Solved Sample Paper 2 — Question 17

(a) As we know,

$\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \\[0.5em]$

Total distance travelled by the sound in going and then coming back = 2d

Given,

d = 22 m

V = 330 ms^-1

Substituting the values in the formula above, we get,

$330 = \dfrac{2 \times 22}{t} \\[0.5em] \Rightarrow t = \dfrac{44}{330} \\[0.5em] \Rightarrow t = \dfrac{4}{30} \\[0.5em] \Rightarrow t = 0.13 \text{ s} \\[0.5em]$

Hence, the time after which he receives the reflected sound = 0.13 seconds

(b) As we know,

$\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \\[0.5em]$

t = 0.12 s

V = 330 ms^-1

d = ?

Substituting the values in the formula above, we get,

$330 = \dfrac{2d}{0.12} \\[0.5em] \Rightarrow d = \dfrac{330 \times 0.12}{2} \\[0.5em] \Rightarrow d = 19.8 \text{ m} \\[0.5em]$

Hence, distance to receive the sound in 0.12 s = 19.8 m