Question

The height of a mercury barometer is $75cm$ at sea level and $50cm$ 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. $250m$
• B. $2.5km$
• C. $1.25km$
• D. $750m$

Answer 1

Answer: (B)

Difference of pressure between sea level and the top of hill
$\Delta P=\left(h_1-h_2\right)\times {\left(\rho \right)}_{Hg}\times g=\left(75-50\right)\times 10^{-2}\times {\left(\rho \right)}_{Hg}\times g$ .....(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
$h\times {\left(\rho \right)}_{air}\times g=\left(75-50\right)\times 10^{-2}\times {\left(\rho \right)}_{Hg}\times g$
$\therefore$ $h=25\times 10^{-2}\left(\frac{{\left(\rho \right)}_{Hg}}{{\left(\rho \right)}_{air}}\right)=25\times 10^{-2}\times 10^4=2500m$
$\therefore$ Height of the hill $=2.5km$