Question

The volume of an air bubble increases by x% as it rises from the bottom of a lake to its surface. If the height of the water barometer is H, the depth of the lake is

• A. $\frac{Hx}{(100-x)}$
• B. $\frac{Hx}{(100+x)}$
• C. $\frac{Hx}{100}$
• D. 100H/x

Answer 1

Answer: (C)

$\frac{Hx}{100}$

Answer 2

We have, $P _{1} V _{1}= P _{2} V _{2}$ $V _{2}= V _{1}+\frac{ x }{100} V _{1}$ $P _{1} V _{1}= P _{2}\left( V _{2}+\frac{ x }{100} V _{1}\right)$ $P _{1}= P _{2}\left(1+\frac{ x }{100}\right)$ But, $P _{2}=1 atm$ Then, $P _{1}= P _{2}+\frac{ h }{ H }$ $1+\frac{ h }{ H }=1+\frac{ x }{100}$ $h =\frac{ xH }{100}$