Explanation

(b) Difference of pressure between sea level and the top of hill
$\mathrm{\Delta}P = \left( h_{1} - h_{2} \right) \times \rho_{Hg} \times g = (75 - 50) \times 10^{- 2} \times \rho_{Hg} \times g$ …(i)
and pressure difference due to $h$ metre of air
$\mathrm{\Delta}P = h \times \rho_{air} \times g$ …(ii)
By equating (i) and (ii) we get
$h \times \rho_{air} \times g = (75 - 50) \times 10^{- 2} \times \rho_{Hg} \times g$
$\therefore h = 25 \times 10^{- 2}\left( \frac{\rho_{Hg}}{\rho_{air}} \right) = 25 \times 10^{- 2} \times 10^{4} = 2500m$
$\therefore$ Height of the hill $= 2.5\ km$