First slide

Pressure in a fluid-mechanical properties of fluids

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

Moderate

A

250 m
B

2.5 km
C

1.25 km
D

750 m

Solution

Difference of pressure between sea level and the top of hill
$ΔP = (h_{1} - h_{2}) \times ρ_{Hg} \times g = (75 - 50) \times 10^{- 2} \times ρ_{Hg} \times g$ ………..(i)
and pressure difference due to h meter of air
$ΔP = h \times ρ_{air} \times g$ ………..(ii)
By equating (i) and (ii) we get
$\begin{array}{l} h \times ρ_{air} \times g = (75 - 50) \times 10^{- 2} \times ρ_{Hg} \times g \\ ∴ h = 25 \times 10^{- 2} (\frac{ρ_{Hg}}{ρ_{air}}) = 25 \times 10^{- 2} \times 10^{4} = 2500 m (∵ ρ_{h g} = 13600 kg / m^{3} & ρ_{air} = 1.26 kg / m^{3}) \end{array}$
$∴$ Height of the hill = 2.5 km.

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