Ncert Exercises & Solutions

10.19 What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? The surface tension of mercury at that temperature (20 °C) is 4.65 × 10^–1 N m^–1. The atmospheric pressure is 1.01 × 10⁵ Pa. Also, give the excess pressure inside the drop.

Given,

The radius of the mercury drop, r = 3.00 mm = 3 × 10^–3 m

The surface tension of mercury, S = 4.65 × 10^–1 N m^–1

$Excess pressure inside the drop = \frac{2 S}{r} = \frac{2 \times 4.65 \times 10^{- 1}}{3 \times 10^{- 3}} = 310 Pa$

Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure

$= 310 + 1.01 \times 10^{5} = 1.0131 \times 10^{5} Pa \approx 1.01 \times 10^{5} Pa$

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