b
(b)Difference of pressure between sea level and the top of hill
\(\Delta\)P\( = ({h_1} - {h_2}) \times {\rho _{Hg}} \times g\)\( = (75 - 50) \times {10^{ - 2}} \times {\rho _{Hg}} \times g\) …\((i)\)
and pressure difference due to h meter of air
\(\Delta\)P =\(h \times {\rho _{air}} \times g\) …\((ii)\)
By equating \((i)\) and \( (ii)\)  we get
\(h \times {\rho _{air}} \times g = (75 - 50) \times {10^{ - 2}} \times {\rho _{Hg}} \times g\)
\(\therefore \;h = 25 \times {10^{ - 2}}\left( {\frac{{{\rho _{Hg}}}}{{{\rho _{air}}}}} \right)\)\( = 25 \times {10^{ - 2}} \times {10^4} = 2500\,m\)
Height of the hill \(= 2.5 km.\)