A bubble is at the bottom of the lake of depth $h$. As the bubble comes to sea level, its radius increases three times. If atmospheric pressure is equal to $l\ $metre of water column, then $h$ is equal to

Question

Dr. Abubacker · Accepted Answer

(a) From Boyle’s law
$pV =$ constant
$	herefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ p_{1}V_{1} = p_{2}V_{2}$
Here, $p_{1} = (h + l),\ \ V_{1} = \frac{4}{3}\pi r^{3}$
$p_{2} = l,\ \ V_{2} = \frac{4}{3}\pi(3{r)}^{3}$
$	herefore\ \ \ \ \ \ \ \ \ \ (h + l)\frac{4}{3}\pi r^{3} = l 	imes \frac{4}{3}\pi(3r)^{3}$
or$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h + l = 27l$
$	herefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h = 26l$
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Q. A bubble is at the bottom of the lake of depth $h$. As the bubble comes to sea level, its radius increases three times. If atmospheric pressure is equal to $l\ $metre of water column, then $h$ is equal to

Explanation

Reviewed by Dr. Abubacker

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