# Q.8.    Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.(Given: Speed of sound in air is  $346\ m/s$; Speed of sound in aluminium is $6240\ m/s$ )

Here, let us assume the length of the rod be $'l'$.

Then,

Time taken by the sound wave in air:

$\left ( Given: v_{air} = 346\ m/s \right )$

$Time = \frac{Distance}{Speed}$

$t_{air} = \frac{l}{346} seconds$

Time taken by the sound wave in Aluminium to reach from one end to the other end,

$\left ( Given: v_{aluminium} = 6420\ m/s \right )$

$t_{aluminium} = \frac{l}{6420} seconds$

Therefore, the ratio of the time taken by the sound wave in air and in aluminium will be:

$\frac{t_{air}}{t_{aluminium}} = \frac{\frac{l}{346}}{\frac{l}{6420}} = \frac{6420}{346} = \frac{18.55}{1}$

Hence, the ratio is $18.55:1$.

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