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

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

Answer 1

The correct option is B

2.5 km

Difference of pressure between sea level and the top of hill

$△P=(h_1-h_2)\times {\rho }_{Hg}\times g=(75-50)\times {10}^{-2}\times {\rho }_{Hg}\times g...\left(i\right)$

and pressure difference due to h meter of air

$△P=h\times {\rho }_{air}\times g...............\left(ii\right)$

By equating (i) and (ii) we get

$h\times {\rho }_{air}\times g=(75-50)\times {10}^{-2}\times {\rho }_{Hg}\times g$

$\therefore h=25\times {10}^{-2}⟮\frac{{\rho }_{Hg}}{{\rho }_{air}}⟯=25\times {10}^{-2}\times {10}^4=2500m$

$\therefore Heightofthehill=2.5km.$