Difference of pressure between sea level and the top of hill

Δ Δ P $= (h_1 - h_2) \times \rho_{\mathrm{Hg}} \times g = (75 - 50) \times 10^{-2} \times \rho_{\mathrm{Hg}} \times g$ …(i)

and pressure difference due to h meter of air

Δ Δ P = $h \times \rho_{air} \times g$ …(ii)

By equating (i) and (ii) we get

$h \times \rho_{\text{air}} \times g = (75 - 50) \times 10^{-2} \times \rho_{\text{Hg}} \times g$ $\therefore h = 25 \times 10^{-2} \left(\frac{\rho_{\text{Hg}}}{\rho_{\text{air}}}\right) = 25 \times 10^{-2} \times 10^{4} = 2500 \text{m}$

∴ ∴ Height of the hill = 2.5 km.