Question

A large bubble rises from the bottom of the lake to
the surface and its radius becomes 4 times, find
the depth of lake if atmospheric pressure is equal
to that of column of water of height H.
(1) 2 H
(2) 60 H
(3) 55 H
(4) 63 H​

Answer 1

Answer:

Depth (h) = H(n3-1)

= H( 4 cube -1)

=H(64-1)

= 63H is the answer

Answer 2

The depth of the lake is 63H.

Explanation:

We can consider the process is isotherm ally so use Boyle`s law as,

$P_{1} V_{1} =P_{2} V_{2}$

Here, P_{1} is pressure at depth and P_{2} is pressure at the surface of water and V is the volume.

As the radius of bubbles 4 times, so the volume of the bubbles at the surface becomes,

$V_{2} =64V_{1}$

Now the atmospheric pressure is given by

$P_{atm} =P_{1} =d\times g\times H$

The pressure at depth h is given by

$P=P_{2} =P_{atm} +dgh$

Substitute these values in Boyle`a law,we get

$(dgH+dgh)V_{1} =(dgH)64V_{1}$

$h=63H$

Thus, the depth of the lake is 63H.

#Learn More:

Topic : Boyle`s law

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