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Given 

Speed of sound (u) = 346 m/s 

Time taken for hearing the echo (t) = 4s 

Now

speed = \(\frac{distance}{time}\)

⇒ 346 = \(\frac{distance}{4}\)

⇒ Distance travelled by the sound = 346 ×4 = 1384 m.

In 4 s, sound has to travel twice the distance between the cliff and aditi. Therefore, the distance between the cliff and aditi is =\(\frac{1384}{2}\) = 692 m

Distance of cliff from aditi = 692 m.

Ultrasonic waves are the sound waves with frequency more than 20 kHz. Animals like dolphins, bats and dogs produce ultrasound.

The ceilings of concert halls are curved because sound after reflection from the ceiling it reaches all the comers of the hall and is audible to each person in the hall.

A sound persists in our ear for 0.1 s. At temperature 22℃ in air and speed of sound 344 ms^-1 , the total distance covered by sound between the point of generation and reflecting surface should be at least (344 m/s) x 0.1 s = 34.4 m. Thus, the minimum distance between the point of generation and reflecting surface should be half of it i.e. 17.2 m. So, we cannot hear an echo in a small room as the minimum distance for the echo will not be satisfied. But we can hear an echo in a long gallery and big halls.

Stethescope is a medical instrument used for listening to sounds produced in the body. In stethescopes, the sounds reach doctor’s ears by multiple reflections that happen in the connecting tube.

We know, distance = speed × time 

2d = speed of ultrasound × time 

2d = 1531 × 3.42

∴ d = \(\frac{5236}{2}\) = 2618 m

Thus, the distance of the seabed from the ship is 2618 m or 2.618 km.

The distance of first cliff from the person, 2 × D1 = velocity x time

D1 = 320 × 4 / 2 = 640 m

Distance of the second cliff from the person, D2 = 320 × 6 / 2 = 960 m

Distance between cliffs = D1 + D2 = 640 + 960 = 1600 m

Distance of hill from the man

D1 = velocity x time/ 2 = v × 5 / 2----------------- (equation 1)

Now, D1 − 310 = v × 3 / 2--------------------------- (equation 2)

By subtracting equation 2 form equation 1 ,we get310 = v x (5/2 − 3/2)

So, v = 310m/s

Correct Answer is: (d) λ/8

Let a = amplitude due to each wave.

I₀ = k(2a)² = 4ka². 

Let ϕ = phase difference to obtain the intensity I₀ /2.

Amplitude = a' = √(a² + a² + 2a² cos ϕ) =2acos (ϕ/2).

∴ I₀ /2 = k[4a²cos²(ϕ/2)] = I₀cos² (ϕ/2).

or (ϕ/2) = 1/√2 

or ϕ/2 = π/4

or ϕ = π/2 = 2π/λ Δ, where Δ = path difference.

∴ Δ = λ/4 = 2x, where x = displacement of the sliding tube.

∴ x = λ/8.

Correct answer is (D) solid

Correct answer is (D) Absorb sound

Frequency is the no.of vibrations produced per second = 128 Hz 

Wavelength, λ = 2.7 m 

Speed = f λ 

= 345.6 m/s

Velocity, v = 340 m/s 

Frequency, f = 512 Hz 

Wavelength, λ = ? 

= 0.66 m

The intensity at any point of the medium is the amount of sound energy passing per second normally through unit area at that point.

The loudness of a sound depends on the energy conveyed by the sound wave near the eardrum of the listener. Loudness, being a sensation, also depends on the sensitivity of the ears of the listener. Thus the loudness of sound of a given intensity may differ from listener to listener. Further, two sounds of the same intensity but of different frequencies may differ in loudness even to the same listener because of the sensitivity of ears is different for different frequencies. So, loudness is a subjective quantity while intensity being a measurable quantity is an objective quantity for the sound wave.

Relationship between loudness L and intensity I is given as: L = K log I, where K is a constant of proportionality.

The intensity at any point of the medium is the amount of sound energy passing per second normally through unit area at that point. Its unit is microwatt per metre².

Because the board provides comparatively a large area and forces a large volume of air to vibrate and thereby increases the sound energy reaching our ears.

No, the ruler vibrates very slowly, which has frequency which is not audible.

In medical science, echo method of ultrasonic waves is used for the imaging of human organs such as liver, gall bladder, uterus, womb; which is called ultrasonography.

Longitudinal waves

The volume of the TV or radio in the house should be restricted to those watching the programmes.

During the recess in the school, we hear.

1. laughing 

2. students running 

3. talking loudly 

4. ringing of the bell 

5. stamping on the staircase 

6. shouting

Sound is said to be propagated when sound waves spread in all directions from a source of sound.

The substance around a source of sound through which sound waves spread is called the medium of propagation of sound.

Frequency, amplitude and Phase are the three characteristics which are used to describe oscillations (or vibrations).

a) Longitudinal wave 

b) Transverse wave

Solid: table 

Liquid: water 

Gas: air

Correct option is A) infrasonic

Hertz is the SI unit of frequency.

D) wave form

False The frequency is measured in second.

False The quality of a sound depends on the amplitude of wave.