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 the air and in aluminium to reach the second child.

Let length of the aluminium rod be l.

Speed of sound in air (v_air) at 25°C = 346 ms^-1

Speed of sound in aluminium (v_alu) at 25°C = 6420 ms^-1

We know,

Speed = $\dfrac{\text{Distance}}{\text{Time}}$

or

Time (t) = $\dfrac{\text{Distance}}{\text{Speed}}$

Substituting we get,

t_air = $\dfrac{\text{l}}{\text{346}}$

t_alu = $\dfrac{\text{l}}{\text{6420}}$

So, the ratio of time taken by sound in air and aluminium is

$\dfrac{\text{t}$\text{air}}{\text{t}\text{alu}} = $\dfrac{\dfrac{\text{l}}{\text{346}}}{\dfrac{\text{l}}{\text{6420}}}$ = $\dfrac{6420}{346}$ = 18.55

Hence, the ratio of time taken by sound in air and aluminium = 18.55