Question

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