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 10⁴. The height of the hill is

• A. (a) 250 m
• A. (b) 2.5 km
• A. (c) 1.25 km
• A. (d) 750 m

Answer 1

(b)

Difference of pressure between sea level and the top of hill

∆P $= (h_{1} - h_{2}) \times \rho_{Hg} \times g = (75 - 50) \times 10^{- 2} \times \rho_{Hg} \times g$ ......(i)

and pressure difference due to h meter of air

∆P =$h \times \rho_{air} \times g$ ......(ii)

By equating (i) and (ii) we get

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

$\therefore\mspace{6mu} h = 25 \times 10^{- 2}\left( \frac{\rho_{Hg}}{\rho_{air}} \right) = 25 \times 10^{- 2} \times 10^{4} = 2500m$

∴ Height of the hill = 2.5 km.