Question

An air bubble released at the bottom of a lake, rises and on reaching the top, its radius found to be doubled. If the atmospheric pressure is equivalent to H metre of water column, find the depth of the lake (Assume that the temperature of water in the lake is uniform)​

Answer 1

AnSwer :-

Volume of the air bubble at the bottom of the lake

$\sf (V_1) = \dfrac{4}{3}\pi (2r)^{3}$

Volume of the air bubble at the surface of the lake

$\sf (V_2) = \dfrac{4}{3}\pi (2r)^{3}$

Pressure at the surface of the lake (P₂) = H meter of water column. if 'h' is the depth of the lake (P₁) = (H + h) metre of water column.

since the temperature of the lake is uniform

According to Boyle's law, .i.e.,

$\sf p_{1} V_{1} = p_{2}V_{2}$

$\sf (H + h) \bigg(\dfrac{4}{3}\pi r^{3} \bigg) = H \bigg[ \dfrac{4}{3} \pi (2r)^{3}\bigg]$

$\sf (H + h) = 8H$

$\sf h = 7H$