Difference of pressure between sea level and the top of the hill

$\mathrm{\Delta }\text{P}=\left({\text{h}}_1-{\text{h}}_2\right)\times {\mathrm{\rho }}_{\text{Hg}}\times \text{g}=\left(75-50\right)\times {10}^{-2}\times {\mathrm{\rho }}_{\text{Hg}}\times \text{g}$      .......(i)

and pressure difference due to h metres of air $\mathrm{\Delta P}=\mathrm{h}\times {\mathrm{\rho }}_{\mathrm{air}}\times \mathrm{g}$     ......(ii)

By equating (i) and (ii) we get $\mathrm{h}\times {\mathrm{\rho }}_{\mathrm{air}}\times \mathrm{g}=(75-50)\times {10}^{-2}\times {\mathrm{\rho }}_{\mathrm{Hg}}\times \mathrm{g}$

$\therefore \text{h}=25\times {10}^{-2}\left(\frac{{\mathrm{\rho }}_{\text{Hg}}}{{\mathrm{\rho }}_{\text{air}}}\right)=25\times {10}^{-2}\times {10}^4=2500\text{m}$

∴Height of the hill =$2.5 \mathrm{km}$.