What is the pressure inside the drop of mercury of radius $3.00 \;mm$ at room temperature? Surface tension of mercury at that temperature $\left(20\,^{\circ} C \right)$ is $4.65 \times 10^{-1}\; N m ^{-1} .$ The atmospheric pressure is $1.01 \times 10^{5}\; Pa$. Also give the excess pressure inside the drop.
What is the pressure inside the drop of mercury of radius $3.00 \;mm$ at room temperature? Surface tension of mercury at that temperature $\left(20\,^{\circ} C \right)$ is $4.65 \times 10^{-1}\; N m ^{-1} .$ The atmospheric pressure is $1.01 \times 10^{5}\; Pa$. Also give the excess pressure inside the drop.
Radius of the mercury drop, $r=3.00 mm =3 \times 10^{-3} m$
Surface tension of mercury, $S=4.65 \times 10^{-1} N m ^{-1}$
Atmospheric pressure, $P_{0}=1.01 \times 10^{5} Pa$
Total pressure inside the mercury drop
$=$ Excess pressure inside mercury $+$ Atmospheric pressure $=\frac{2 S}{r}+P_{0}$
$=\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}}+1.01 \times 10^{5}$
$=1.0131 \times 10^{5}=1.01 \times 10^{5} Pa$
Excess pressure $=\frac{2 S}{r}=\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}}$
$=310\, Pa$
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