A pebble is dropped into a well of depth h. The splash is heard after time t. If c be the velocity of sound, then

Question

A pebble is dropped into a well of depth h. The splash is heard after time t. If c be the velocity of sound, then (A) t=(g h2/c v2) (B) t=(c+v/g) (C) t=(c-v/g) (D) t=√(2 h/g)+(h/c)

Tardigrade Admin · Accepted Answer

To go down, the stone takes a time t1, Using, s=u t+a t2 ; h=(0 ⋅ t1)+(1/2) g t12 ⇒ t1=√(2 h/g) The sound takes a time t2=(h/c) to reach back to the ear Total time taken =(t1+t2)=√(2 h/g)+(h/c)

Q. A pebble is dropped into a well of depth $h$ . The splash is heard after time $t$ . If $c$ be the velocity of sound, then

$t = \frac{g h ^{2}}{c v ^{2}}$

$t = \frac{c + v}{g}$

$t = \frac{c - v}{g}$

$t = \frac{2 h}{g} + \frac{h}{c}$

To go down, the stone takes a time $t_{1}$ ,
Using, $s = u t + a t^{2}; h = (0 \cdot t_{1}) + \frac{1}{2} g t_{1}^{2}$
$\Rightarrow t_{1} = \frac{2 h}{g}$
The sound takes a time $t_{2} = \frac{h}{c}$ to reach back to the ear
Total time taken $= (t_{1} + t_{2}) = \frac{2 h}{g} + \frac{h}{c}$

Q. A pebble is dropped into a well of depth h. The splash is heard after time t. If c be the velocity of sound, then

t=cv2gh2​

t=gc+v​

t=gc−v​

t=g2h​​+ch​