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Q.
Surface tension of mercury is $ 0.465 \,N\, m^{-1} $ . The excess pressure inside a mercury drop of diameter $ 6 \,mm $ is
Mechanical Properties of Fluids
Solution:
Here, $ D = 6\, mm $
$ r=\frac{D}{2}=\frac{6 \,mm}{2}=3\,mm=3\times10^{-3}\,m $
Excess pressure inside the mercury drop is
$ P=\frac{2S}{r}=\frac{2\times0.465\,N\,m^{-1}}{3\times10^{-3}\,m} $
$= 310\,Pa $