MHT CET · Physics · Mechanical Properties of Fluids
Let ' \(\mathrm{R}_1\) ' and ' \(\mathrm{R}_2\) ' are radii of two mercury drops. A big mercury drop is formed from then under isothermal conditions. The radius of the resultant drop is
- A \(\sqrt{\mathrm{R}_1^2+\mathrm{R}_2^2}\)
- B \(\left(\mathrm{R}_1^3+\mathrm{R}_2^3\right)^{\frac{1}{3}}\)
- C \(\sqrt{\mathrm{R}_1^2-\mathrm{R}_2^2}\)
- D \(\frac{R_1+R_2}{2}\)
Answer & Solution
Correct Answer
(B) \(\left(\mathrm{R}_1^3+\mathrm{R}_2^3\right)^{\frac{1}{3}}\)
Step-by-step Solution
Detailed explanation
The volume of the bigger drop will be equal to the sum of the volumes of the smaller drops
\(
\begin{aligned}
& \frac{4}{3} \pi \mathrm{R}^3=\frac{4}{3} \pi \mathrm{R}_1^3+\frac{4}{3} \pi \mathrm{R}_2^3 \\
& \therefore \mathrm{R}=\left(\mathrm{R}_1^3+\mathrm{R}_2^3\right)^{1 / 3}
\end{aligned}
\)
\(
\begin{aligned}
& \frac{4}{3} \pi \mathrm{R}^3=\frac{4}{3} \pi \mathrm{R}_1^3+\frac{4}{3} \pi \mathrm{R}_2^3 \\
& \therefore \mathrm{R}=\left(\mathrm{R}_1^3+\mathrm{R}_2^3\right)^{1 / 3}
\end{aligned}
\)
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