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. ${\left(\frac{H+x}{100}\right)}^2$
• B. $\frac{Hx}{(100+x)}$
• C. $\frac{Hx}{100}$
• D. $\frac{100H}{x}$

Answer 1

The correct option is C $\frac{Hx}{100}$

We have,

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

$V_2=V_1+\frac{x}{100}{V}_1$

$P_1{V}_1=P_2(V_2+\frac{x}{100}{V}_1)$

$P_1=P_2(1+\frac{x}{100})$

But,

$P_2=1atm$

Then,

$P_1=P_2+\frac{h}{H}$

$1+\frac{h}{H}=1+\frac{x}{100}$

$h=\frac{xH}{100}$