Question

The height of a mercury barometer is $75\text{cm}$ at sea level and $50\text{cm}$ at the top of a hill. Ratio of density of mercury to that of air is ${10}^4$. The height of the hill is

• A. $1.25\text{km}$
• B. $2.5\text{km}$
• C. $250\text{m}$
• D. $750\text{m}$

Answer 1

The correct option is B $2.5\text{km}$

Let the height of the hill be ${}^{\prime }{h}^{\prime }$.
Pressure difference between sea level and the top of hill
$\Delta p=(h_1-h_2)\times {\rho }_{Hg}\times g$
$\Delta p=(75-50)\times {10}^{-2}\times {\rho }_{Hg}\times g$ ...$\left(i\right)$
and pressure difference due to $‘h^{\prime }$ metre of air column,
$\Delta p=h\times {\rho }_{\text{air}}\times g$ ..... $\left(ii\right)$
By equating Eqs. $\left(i\right)$ and $\left(ii\right)$
$h\times {\rho }_{\text{air}}\times g=(75-50)\times {10}^{-2}\times {\rho }_{Hg}\times g$
$\Rightarrow h=25\times {10}^{-2}\times \left(\frac{{\rho }_{Hg}}{{\rho }_{\text{air}}}\right)$
$\Rightarrow h=25\times {10}^{-2}\times {10}^4\text{m}=2500\text{m}$
So, height of the hill is equal to the column of air of height $h=2500\text{m}$
$\therefore$ Height of the hill $=2.5\text{km}$