A big water drop is divided into 8 equal droplets. $\Delta \mathrm{P}_{\mathrm{S}}$ and $\Delta…

A big water drop is divided into 8 equal droplets. $\Delta \mathrm{P}_{\mathrm{S}}$ and $\Delta \mathrm{P}_{\mathrm{B}}$ be the excess pressure inside a smaller and bigger drop respectively. The relation between $\Delta \mathrm{P}_{\mathrm{S}}$ and $\Delta \mathrm{P}_{\mathrm{B}}$ is
  1. $\Delta \mathrm{P}_{\mathrm{B}}=\Delta \mathrm{P}_{\mathrm{S}}$
  2. $\Delta \mathrm{P}_{\mathrm{B}}=\frac{1}{2} \Delta \mathrm{P}_{\mathrm{S}}$
  3. $\Delta \mathrm{P}_{\mathrm{B}}=\frac{1}{4} \Delta \mathrm{P}_{\mathrm{S}}$
  4. $\Delta \mathrm{P}_{\mathrm{B}}=2 \Delta \mathrm{P}_{\mathrm{S}}$

Solution

Volume of 8 smaller drop $=$ Volume of the bigger drop $\begin{aligned} & \therefore 8 \times \frac{4}{3} \pi r^3=\frac{4 \pi}{3} R^3 \\ & \therefore 2 r=R \text { or } r=\frac{R}{2} \end{aligned}$ Excess pressure $\Delta \mathrm{P}_{\mathrm{s}}=\frac{2 \mathrm{~T}}{\mathrm{r}}$ and $\Delta \mathrm{P}_{\mathrm{B}}=\frac{2 \mathrm{~T}}{\mathrm{R}}$ $\therefore \frac{\Delta \mathrm{P}_{\mathrm{B}}}{\Delta \mathrm{P}_{\mathrm{S}}}=\frac{\mathrm{r}}{\mathrm{R}}=\frac{1}{2}$

Asked in: MHT CET 2021 (21 Sep Shift 2)

Practice more Mechanical Properties of Fluids questions on Aicharya