Question

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 $10^{4}$ . Then the height of the hill is

• A. $250 \, m$
• B. $2.5 \, km$
• C. $1.25 \, km$
• D. $750 \, m$

Answer 1

Answer: (B)

Difference of pressure between sea level and the top of hill
$\Delta P=\left(h_{1} - h_{2}\right)\times \left(\rho \right)_{H g}\times g=\left(75 - 50\right)\times 10^{- 2}\times \left(\rho \right)_{H g}\times g$ .....(i)
and pressure difference due to h meter of air
$\Delta P=h\times \rho _{a i r}\times g$ .....(ii)
By equating (i) and (ii) we get
$h\times \left(\rho \right)_{a i r}\times g=\left(75 - 50\right)\times 10^{- 2}\times \left(\rho \right)_{H g}\times g$
$\therefore $ $h=25\times 10^{- 2}\left(\frac{\left(\rho \right)_{H g}}{\left(\rho \right)_{a i r}}\right)=25\times 10^{- 2}\times 10^{4}=2500 \, m$
$\therefore $ Height of the hill $=2.5 \, km$