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.

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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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