Question

The height of mercury barometer is $75cm$ at sea level and $50cm$ at top of a hill. Ratio of density of mercury to that of air is $10^4$ . 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 the hill
$\Delta P=\left((h{)}_1-(h{)}_2\right)\times (\rho {)}_{Hg}\times g$
$=(75-50)\times (10{)}^{-2}\times (\rho {)}_{Hg}\times g\dots \dots .$ (i)
and pressure difference due to $h$ metres of air
$\Delta P=h\times {\rho }_{\text{air }}\times g\dots \dots$ (ii)
By equating (i) and (ii) we get
$h\times (\rho {)}_{air}\times g=(75-50)\times (10{)}^{-2}\times (\rho {)}_{Hg}\times g$
$\therefore h=25\times (10{)}^{-2}\left(\frac{(\rho {)}_{Hg}}{(\rho {)}_{air}}\right)$
$=25\times (10{)}^{-2}\times (10{)}^4=2500m$
$\therefore$ Height of the hill $=2.5km$.