MHT CET · Physics · Mechanical Properties of Fluids

Let \({ }^{\prime} \mathrm{R}_{1}\) ' and \({ }^{\prime} \mathrm{R}_{2}\) ' are radii of two mercury drops. A big mercury drop is formed from
them under isothermal conditions. The radius of the resultant drop is

A \(\mathrm{R}=\sqrt{\mathrm{R}_{1}^{2}-\mathrm{R}_{2}^{2}}\)
B \(R=\frac{R_{1}+R_{2}}{2}\)
C \(\mathrm{R}=\sqrt{\mathrm{R}_{1}^{2}+\mathrm{R}_{2}^{2}}\)
D \(\mathrm{R}=\left(\mathrm{R}_{1}^{3}+\mathrm{R}_{2}^{3}\right)^{\frac{1}{3}}\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\mathrm{R}=\left(\mathrm{R}_{1}^{3}+\mathrm{R}_{2}^{3}\right)^{\frac{1}{3}}\)

Step-by-step Solution

Detailed explanation

The Volume of the bigger drop is equal to the sum of the volumes of the smaller drops.
\(\begin{aligned}
& \frac{4}{3} \pi R^{3}=\frac{4}{3} \pi R_{1}^{3}+\frac{4}{3} \pi R_{2}^{3} \\
\therefore \quad & R=\sqrt[3]{R_{1}^{3}+R_{2}^{3}}
\end{aligned}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Let \({ }^{\prime} \mathrm{R}_{1}\) ' and \({ }^{\prime} \mathrm{R}_{2}\) ' are radii of two mercury drops. A big mercury drop is formed fromthem under isothermal conditions. The radius of the resultant drop is

Step-by-step Solution

Let \({ }^{\prime} \mathrm{R}_{1}\) ' and \({ }^{\prime} \mathrm{R}_{2}\) ' are radii of two mercury drops. A big mercury drop is formed from
them under isothermal conditions. The radius of the resultant drop is