An air bubble released at the bottom of a lake, rises and on reaching

1.	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» rong>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., $\sf p_{1} V_{1} = p_{2}V_{2}$ $\sf (H + h) \bigg(\dfrac{4}{3}\pi r^{3} \bigg) = H \bigg[ \dfrac{4}{3} \pi (2r)^{3}\bigg]$ $\sf (H + h) = 8H$ $\sf h = 7H$

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»

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.,

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

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

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

$\sf h = 7H$

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