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The height of a mercury barometer is $75 \mathrm{~cm}$ at sea level and $50 \mathrm{~cm}$ at the top of a hill. Ratio of density of mercury to that of air is $10^4$. The height of the hill is
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$2.5 \mathrm{~km}$
Difference of pressure between sea level and the top of hill
$\Delta P=\left(h_1-h_2\right) \times \rho_{H_g} \times g=(75-50) \times 10^{-2} \times \rho_{H_g} \times g$...(i)
and pressure difference due to $h$ meter of air
$\Delta P=h \times \rho_{\text {air }} \times g$...(ii)
By equating (i) and (ii) we get
$h \times \rho_{a b} \times g=(75-50) \times 10^{-2} \times \rho_{H_g} \times g$
$\therefore h=25 \times 10^{-2}\left(\frac{\rho_{H_g}}{\rho_{\text {aiv }}}\right)=25 \times 10^{-2} \times 10^4=2500 \mathrm{~m}$
$\therefore$ Height of the hill $=2.5 \mathrm{~km}$.
$\Delta P=\left(h_1-h_2\right) \times \rho_{H_g} \times g=(75-50) \times 10^{-2} \times \rho_{H_g} \times g$...(i)
and pressure difference due to $h$ meter of air
$\Delta P=h \times \rho_{\text {air }} \times g$...(ii)
By equating (i) and (ii) we get
$h \times \rho_{a b} \times g=(75-50) \times 10^{-2} \times \rho_{H_g} \times g$
$\therefore h=25 \times 10^{-2}\left(\frac{\rho_{H_g}}{\rho_{\text {aiv }}}\right)=25 \times 10^{-2} \times 10^4=2500 \mathrm{~m}$
$\therefore$ Height of the hill $=2.5 \mathrm{~km}$.
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