Question

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

• A. 310 Pa
• B. 410 Pa
• C. 510 Pa
• D. 610 Pa

Answer 1

Answer: (A)

310 Pa

Here, $D=6mm$
$r=\frac{D}{2}=\frac{6mm}{2}=3mm=3\times {10}^{-3}$
Excess pressure inside the mercury drop is
$P=\frac{2S}{r}=\frac{2\times 0.465N{m}^{-1}}{3\times {10}^{-3}m}=310Pa$