CBSE Class 7 Science Advanced Physics Sound MCQs Set 01

Answer: (c) Both (a) and (b)

Explanation:
1. The prongs move to and fro about a mean position, so the motion is oscillatory.
2. The motion also repeats itself after equal intervals of time, so it is periodic.
3. Hence the vibrations are both oscillatory and periodic.

Teacher's Note:
1. A motion can be oscillatory and periodic at the same time.
2. Think of a pendulum: it swings about a fixed point and repeats at regular intervals.

Answer: (c) between 20 Hz and 20 kHz

Explanation:
1. Below 20 Hz the sound is infrasonic and above 20 000 Hz (20 kHz) it is ultrasonic; humans hear neither.
2. The audible range is therefore between 20 Hz and 20 kHz.

Teacher's Note:
1. Remember the audible range as 20 Hz to 20 000 Hz.
2. Infrasonic is below the range; ultrasonic is above it.

Answer: (c) 0.1

Explanation:
1. Frequency f = 1 ÷ T.
2. f = 1 ÷ 10 = 0.1 Hz.

Teacher's Note:
1. Frequency and time period are reciprocals of each other.
2. A long time period means a small frequency.

Answer: (b) in the second case is more.

Explanation:
1. A shorter vibrating string vibrates faster, so its frequency is higher.
2. The length is decreased in the second case, so the frequency is more in the second case.

Teacher's Note:
1. Shorter string means higher frequency and higher pitch; this is how a violinist changes notes.
2. The tension (1 kgwt) stays the same in both cases.

Answer: (d) All the above

Explanation:
1. An echo is a reflected sound.
2. Sonar uses reflected ultrasound to find fish shoals, and bats find insects from the reflected ultrasound.
3. All three use reflection of sound, so the answer is all the above.

Teacher's Note:
1. Whenever a sound bounces back from an obstacle, reflection is at work.
2. Sonar, echo and bat navigation are the standard examples.

Answer: (b) solids

Explanation:
1. The order of speed is solids > liquids > gases.
2. So the velocity is maximum in solids.

Teacher's Note:
1. Particles in solids are packed closely, so the disturbance passes on quickly.
2. Gases have the slowest sound.

Answer: (d) vacuum

Explanation:
1. Sound needs a material medium made of particles.
2. Air, glass and iron are media, but a vacuum has no particles.

Teacher's Note:
1. The bell jar experiment proves this.
2. Astronauts on the Moon cannot talk directly for the same reason.

Answer: (a) ultrasonic sound

Explanation:
1. Bats send out sound of frequency above 20 000 Hz and locate prey from its echo.
2. This is ultrasonic sound.

Teacher's Note:
1. Bat navigation is a favourite example of ultrasound.
2. Humans cannot hear these sounds.

Answer: (b) decreases

Explanation:
1. f = 1 ÷ T, so the frequency is inversely related to the time period.
2. When T increases, f decreases.

Teacher's Note:
1. Inverse relation: one goes up, the other goes down.
2. Doubling T halves f.

Answer: (c) V_g < V_ℓ < V_s

Explanation:
1. Sound is slowest in gases, faster in liquids and fastest in solids.
2. So V_g < V_ℓ < V_s, which is option (c).

Teacher's Note:
1. Option (a) gives the reverse order.
2. Typical values: air 330 m s⁻¹, water about 1500 m s⁻¹, steel several thousand m s⁻¹.

Answer: (b) vacuum

Explanation:
1. Sound cannot travel in a vacuum, so its velocity there is zero, the least possible.
2. So the velocity is minimum in a vacuum.

Teacher's Note:
1. Among real media air is slowest, but a vacuum carries no sound at all.
2. Zero is lower than any speed in a medium.

Answer: (a) its amplitude

Explanation:
1. The greatest distance moved from the mean position is the definition of amplitude.

Teacher's Note:
1. Amplitude is measured from the rest position to the extreme, not from one extreme to the other.
2. Its unit is metre or centimetre.

Answer: (a) All vibrations produce sound.

Explanation:
1. Every vibrating body produces sound waves in the medium around it.
2. The human ear can respond only to frequencies from 20 Hz to 20 000 Hz, but this limits what we hear, not what is produced.
3. So the statement that all vibrations produce sound is the one the book treats as correct.

Teacher's Note:
1. The printed question lacks option (b); it has been supplied from the identical question in Assessment Test 1.
2. Production of sound and hearing of sound are two different things.

Answer: (a) its amplitude

Explanation:
1. Louder sounds come from vibrations of larger amplitude.
2. Frequency decides the pitch, not the loudness.
3. So loudness depends on amplitude.

Teacher's Note:
1. Do not mix up loudness (amplitude) with pitch (frequency).
2. More energy means more amplitude and more loudness.

Answer: (b) a wind instrument

Explanation:
1. In a harmonium air is blown over reeds, which vibrate and give the notes.
2. It is therefore a wind (reed) instrument.

Teacher's Note:
1. Wind instruments produce sound from a vibrating air column or reeds.
2. Flute, oboe and harmonium belong to this group.

Answer: (b) 1/10

Explanation:
1. The bel is a larger unit; 1 bel = 10 decibel.
2. So 1 decibel = 1/10 bel.

Teacher's Note:
1. The prefix deci means one tenth.
2. Loudness is usually quoted in decibels (dB).

Answer: (d) painful to the ear

Explanation:
1. 15 B to 16 B equals 150 dB to 160 dB.
2. Sound above about 90 dB is painful and harmful, so this is painful to the ear.

Teacher's Note:
1. The letter B stands for bel, which is 10 dB; do not read it as dB.
2. Convert to decibels before judging the loudness.

Answer: (c) BDAECFG

Explanation:
1. First set up the pendulum (B), then pull and release the bob (D).
2. Start the watch and count (A), stop after 20 oscillations to get t₁ (E), repeat to get t₂ (C), average (F), then calculate T and f (G).
3. The sequence is B, D, A, E, C, F, G, which is BDAECFG.

Teacher's Note:
1. Think of it as setting up, doing, repeating and calculating.
2. t₁ must be obtained before the repeat for t₂.

Answer: (b) ABDCFE

Explanation:
1. Place the persons (A), fire the cracker and start the watch (B), stop the watch on hearing (D), note the time (C), measure the distance (F), then find the speed (E).
2. The sequence is ABDCFE.

Teacher's Note:
1. Time and distance must both be known before the speed is calculated.
2. The speed formula is always the last step.

Answer: (d) All the above

Explanation:
1. Ultrasonic baths clean dishes and instruments, ultrasound scanning is used in medical diagnosis, and ultrasound shows cracks inside metals.
2. So the answer is all the above.

Teacher's Note:
1. Ultrasound has many uses because it can be sent in narrow beams.
2. Learn at least the medical, industrial and cleaning uses.

Answer: (c) 1/3

Explanation:
1. Time = distance ÷ speed.
2. t = 110 ÷ 330 = 1/3 s.

Teacher's Note:
1. Divide the distance by the speed to get the time.
2. 110 is exactly one third of 330.

Answer: (a) 440

Explanation:
1. Distance = velocity × time.
2. d = 220 × 2 = 440 m.

Teacher's Note:
1. Use distance = speed × time.
2. Keep the units in metres and seconds.

Answer: (b) 247.5

Explanation:
1. The sound goes to the tower and back, so the distance covered is 2d.
2. 2d = 330 × 1.5 = 495 m.
3. d = 495 ÷ 2 = 247.5 m.

Teacher's Note:
1. For echoes always divide by 2, since the sound travels the distance twice.
2. Option (a) is the total path, not the distance of the tower.

Answer: (a) the speed of light is more than the speed of sound.

Explanation:
1. Light travels at 3 × 10⁸ m s⁻¹ and sound at about 330 m s⁻¹.
2. Both start together, but light arrives first because its speed is much greater.

Teacher's Note:
1. Count the seconds between flash and thunder to find how far the cloud is.
2. Sound does not travel in a vacuum at all.

Answer: (c) cork

Explanation:
1. Hard, smooth surfaces such as metal, plywood and still water reflect sound.
2. Soft, porous cork absorbs it, so cork is the best absorber.

Teacher's Note:
1. Soft and porous means good absorber of sound.
2. Hard and smooth means good reflector.

Answer: (a) frequency

Explanation:
1. Pitch is how shrill or flat a sound is, and it depends on frequency.
2. So pitch is determined by the frequency.

Teacher's Note:
1. Frequency gives pitch; amplitude gives loudness.
2. A higher frequency gives a shriller sound.

Answer: (a) by 'A' will be more.

Explanation:
1. Pitch increases with frequency.
2. Fork A has 512 Hz, which is greater than 212 Hz of fork B.
3. So the pitch of A is higher.

Teacher's Note:
1. Compare the frequencies; the bigger one is the higher pitch.
2. Both lie within the audible range.

Answer: (b) 2

Explanation:
1. A seconds pendulum is defined as a pendulum whose time period is 2 s (one second for each swing).
2. So T = 2 s.

Teacher's Note:
1. The name comes from the fact that it ticks once every second, one tick per half oscillation.
2. Its frequency is 0.5 Hz.

Answer: (a) 0.01 s

Explanation:
1. T = 1 ÷ f = 1 ÷ 100 = 0.01 s.
2. The unit of time period is the second, so options with s⁻¹ are wrong.

Teacher's Note:
1. Check the unit as well as the number.
2. Time period is in seconds, frequency in hertz.

Answer: (d) zero

Explanation:
1. Sound cannot travel through a vacuum.
2. Hence its velocity there is zero. (3 × 10⁸ m s⁻¹ is the speed of light.)

Teacher's Note:
1. Do not confuse sound with light, which does travel through a vacuum.
2. 332 and 330 m s⁻¹ are values in air.

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

Explanation:
1. One full oscillation goes: mean to one extreme, back to mean, to the other extreme, and back to mean.
2. Mean to extreme is one of these four equal parts, so it takes T/4.

Teacher's Note:
1. Count the four quarters in a complete oscillation.
2. Mean to extreme and back is T/2.

Answer: (b) Infrasonic vibrations are used for homogenizing milk.

Explanation:
1. A Galton whistle gives ultrasound and dolphins use ultrasound, so (a) and (c) are true.
2. Milk is homogenised with ultrasonic vibrations, not infrasonic, so statement (b) is wrong.

Teacher's Note:
1. Read the question carefully: you have to find the wrong statement.
2. Infrasonic is below 20 Hz; ultrasonic is above 20 000 Hz.

Answer: (a) increases

Explanation:
1. Loudness is directly related to amplitude.
2. A larger amplitude means a louder sound.

Teacher's Note:
1. Hitting a drum harder gives larger vibrations and a louder sound.
2. Pitch stays the same when only the amplitude changes.

Answer: (c) percussion instrument

Explanation:
1. A tabla has stretched leather membranes that are struck.
2. This makes it a percussion instrument.

Teacher's Note:
1. Percussion instruments have stretched membranes; ghana vadya are solid instruments.
2. Mridangam and drums are in the same group.

Answer: (c) decibels

Explanation:
1. Seconds measure time, metres measure length and hertz measure frequency.
2. Loudness is measured in decibels.

Teacher's Note:
1. dB is the symbol for decibel.
2. Match each quantity with its unit.

Answer: (a) not audible

Explanation:
1. Humans can sense sounds above about 10 dB.
2. 5 dB to 8 dB is below that, so it is not audible.

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

Answer: (a) increases

Explanation:
1. The time period of a simple pendulum depends on its length: a longer pendulum swings more slowly.
2. So T increases.

Teacher's Note:
1. The formula is T = 2π√(ℓ/g).
2. The mass of the bob does not matter.

Answer: (a) ABCDEF

Explanation:
1. Set up the plank and pipes (A), place the clock (B), listen at Q (C), measure the angles (D), repeat with another inclination (E), and conclude (F).
2. The order is ABCDEF.

Teacher's Note:
1. The conclusion always comes last.
2. Repeating the experiment (E) comes after the first measurement (D).

Answer: (d) BDAC

Explanation:
1. First the ultrasound is sent (B), then the reflected sound is received (D), then the time is measured (A), and finally the depth is found from v = 2d/t (C).
2. The order is BDAC.

Teacher's Note:
1. Send, receive, time, calculate.
2. The time can only be measured after the echo has returned.

Answer: (b) 60°

Explanation:
1. The angle of incidence is measured from the normal, not from the surface.
2. Angle with the normal = 90° − 30° = 60°.
3. By the law of reflection, the angle of reflection is 60°.

Teacher's Note:
1. Angles of incidence and reflection are always measured with the normal.
2. Subtract the given angle from 90° to find the angle with the normal.