Question

A firecracker exploding on the surface of a lake is heard as two sounds at a time interval \[t\] apart by a man on a boat close to the water surface. Sound travels by a speed \[v\] in air. The distance from the exploding firecracker to the boat is:
A. \[\dfrac{{uvt}}{{u + v}}\]
B. \[\dfrac{{t\left( {u + v} \right)}}{{uv}}\]
C. \[\dfrac{{t\left( {u - v} \right)}}{{uv}}\]
D. \[\dfrac{{uvt}}{{u - v}}\]

Answer 1

Hint:It is given in the problem that a firecracker exploding on the surface of a lake is heard as two sounds. These two sounds are from the same source and originate at same time. But, they travel through two different mediums; water and air and are heard at different times because speed of sound in water is greater than speed of sound in air.

Formula Used: The distance covered by the sound wave to travel through water will be: \[{d_w} =
u{t^\prime }\]
The distance covered by the sound wave to travel through air will be: \[{d_a} = v(t' + t)\]

Complete step by step solution: Sound travels by different velocities in water and air. The speed of sound in air is given as \[v\]. Let the speed of sound in water be \[u\]. The exploding firecracker is heard as two sounds a time interval \[t\] apart. This is because the sound travels faster in water than in air.
Therefore, the sound wave which is travelling by water will reach before the one travelling through air by an interval \[t\].
Thus, the distance covered by the sound wave to travel through water will be
\[{d_w} = u{t^\prime }\] \[ \to (1)\]
where, \[u\]is the velocity of sound in water and
\[t'\]is the time taken by a sound wave to travel through water.
And, the distance covered by the sound wave to travel through air will be
\[{d_a} = v(t' + t)\] \[ \to (2)\]
where, \[v\]is the velocity of sound in water and
\[t' + t\] is the time taken by sound wave to travel through water
The exploding firecracker and the boat are at the same distance. So, \[{d_w} = {d_a} = d\].
Therefore,\[ut' = v(t' + t)\].
Solving for \[t\].
\[ut' = v(t' + t) \to ut' = vt' + vt \to (u - v)t' = vt \to t' = \dfrac{{vt}}{{(u - v)}}\]
Substituting the value of \[t\]in equation (1)
\[d = u\left( {\dfrac{{vt}}{{(u - v)}}} \right) \to \dfrac{{uvt}}{{u - v}}\]

Hence, option (D) is the correct answer.

Note:The velocity of sound in water may vary due to pressure and density. Here in the problem, the firecracker is exploding on the surface of a lake; therefore the factor of pressure is not taken into account as pressure at the surface of water is very less. The pressure grows as deep one goes inside the water.