Search any question & find its solution
Question:
Answered & Verified by Expert
When a big drop of water is formed from $n$ small drops of water, the energy loss is $3 E$, where, $E$ is the energy of the bigger drop. If $R$ is the radius of the bigger drop and $r$ is the radius of the smaller drop, then number of smaller drops $(n)$ is
Options:
Solution:
1178 Upvotes
Verified Answer
The correct answer is:
$\frac{4 R^2}{r^2}$
The energy of $n$ small drop - the energy of the bigger drop = Energy loss by bigger drop
$$
\begin{gathered}
n \times 4 \pi r^2 \times T-4 \pi r^2 \times T=3 \times 4 \pi R^2 \times T \\
n \times 4 \pi r^2 \times T=12 \pi R^2 \times T+4 \pi R^2 \times T \\
n=\frac{16 \pi R^2 \times T}{4 \pi r^2 \times T} \\
n=4 \frac{R^2}{r^2}
\end{gathered}
$$
$$
\begin{gathered}
n \times 4 \pi r^2 \times T-4 \pi r^2 \times T=3 \times 4 \pi R^2 \times T \\
n \times 4 \pi r^2 \times T=12 \pi R^2 \times T+4 \pi R^2 \times T \\
n=\frac{16 \pi R^2 \times T}{4 \pi r^2 \times T} \\
n=4 \frac{R^2}{r^2}
\end{gathered}
$$
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.