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1. These are basically curved surfaces (concave), which are used in auditoria and halls to improve the quality of sound. This board is placed such that the speaker is at the focus of the concave surface. 

2. The sound of the speaker is reflected towards the audience thus improving the quality of sound heard by the audience.

Time interval t = 1.6 s 

Velocity of sound in water = 1400 m/s. 

Depth of the sea = \(\frac{v \times t}{2}\)=\(\frac{1400 \times 1.6}{2}\) = 1120 m 

Depth of the sea = 1120 m

Distance of the observer = 400 m 

Frequency of sound n = 600 Hz 

Let velocity of sound be v = 330 \(\frac{m}{s}\)

Wavelength λ = \(\frac{v}{n}\) = \(\frac{330}{600}\)

= 55 × 10^-2 = 0.55 m

Distance between two successive compressions is \(\frac{λ}{2}\)

Time period successive compressions is = \(\frac{1}{\frac{λ}{2}}\)

= \(\frac{2}{λ}\) = \(\frac{2}{0.55}\)

T = \(\frac{2}{0.55}\) = 3.6363 second

Time period = 3.6363 second.

(d) pressure of the gas

(c) 51.2π m/s

(b) Distance between the listener and the source. 

Apparent frequency in Doppler’s effect depends on frequency of the source, direction and velocity of source and listener. It does not depend on the distance between the listener and the source.

(b) If both the assertion and the reason are true but the reason is not the correct explanation of the assertion.

Velocity of sound at 0°C is V₀ = 331 ms^-1 

Let the sound at 46°C is be Vt

\(\frac{v_t}{v_0}\) = \(\sqrt\frac{273 +T}{273}\)

where temperature is T = 46°C

= \(\frac{V_t}{331}\) = \(\sqrt{\frac{273 + 46}{273}}\)

= \(\sqrt{\frac{319}{273}}\) = \(\sqrt{1.168}\)

= 1.080

V_t= 331 × 1.080 = 357.48 m/s. 

∴ Velocity of sound in air at that temperature = 357.48 m/s.

(a) V_T = (V₀ + 0.61 T)

(b) compressions and rarefactions

(b) ultrasonic waves

(i) – (D) 

(ii) – (A) 

(iii) – (B) 

(iv) – (E)

(c) ultrasonic wave

A. (iii) 

B. (iv) 

C. (ii) 

D. (i)

(c) audible waves

Correct answer is (c) 0.02752 m

In SONAR, by measuring the change in the frequency between the sent signal and received signal, the speed of marine animals and submarines can be determined.

In RADAR, radio waves are sent, and the reflected waves are detected by the receiver of the RADAR station. From the frequency change, the speed and location of the aeroplanes and aircrafts are tracked.

The frequency of radio waves emitted by a satellite decreases as the satellite passes away from the Earth. By measuring the change in the frequency of the radio waves, the location of the satellites is studied.

When ever there is a relative motion between a source and a listener, the frequency of the sound heard by the listener is different from the original frequency of sound emitted by the source. This is known as “Doppler effect”.

(d) All the above

A megaphone is a horn-shaped device used to address a small gathering of people. Its one end is wide and the other end is narrow. When a person speaks . at the narrow end, the sound of his speech is concentrated by the multiple reflections from the walls of the tube. Thus, his voice can be heard loudly over a long distance.

An echo is the sound reproduced due to the reflection of the original sound from various rigid surfaces such as walls, ceilings, surfaces of mountains, etc…

An echo can only be heard if it reaches the ear after 0.1 secs. 

Time taken = \(\frac{Total\,distance}{Velocity}\)

As on a hotter day, the velocity of sound is more. So if the time taken by the echo is less than 0.1 sec than it won’t be heard.