The height of a mercury barometer is 75 cm at sea level and 50 cm at the top of a hill. Ratio of density of mercury to that of air is 104. The height of the hill is

Difference of pressure between sea level and the top of hill DP\[=({{h}_{1}}-{{h}_{2}})\times {{\rho }_{HG}}\times G\]\[=(75-50)\times {{10}^{-2}}\times {{\rho }_{Hg}}\times g\]   ?(i)                                 and pressure difference due to h meter of air                                  DP =\[h\times {{\rho }_{air}}\times g\]                                   ?(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.