Aug 24, 2017

Let the volume of the air bubble at the bottom of the lake be ${V}_{b}$. So its volume at the surface ${V}_{s} = 2 {V}_{b}$

Let the depth of the lake be $h c m$

So pressure at the bottom of the lake

${P}_{b} = \text{pressure of water of height h+ atmospheric prssure}$

${P}_{b} = h \times {d}_{w} \times g + 75 \times {d}_{H g} \times g$

Pressure at the surface

${P}_{s} = 75 \times {d}_{H g} \times g$

Now applying Boyle's law we get

${P}_{b} \times {V}_{b} = {P}_{s} \times {V}_{s}$

$\implies \left(h \times {d}_{w} \times g + 75 \times {d}_{H g} \times g\right) {V}_{b} = \left(75 \times {d}_{H g} \times g\right) \times 2 {V}_{b}$

$\implies h + 75 \times {d}_{H g} / {d}_{w} = 75 \times {d}_{H g} / {d}_{w} \times 2$

$\implies h = 75 \times {d}_{H g} / {d}_{w}$

$\implies h = 75 \times \frac{40}{3} = 1000 c m = 10 m$