The volume of an air bubble becomes three times as it rises from the bottom of a lake to its surface.Assuming atmospheric pressure to be 75 cm of Hg and the density of water to be 1/10 of the densityof mercury,the depth of the lake is

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Q. The volume of an air bubble becomes three times as it rises from the bottom of a lake to its surface. Assuming atmospheric pressure to be 75 cm of Hg and the density of water to be 1/10 of the density of mercury, the depth of the lake.

The volume of an air bubble becomes three times as it rises from the bottom of a lake to its surface. Assuming atmospheric pressure to be 75 cm of Hg and the density of water to be 1/10 of the density of mercury, the depth of the lake is

Q. The volume of an air bubble becomes three times as it rises from the bottom of a take to its surface. Assuming atmospheric pressure to be

75 c m

H g

and the density of water to be

\frac{1}{10}

of the density of mercury, the depth of the take is :

Q. The volume of an air bubble is doubled as it rises from the bottom of a lake to its surface. If the atmospheric pressure is

H m

of mercury & the density of mercury is

n

times that of lake water. The depth of the lake is given as

x n H

. Find

x

Q. An air bubble of volume

V_{0}

is released by a fish at a depth

h

in a lake. The bubble rises to the surface. Assume constant temperature and standard atmospheric pressure

P

above the lake. The volume of the bubble just before touching the surface will be (density of water is

ρ)

(

Assume that the radius of bubble is neglegible

)

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