A pebble is dropped freely in a well from its top. It takes $20 s e c$ for the pebble to reach the water surface in the well. Taking $g = 10 m s^{- 2}$ and speed of sound $= 330 m s^{- 1}$ , find : (i) the depth of water surface, (ii) the time when echo is heard after the pebble is dropped.

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Solution

$h = \frac{1}{2} \times 10 \times (20)^{2} = 5 \times 400 = 2000 m$
Now, $t_{1} =$ time of falling $= 20 s e c$
$t_{2} =$ time taken by sound
$= \frac{H}{V_{s o n i c}} = \frac{2000}{330} = 6.06 s$
$∴$ Time after dropping the pebble at which echo is heard $= 26.06 s e c$

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A pebble is dropped freely in a well from its top. It takes 20sec for the pebble to reach the water surface in the well. Taking g=10m s−2 and speed of sound =330m s−1, find : (i) the depth of water surface, (ii) the time when echo is heard after the pebble is dropped.

h=12×10×(20)2=5×400=2000mNow, t1= time of falling =20sect2= time taken by sound=HVsonic=2000330=6.06s∴ Time after dropping the pebble at which echo is heard =26.06sec

A pebble is dropped freely in a well from its top. It takes $20 s e c$ for the pebble to reach the water surface in the well. Taking $g = 10 m s^{- 2}$ and speed of sound $= 330 m s^{- 1}$ , find : (i) the depth of water surface, (ii) the time when echo is heard after the pebble is dropped.

$h = \frac{1}{2} \times 10 \times (20)^{2} = 5 \times 400 = 2000 m$
Now, $t_{1} =$ time of falling $= 20 s e c$
$t_{2} =$ time taken by sound
$= \frac{H}{V_{s o n i c}} = \frac{2000}{330} = 6.06 s$
$∴$ Time after dropping the pebble at which echo is heard $= 26.06 s e c$