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 $10^4$. The height of the hill is ....... $km$
Diffcult
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(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 $= 2.5 km.$
$\Delta$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 $= 2.5 km.$
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