The excess pressure inside mercury drop of diameter 4 mm is (Take surface tension of mercury is 0.465 N/m)

Options

A. 410 Pa
B. 465 Pa
C. 610 Pa
D. 310 Pa

Answer

B

Explanation

The radius of mercury drop,
`r=(D)/(2)=(4)/(2)=2mm` [given, D=4 mm]
`=2xx10^(-3)m`
`therefore` Excess pressure inside drop is
`P=(2T)/(r)=(2xx0.465N//m)/(2xx10^(-3)m)=465Pa`