Question

Two rods P and Q are considered such that Young's modulus of elasticity of rod P is twice that of the other while the density of rod Q is eighteen times that of rod P. If a sound wave is allowed to transverse 2 m distance through each rod, then by how many times the time taken through P is less than that of Q ?

• A. 2 times
• B. 3 times
• C. 4 times
• D. 5 times

Answer 1

The correct option is B 3 times

Let time taken to traverse 2 m distance by rods P and Q able $t_p$ and $t_Q$ respectively.
Velocity of sound in solids $=\sqrt{\frac{Y}{d}}$
Given $Y_P=2Y_Q$
$d{P}_Q=18d_P$
$\therefore \frac{V_P}{V_Q}=\sqrt{\frac{Y_P}{Y_Q}\times \frac{d_Q}{d_P}}$
$=\sqrt{\frac{1}{2}\times \frac{18}{1}}=\sqrt{9}=3$
$V_P=3V_Q$
$V\alpha \frac{1}{t}$ (for same distance)
$\therefore$ Time taken by rod P is less than that of Q and by 3 times.