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1. Wavelength (l): The distance traveled by the wave in a time equal to its period is called wavelength. 

2. Wave velocity (v): It is the distance traveled by a wave in one second.

(a) 5m, (b) 5m, (c) 50Hz, (d) 250ms^-1, (e) 500p ms^-1.

Given v = √(rP/ρ)…...(1)

According ideal gas law, P = (ρRT/M),where

ρ is the density, T is the temperature, M is the Molecular mass of the gas;

R – universal gas constant.

Substituting for P in (1) we get

v =√(rRT/M)

This shows that v is

1. Independent of pressure

2. Increases with temperature i.e. v ∝ √T

3. We know that the molecular mass of water (18) is less than that of N₂ (28) and O₂ (32). therefore as humidity increases, the effective molecular mass of air decreases and hence velocity increases.

Tension in the string. T = 200;

Mass of string, M = 2.50 g;

Length l = 20 m

Mass per unit length, m = M/L = 2.5/20

= 0.125 kg m^-1

Velocity, v = √{T/m} = √{200/0.125}

= 40 ms^-1

time taken by the disturbance to reach other end,

t = l/v = 20/40 = 0.5 s.

Mass, M = 2.10 kg

Length, l = 12;

v = 343 ms^-1

∴ Mass per unit length, m = M/ρ = 2.10/12

= 0.175 kg m^-1

∴ using √{T/m} = u ⇒ T = mv₂

or, T = 0.175 x (343)²

= 2.06 x 10⁴ N

Here,

h = 300 m

g = 9.8 ms^-2

and v = 340 ms^-1

Let t₁ be the time taken by the stone to reach at the surface of the pond.

Then, using s = ut + 1/2 at²

⇒ h = 0 x t + 1/2 x gt₁²

t₁ = √{(2 x 300)/(9.8)}  (Here, h = 300 m)

Also, if t₂ be the time taken by the sound to reach at a height h, then

t₂ = h/v = 300/342 = 0.88 s

∴ Total time after which sound of splash is heard

= t₁ + t₂

= (7.82 + 0.88) s

= 8.7 s

Roaring of a lion produces a sound of low pitch and high intensity whereas buzzing of mosquitoes produces a sound of high pitch and low intensity and hence the two sounds can be differentiated.  

Correct answer is (b) 117

(d) 100 Hz and 6 m

(d) 100 Hz and 6 m

(a) its phase changes by 180° , but velocity does not change

Correct answer is (b) v_m > v_h > v_a

(a) human beings cannot hear

Because as speed of sound in water is roughly four times the sound in air, hence refractive index u = Sin i/Sin r = Va/Vw = 1/4 = 0.25

For, refraction r_max = 900, i_max=140. Since i_max ≠ r_max hence, sounds gets reflected in air only and person deep inside the water cannot hear the sound.  

No, because the emission of light is a random and rapid phenomenon and instead of beats we get uniform intensity. 

For the formation of distinct beats, frequencies of two sources of sound should be nearly equal, i.e., difference in frequencies of two sources must be small, say less than 10. The impression of sound heard by our ears persists on our mind for 1/10th of a second. If another sound is heard before (1/10) second passes, the impressions of the two sounds mix up and our mind cannot distingush between the two.

In order to hear distinct beats, time-interval between two successive beats must be greater than 1/10 second. Therefore, frequency of beats must be less than 10, i.e., number of beats per second, which is equal to difference in frequencies of two sources, must be less than 10. Hence, the two sources should be of nearly equal frequencies.

Fundamental frequency of a closed organ pipe is, v = v - 4l, where v is the velocity of sound in air. When air is replaced by gas heavier than air, then velocity of sound in gas will decrease. (Because, v ∝ 1/√ρ, where ρ is the density of the gas)

Hence, the fundamental frequency will also decrease.

Here,

v = 1000 kHz

= 1000 x 10³ Hz = 10⁶ Hz

Velocity of sound in air, v_a = 340 ms^-1

Velocity of sound in water, v_w = 1486 ms^-1

(a) For reflected sound, medium remains the same

∴ Wavelength, λ = v_a/v = 340/10⁶

= 3.4 x 10^-4 m

(b) For the same transmitted into water,

Wavelength, λ = v_ω/v = 340/10⁶

= 1.49 x 10^-3 m

v = 1.7 km s^-1

= 1.7 x 1000 ms^-1

= 1700 ms^-1

Frequency, v = 4.2 MHz

= 4.2 x 10⁶ Hz

∴ λ = ν/v = {1700}/{4.2 x 10⁶}

= 4.05 x 10⁴ m

Pressure waves are the longitudinal waves which involve changes in volume and pressure during their propagation through a medium.

π/2 radians.

Y₁ = a sin (ω t – kx).

Y₂ = a cos (ω t – kx).

= a sin(ω t – kx + π /2).

∴ Phase difference between Y₁ and Y₂ is π/2 rad.

(1) Determination of velocity of a star or a galaxy:- When the light emitted by a star is examined, the spectrum is found to consist of many colours (i.e., spectral lines). If the star approaches the earth, the shift of spectral lines of the spectrum takes place towards the violet end of the spectrum. This is called violet or blue shift and indicates the decrease in wavelength or increase in frequency.

If the star recedes away from the earth, the shift of spectral lines takes place towards the red end of the spectrum. The shift is called red shift and indicates the increase in wavelength or decrease in frequency.

Suppose as star (i.e., source) of light is moving away from the earth (i.e., observer) with velocity v, the apparent frequency of the light waves emitted by the star is

v' = ({c}/{c + v})v,    ....(i)

where c is the velocity of light

Now v = c/λ and v' = c/λ'

Hence, c/λ' = {c}/{c + v} x c/λ

or, λ'/λ = {c + v}/{c} = 1 + v/c

or, λ'/λ - 1 = v/c or {λ' - λ}/{λ} = v/c

or, Δλ/λ = v/c, where Δλ = λ' = λ

or, v = c x Δλ/λ'

where c = 3 x 10⁸ ms^-1    ...(ii)

Knowing the change in wavelength (Δλ), we can calculate the velocity (v) of the star.

(2) Radar (Radio Detection and Ranging):- It consists of a radio transmitter which sends pulses of electromagnetic radiation and a radio receiver, which receives the echoes due to the reflection of these radiation from the objects in their way.

According to Doppler's effect, if the object (say aircraft) is approaching the radar, then the reflected waves shall have a higher frequency (or lower wavelength) than the transmitted waves. If the object is receding away from the radar, then the reflected waves will have a lower frequency (or higher wavelength) than the transmitted waves.

(3) Sonar (Sound Navigation and Ranging):- In case of sonar, the waves, (i.e., ultrasonic waves) are sent from the ship towards the bed of the sea. If the submarine is approaching the ship, the reflected waves from the submarine will have higher frequency than the transmitted waves and vice-versa.

(4) Doppler's effect is also used to determine the speed of rotation of the sun.It has been calculated as 2 km s^-1.

Correct answer is (c) \(\frac{1}{6}\)