The sound from a very high burst of fireworks takes 5 s to arrive at the observer. The burst occurs 1662 m above the observer and travels vertically through two stratifier layers of air, the top one of thickness
$d_{1}$ at $0^{\circ} C$ and the bottom one of thickness $d_{2}$ at $20^{\circ} C$ . Then (assume velocity of sound at $0^{\circ} C$ is 330 m/s)

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Solution

The correct option is D
Time taken is given by
$T = t_{1} + t_{2} = \frac{d_{1}}{v_{1}} + \frac{d_{2}}{v_{2}} V_{1} = v_{0}^{\circ} c = 330 m / s v_{2} = (330 + .06 t) = 342 m / s d = 1662 m ∴ T = \frac{d_{1}}{330} + \frac{(d - d_{1})}{342} = 5 s \frac{d_{1} (342 - 330)}{330 \times 342} + \frac{d}{342} = 5 s 12 d_{1} = 5 (342 \times 330) - 330 \times 1662 d_{1} = 1320 m d_{2} = 342 m$

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