Question

Surface tension of mercury is $0.465 \,N\, m^{-1}$ . The excess pressure inside a mercury drop of diameter $6 \,mm$ is

• A. $310 \, pa$
• B. $410 \, pa$
• C. $510 \, pa$
• D. $610 \, pa$

Answer 1

Answer: (A)

$310 \, pa$

Answer 2

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 $P=\frac{2S}{r}=\frac{2\times0.465\,N\,m^{-1}}{3\times10^{-3}\,m}$ $= 310\,Pa$