
1. The softest sound a human can sense is about 10 dB.
2. 5 to 8 dB is below it, so not audible.

Teacher's Note:
1. Compare with the 10 dB lower limit.
2. Painful sounds are above about 90 dB.

Answer: (b) vacuum

Explanation:
1. Sound does not travel through a vacuum, so its velocity there is zero, the minimum.
2. The answer is vacuum.

Teacher's Note:
1. Of the real media, air is the slowest, but the vacuum has no sound at all.
2. This repeats Test Your Concepts Question 21.

Answer: (a) the amplitude of vibrating body is more.

Explanation:
1. More energy makes the body move farther from its mean position.
2. So the amplitude is more.

Teacher's Note:
1. Energy and amplitude go together.
2. Options (b) and (c) contradict this.

Answer: (d) \( \dfrac{\text{time period}}{4} \)

Explanation:
1. A complete oscillation has four equal quarter movements.
2. Mean position to an extreme is one quarter, so the time is T/4.

Teacher's Note:
1. Visualise the four parts of one oscillation.
2. The printed options are numbered (1) to (4); they are shown here as (a) to (d).

Answer: (a) wind or reed instruments.

Explanation:
1. Hollow pipes with side holes in which an air column vibrates are wind or reed instruments.
2. Stringed instruments have strings and drums have membranes.

Teacher's Note:
1. The holes let the player change the length of the vibrating air column.
2. The flute is a good example.

Answer: (c) 363 m

Explanation:
1. The clap of P₁ reaches P₂ in 330 ÷ 330 = 1 s.
2. The clap of P₃ must arrive at least 0.1 s later, that is, after 1.1 s.
3. Distance = 330 × 1.1 = 363 m.

Teacher's Note:
1. This repeats Concept Application Question 48.
2. Add the 0.1 s persistence time to the first sound's time.

Answer: (b) The frequency of vibration determines the pitch of any sound.

Explanation:
1. Pitch depends on frequency, so (b) is correct.
2. Amplitude does not change pitch, a woman's voice is higher than a man's, and a tabla is not shriller than a flute.
3. The answer is (b).

Teacher's Note:
1. Frequency fixes pitch.
2. This repeats Concept Application Question 56.

Answer: (c) A is true, but R is false.

Explanation:
1. Assertion: stretched membranes of percussion instruments give different notes at different tensions, which is true.
2. Reason: flute, bassoon and oboes are wind instruments, not drum instruments, so R is false.
3. A is true but R is false: (c).

Teacher's Note:
1. Check each statement separately before judging the explanation.
2. Drums, tabla and mridangam are percussion instruments.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Both instruments have the same pitch (same length and thickness) and loudness, yet sound different.
2. The reason is the difference in quality (timbre), which is exactly what R states.
3. Both are true and R explains A: (a).

Teacher's Note:
1. Quality tells apart two sounds with the same pitch and loudness.
2. Link the example to the property.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The assertion describes a real situation and is true.
2. The reason as printed mixes up the definitions; the intended point is that music is pleasant and noise unpleasant. Taken as a general definition it is acceptable.
3. The true reason for A is that whether a sound is music or noise depends on the listener, and R does not state that. So both are true but R is not the explanation: (b).

Teacher's Note:
1. The book's hint says R is correct but is not the reason.
2. The same sound can be music for one person and noise for another.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The pitch of a vibrating string can be changed, which is true.
2. It can be done by changing length, tension or thickness, which is also true and shows how the pitch changes.
3. So (a): R explains A.

Teacher's Note:
1. This is how a violin or guitar is tuned.
2. Shorter, tighter or thinner means higher pitch.