A stone is dropped into a well. If the depth of water below the top be h and the velocity of sound is v, then the splash in water is heard after T second. Then

Question

A stone is dropped into a well. If the depth of water below the top be h and the velocity of sound is v, then the splash in water is heard after T second. Then (A)  T=√(2h/g)+(h/v)  (B)  T=2√(2h/g)  (C)  T=(2h/v)  (D)  T=√(2h/g)× (h/v)

Tardigrade Admin · Accepted Answer

Required time = time taken by stone to reach the water surface + time taken by sound to reach the top. =√(2h/g)+(η /v)

Q. A stone is dropped into a well. If the depth of water below the top be h and the velocity of sound is v, then the splash in water is heard after T second. Then

$T = \frac{2 h}{g} + \frac{h}{v}$

$T = 2 \frac{2 h}{g}$

$T = \frac{2 h}{v}$

$T = \frac{2 h}{g} \times \frac{h}{v}$

Required time = time taken by stone to reach the water surface + time taken by sound to reach the top.
$= \frac{2 h}{g} + \frac{η}{v}$

Q. A stone is dropped into a well. If the depth of water below the top be h and the velocity of sound is v, then the splash in water is heard after T second. Then

T=g2h​​+vh​

T=2g2h​​

T=v2h​

T=g2h​​×vh​

Required time = time taken by stone to reach the water surface + time taken by sound to reach the top. =g2h​​+vη​

$T = \frac{2 h}{g} + \frac{h}{v}$

$T = 2 \frac{2 h}{g}$

$T = \frac{2 h}{v}$

$T = \frac{2 h}{g} \times \frac{h}{v}$

Required time = time taken by stone to reach the water surface + time taken by sound to reach the top.
$= \frac{2 h}{g} + \frac{η}{v}$