What is the pressure inside the drop of mercury of radius 3.00mm … - Vidyadip

Question

What is the pressure inside the drop of mercury of radius $3.00mm$ at room temperature? Surface tension of mercury at that temperature $(20^\circ C) is 4.65 \times 10^{–1}N m^{–1}$. The atmospheric pressure is $1.01 × 105Pa$. Also give the excess pressure inside the drop.

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Answer

Radius of the mercury drop, $\mathrm{r}=3.00 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}$ Surface tension of mercury, $\mathrm{S}=4.65 \times 10^{-1} \mathrm{~N} \mathrm{~m}^{-1}$ Atmospheric pressure, $\mathrm{P}_0=1.01 \times 10^5 \mathrm{~Pa}$ Total pressure inside the mercury drop. $=$ Excess pressure inside mercury + Atmospheric pressure $=2 \mathrm{~S} / \mathrm{r}+\mathrm{P}_0=\left[\frac{2 \times 4.65 \times 10^{-1}}{\left(3 \times 10^{-3}\right)}\right]+1.01 \times 10^5=1.0131 \times 10^5=1.01 \times 10^5 \mathrm{~Pa}$ Excess pressure $=2 \mathrm{~S} / \mathrm{r}$ $=\left[\frac{2 \times 4.65 \times 10^{-1}}{\left(3 \times 10^{-3}\right)}\right] 310 \mathrm{pa}$

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