Question

Two children are at the opposite end of an iron pipe. One strucks the end of the pipe with a stone. Calculate the ratio of the time taken by sound waves in air and in iron to reach the other child. Use speed of sound in (i) air $=$ 344 m s${}^{-1}$ and (ii) iron $=$ 5130 m s${}^{-1}$.

• A. 15 : 10
• B. 15 : 1
• C. 15 : 2
• D. 15 : 4

Answer 1

The correct option is B 15 : 1

Given velocity, $V_1$ (in air) $=344m{s}^{-1}$,
$V_2$ (an iron) $=$ 5130 m s${}^{-1}$
Let $l\left(m\right)$ be the length of the pipe and $t_1$ and $t_2$ are time taken by sound in air and in iron respectively to reach the other end of the pipe $t=\frac{l}{v}$
$\therefore t_1=\frac{l}{v_1}=\frac{l}{344}s$ and $t_2=\frac{l}{v_2}=\frac{l}{5130}s$
Ratio
$\frac{t_1\left(inair\right)}{t_2\left(iniron\right)}=\frac{\left(\frac{l}{344}\right)}{\left(\frac{l}{5130}\right)}$
$=\frac{5130}{344}=14.9\approx 15:1$