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Mechanical Properties of Fluids

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Question : 19 of 31
Marks: +1, -0
What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65×10−1 N m−14.65 \times 10^{-1} \, \mathrm{N\,m}^{-1}. The atmospheric pressure is 1.01×105 Pa1.01 \times 10^{5} \, \mathrm{Pa}. Also give the excess pressure inside the drop.
Solution:  
Here, radius of drop, R=3.0 mm=3.0×10−3 mR=3.0\,\mathrm{mm}=3.0\times10^{-3}\,\mathrm{m}
Surface tension of mercury, T=4.65×10−1 N m−1T=4.65\times10^{-1}\,\mathrm{N\,m}^{-1}
Pressure outside the mercury drop,
P0=P_0= atmospheric pressure =1.01×105 Pa=1.01\times10^{5}\,\mathrm{Pa}
If PiP_iis pressure inside the drop, then excess of pressure inside the mercury drop,
Pi−P0=2TRP_i-P_0=\frac{2T}{R}
=2×4.65×10−13.0×10−3=\frac{2 \times 4.65 \times 10^{-1}}{3.0 \times 10^{-3}}
=310 N m−2=310\,\mathrm{N\,m}^{-2}
Hence, pressure inside the mercury drop,
Pi=P0+2TRP_i=P_0+\frac{2T}{R}
=1.01×105+310=1.01\times10^{5}+310
=1.0131×105 Pa=1.0131\times10^{5}\,\mathrm{Pa} .
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