If an air bubbles rises from the bottom of a mercury tank to the top its volume becomes `1(1)/(2)` times. When normal pressure is `76 cm` of `Hg` then the depth of the `Hg` tank is

A bubble rises from the bottom of a lake 90m deep on reaching the surface, its volume becomes (take atmospheric pressure equals to 10 m of water)

As an air bubble rises from the bottom of a lake to the surface, its volume is doubled. Find the depth of the lake. Take atmospheric pressure = 76 cm of Hg.

When a large bubble rises from the bottom of a lake to the surface its radius doubles. If atmospheric pressure is equal to that of column of water height H then the depth of lake is

An air bubble rises the bottom of a lake to the top and increases thereby its volume 15 times . If the atmospheric pressure is 75 cm of hg and density of lake water is 1.02 xx10^(3) kg//m^(3) then find the depth of lake ( Neglect surface tension).

When a large bubble rises from the bottom of a lake to the surface, its radius doubles. The atmospheric pressure is equal to that of a column of water of height H. The depth of the lake is

When an air bubble rise from the bottom of a deep lake to a point just below the water surface, the pressure of air inside the bubble

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

(a) A spherical tank of 1.2m radius is half filled with oil of relaive density 0.8 . If the tank is given a horizontal acceleration of 10m//s^(2) . Calculate the inclination of the oil surface to horizontal and maximum pressure on the tank. (b) 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. Find the depth of the lake.