Physics - Fluid Question with Solution | TestHub

Let $V_1$ and $V_2$ be the volumes of the air bubble at the bottom and surface of the lake respectively.

Since the diameter is doubled, the radius is also doubled. The volume of a sphere is proportional to the cube of its radius.

$V_2=\frac{4}{3}\pi (2r{)}^3=8\left(\frac{4}{3}\pi r^3\right)=8V_1$

Since the temperature remains constant, Boyle's law is applicable:

$P_1{V}_1=P_2{V}_2$

$P_1{V}_1=P_2(8V_1)$

$P_1=8P_2$

Let $H$ be the depth of the lake, ${\rho }_w$ be the density of water, and ${\rho }_m$ be the density of mercury. The relative density of mercury is $\rho$, so ${\rho }_m=\rho {\rho }_w$.

Pressure at the surface of the lake is given by $h$ meter of mercury column:

$P_2=h{\rho }_{m}g=h\rho {\rho }_{w}g$

Pressure at the bottom of the lake is:

$P_1=P_2+H{\rho }_{w}g$

Substituting $P_1$ into the equation $P_1=8P_2$:

$P_2+H{\rho }_{w}g=8P_2$

$H{\rho }_{w}g=7P_2$

Substituting the value of $P_2$:

$H{\rho }_{w}g=7(h\rho {\rho }_{w}g)$

$H=7h\rho$

Answer: 7 h $\rho$