MHT CET · Physics · Mechanical Properties of Fluids
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
- A \(R^2=R_1^2+R_2^2\)
- B \(R=R_1+R_2\)
- C \(R=\frac{R_1+R_2}{2}\)
- D \(R^3=R_1^3+R_2^3\)
Answer & Solution
Correct Answer
(D) \(R^3=R_1^3+R_2^3\)
Step-by-step Solution
Detailed explanation
Total volume remains same,
\(\begin{aligned}
& \frac{4}{3} \pi R^3=\frac{4}{3} \pi R_1^3+\frac{4}{3} \pi R_2^3 \\
& R^3=R_1^3+R_2^3
\end{aligned}\)
\(\begin{aligned}
& \frac{4}{3} \pi R^3=\frac{4}{3} \pi R_1^3+\frac{4}{3} \pi R_2^3 \\
& R^3=R_1^3+R_2^3
\end{aligned}\)
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