Search any question & find its solution
Question:
Answered & Verified by Expert
The radii of two mercury drops are $R_1$ and $R_2$. Under isothermal conditions, a single drop of radius $R$ is formed from them. The relations between $R, R_1$ and $R_2$ is
Options:
Solution:
2234 Upvotes
Verified Answer
The correct answer is:
$R^3=R_1^3+R_2^3$
Total volume remains same,
$\begin{aligned}
& \frac{4}{3} \pi R^3=\frac{4}{3} \pi R_1^3+\frac{4}{3} \pi R_2^3 \\
& R^3=R_1^3+R_2^3
\end{aligned}$
$\begin{aligned}
& \frac{4}{3} \pi R^3=\frac{4}{3} \pi R_1^3+\frac{4}{3} \pi R_2^3 \\
& R^3=R_1^3+R_2^3
\end{aligned}$
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.