Question

Due to the excess pressure of surface tension find the change in the radius (in $\mathring{A}$ ) of a liquid drop. of surface tension $0.075\, N / m$ and Bulk modulus $1.25 \times 10^{8} \,N / m ^{2}$.

Answer 1

$\Delta P=B \frac{\Delta V}{V} $
$\Rightarrow \frac{2 T}{R}=B \frac{\Delta V}{V} ; V=\frac{4}{3} \pi R^{3} ; \frac{\Delta V}{V}=\frac{3 \Delta R}{R}$
$\Rightarrow \frac{2 T}{R}=B\left(\frac{3 \Delta R}{R}\right) $
$\Rightarrow \Delta R=\frac{2 T}{3 B}=\frac{2 \times 0.075}{3 \times 1.25 \times 10^{8}}=4\,\mathring{A}$