Import Question JSON

\text{The pressure difference between the sea level and the top of the hill is due to the weight of the air column of height } h\text{, where } h \text{ is the height of the hill.} \text{This pressure difference can also be calculated from the difference in the height of the mercury column in the barometer.} \text{Let } \Delta P \text{ be the difference in pressure between sea level and the top of the hill.} \Delta P = (h_1 - h_2) \times \rho_{\text{Hg}} \times g \text{where } h_1 = 75 \text{ cm and } h_2 = 50 \text{ cm}. h_1 - h_2 = 75 - 50 = 25 \text{ cm} = 25 \times 10^{-2} \text{ m} \Delta P = (25 \times 10^{-2}) \times \rho_{\text{Hg}} \times g \quad \text{...(i)} \text{The pressure difference due to a } h \text{ meter air column is given by:} \Delta P = h \times \rho_{\text{air}} \times g \quad \text{...(ii)} \text{By equating (i) and (ii), we get:} h \times \rho_{\text{air}} \times g = (25 \times 10^{-2}) \times \rho_{\text{Hg}} \times g \text{Cancelling } g \text{ from both sides:} h \times \rho_{\text{air}} = (25 \times 10^{-2}) \times \rho_{\text{Hg}} h = (25 \times 10^{-2}) \times \left(\frac{\rho_{\text{Hg}}}{\rho_{\text{air}}}\right) \text{We are given that the ratio of the density of mercury to that of air is } 10^4. \frac{\rho_{\text{Hg}}}{\rho_{\text{air}}} = 10^4 \text{Substituting this value:} h = 25 \times 10^{-2} \times 10^4 h = 25 \times 10^2 \text{ m} h = 2500 \text{ m} = 2.5 \text{ km} \text{Height of the hill} = 2.5 \text{ km}