Question

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

• A. H
• B. 2H
• C. 7H
• D. 8H

Answer 1

The correct option is C 7H

We know, atmospheric pressure $P_o$ ​= $\rho gH$
where $\rho$ is the density of water.

Assuming the depth of the lake be d, the pressure at this depth will be, P = $P_o+\rho gd$ = $\rho g(H+d$)

Since, the temperature is constant, we can assume that the process takes place isothermally.
So, applying the Boyle's Law

$P_1{V}_1=P_2{V}_2$...(i)
where , 1 denotes the water at depth d and 2 denotes the surface of water.

So, $P_1=\rho g(d+H)$
at surface, it will be atmospheric pressure i.e $P_2=\rho gH$

Volume $V$ = $\frac{4}{3}\pi r^3$
on doubling the radius, the volume will increase to 8 times
i.e $V_2=8V_1$​
put up the values in equation (i)

Substituting the values, in Boyle's Law,
$P_1=\rho g(d+H)V=\rho gH\times 8V$

$d=8H-H$ = $7H$