Correct Answer - B
Difference of pressure between sea level and the top of hill
`DeltaP =(h_(1)-h_(2))xxrho_(Hg)xxg=(75-50)xx10^(-2)xxrho_(Hg)xxg" "……(i)`
and pressure difference due to h meter of air `DeltaP =hxxrho_("air")xxg" "….(ii)`
By equating (i) and (ii) we get `hxx rho_("air")xx g =(75-50) xx 10^(-2) xx rho_(Hg) xx g`
`therefore h=25xx10^(-2)((rho_(Hg))/(rho_(air))) = 25xx10^(-2)xx10^(4) = 2500 m therefore " Height of the hill" = 2.5 km`.