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प्रश्न
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
उत्तर १
1.01 x 105 Pa; 310 Pa
Radius of the mercury drop, r = 3.00 mm = 3 × 10–3 m
Surface tension of mercury, S = 4.65 × 10–1 N m–1
Atmospheric pressure, P0 = 1.01 × 105 Pa
Total pressure inside the mercury drop
= Excess pressure inside mercury + Atmospheric pressure
`= (2S)/r + P_0`
`= (2xx4.65xx10^(-1))/(3xx10^(-3)) + 1.01 xx 10^5`
= 1.0131 × 105
= 1.01 ×105 Pa
Excess Pressure = 2S/r
`= (2xx4.65xx10^(-1))/(3xx10^(-3)) = 310 Pa`
उत्तर २
Excess pressure = `(2sigma)/R` = (2xx4.65xx10^(-1))/(3xx10^(-3)) = 310 Pa`
Total Pressure = `1.01 xx 10^5 + (2sigma)/R`
=` 1.01 xx 10^5 + 310 = 1.0131 xx 10^5 Pa`
Since data is correct up to three significant figures, we should write total pressure inside the drop as 1.01 x 105 Pa.
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