Question Detail
Question
The speed of sound in water and air is 1450 m/s and 330 m/s respectively. Sound from an explosion on the surface of a lake first reaches me when my head is underwater and 5 s later when my head is above the water. How far away was the explosion?
Correct Answer
Option C is the correct answer.
Detailed Explanation
Let the distance be d. Time in water t_w = d/1450. Time in air t_a = d/330. The difference is 5 s: d/330 - d/1450 = 5 => d(1/330 - 1/1450) = 5 => d((1450-330)/(330×1450)) = 5 => d(1120/478500) = 5 => d = 5 × 478500 / 1120 = 2392500/1120 = 2136.16 m. None of the options match. Recheck: The sound reaches the underwater head first, then the air head 5 s later. The distance is the same. t_air - t_water = 5. d/330 - d/1450 = 5 => d(1/330 - 1/1450) = 5 => d(1120/(330×1450)) = 5 => d = 5 × 478500/1120 = 2136.16 m. None of the options. If the speed in water is 1450 m/s and in air is 330 m/s, and the time difference is 5 s, then the distance is about 2136 m. The closest is 1125.2 m, but that's not correct. There might be a misprint.
Hint
Use time = distance/speed. The difference in arrival times is 5 s.