Sound — Question 5

(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 = 300 km

Converting km to m, we get,

1 km = 1000 m

Therefore, 300 km = 3,00,000 m

Hence, d = 3 x 10⁵ m

V = 3 x 10⁸ ms^-1

Substituting the values in the formula above, we get,

$3 \times 10^8 = \dfrac{2 \times (3 \times 10^5) }{t} \\[0.5em] t = \dfrac{2 \times (3 \times 10^5)}{3 \times 10^8} \\[0.5em] t = 2 \times 10^{-3} \\[0.5em]$

Hence, the signal is received back after 2 x 10^-3 s