Q. A stone is dropped into a well. If the depth of water below the top be h and velocity of sound in air be v, the time after which splash of sound is heard is
A) $\sqrt{\frac{2h}{g}} + \frac{h}{v}$
B) $\sqrt{\frac{2h}{g}} - \frac{h}{v}$
C) $\sqrt{\frac{2h}{g}}$
D) $\sqrt{\frac{2h}{g}} \times \frac{h}{v}$
Explanation
(a) Time taken by stone to reach the level of water $t_{1}$ is obtained from $s = ut + \frac{1}{2}at^{2}$
$h = 0 + \frac{1}{2}gt_{1}^{2},t_{1}\sqrt{\frac{2h}{g}}a$.
Time taken by sound to reach the top of well$t_{2} = \frac{h}{v}$.
Total time $t = t_{1} + t_{2} = \sqrt{\frac{2h}{g}} + \frac{h}{v}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.