What is the pressure inside the drop of mercury of radius $3.00 \; mm$ at room temperature? Surface tension of mercury at that temperature $(20 \; ^{\circ}C)$ is $4.65 \times 10^{-1} \; N m^{-1}$. The atmospheric pressure is $1.01 \times 10^{5} \; Pa$. Also,calculate the excess pressure inside the drop.

(N/A) Given:
Radius of the mercury drop,$r = 3.00 \; mm = 3.00 \times 10^{-3} \; m$
Surface tension of mercury,$S = 4.65 \times 10^{-1} \; N m^{-1}$
Atmospheric pressure,$P_{0} = 1.01 \times 10^{5} \; Pa$
$1$. Excess pressure inside the drop:
The excess pressure inside a liquid drop is given by $\Delta P = \frac{2S}{r}$.
$\Delta P = \frac{2 \times 4.65 \times 10^{-1}}{3.00 \times 10^{-3}} = \frac{0.93}{3.00 \times 10^{-3}} = 0.31 \times 10^{3} = 310 \; Pa$.
$2$. Total pressure inside the drop:
The total pressure inside the drop is the sum of the atmospheric pressure and the excess pressure.
$P_{total} = P_{0} + \Delta P$
$P_{total} = 1.01 \times 10^{5} \; Pa + 310 \; Pa$
$P_{total} = 101000 \; Pa + 310 \; Pa = 101310 \; Pa = 1.0131 \times 10^{5} \; Pa$.