Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2511 Upvotes
Verified Answer
The correct answer is:
$\Delta \mathrm{P}_{\mathrm{B}}=\frac{1}{2} \Delta \mathrm{P}_{\mathrm{S}}$
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 $\Delta \mathrm{P}_{\mathrm{B}}=\frac{2 \mathrm{~T}}{\mathrm{R}}$
$$
\therefore \frac{\Delta \mathrm{P}_{\mathrm{B}}}{\Delta \mathrm{P}_{\mathrm{S}}}=\frac{\mathrm{r}}{\mathrm{R}}=\frac{1}{2}
$$
$$
\begin{aligned}
& \therefore 8 \times \frac{4}{3} \pi r^3=\frac{4 \pi}{3} R^3 \\
& \therefore 2 r=R \text { or } r=\frac{R}{2}
\end{aligned}
$$
Excess pressure $\Delta \mathrm{P}_{\mathrm{s}}=\frac{2 \mathrm{~T}}{\mathrm{r}}$ and $\Delta \mathrm{P}_{\mathrm{B}}=\frac{2 \mathrm{~T}}{\mathrm{R}}$
$$
\therefore \frac{\Delta \mathrm{P}_{\mathrm{B}}}{\Delta \mathrm{P}_{\mathrm{S}}}=\frac{\mathrm{r}}{\mathrm{R}}=\frac{1}{2}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.