What should be the radius of water drop so that excess pressure inside it is $72 \mathrm{Nm}^{-2}$ ? (The…
What should be the radius of water drop so that excess pressure inside it is $72 \mathrm{Nm}^{-2}$ ? (The surface tension of water $7.2 \times 10^{-2} \mathrm{Nm}^{-1}$ )
$1 \mathrm{~mm}$
$2 \mathrm{~mm}$
$8\mathrm{~mm}$
$4 \mathrm{~mm}$
Solution
Excess pressure in a water drop $=\frac{2 T}{R}$
$\begin{aligned}
& \mathrm{T}=7.2 \times 10^{-2} \mathrm{Nm}^{-2} \\
& \therefore 72=\frac{2 \times 7.2 \times 10^{-2}}{\mathrm{R}} \\
& \therefore \mathrm{R}=\frac{2 \times 7.2 \times 10^{-2}}{72}=2 \times 10^{-3} \mathrm{~m}=2 \mathrm{~mm}
\end{aligned}$
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