MHT CET · Physics · Mechanical Properties of Fluids

Two small drops of mercury each of radius ‘ $R$ ’ coalesce to form a large single drop. The ratio of the total surface energies before and after the change is

A $2^{\frac{2}{3}} : 1$
B $\sqrt{2} : 1$
C $2^{\frac{1}{3}} : 1$
D $2 : 1$

Verified Solution

Answer & Solution

Correct Answer

(C) $2^{\frac{1}{3}} : 1$

Step-by-step Solution

Detailed explanation

Surface enegy

\propto

Surface Area

∵ (\frac{4}{3} π R^{3}) \times 2 = \frac{4}{3} π R_{0}^{3}

\Rightarrow R_{0} = R {(2)}^{\frac{1}{3}}

∴ \frac{{(S u r f a c e E n e r g y)}_{b e f o r e}}{{(S u r f a c e E n e r g y)}_{a f t e r}} = \frac{(4 π R^{2}) \times 2}{4 π R^{2} {(2)}^{\frac{2}{3}}}

= \frac{2^{\frac{1}{3}}}{1}

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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Two small drops of mercury each of radius ‘ R ’ coalesce to form a large single drop. The ratio of the total surface energies before and after the change is

Step-by-step Solution

Two small drops of mercury each of radius ‘ $R$ ’ coalesce to form a large single drop. The ratio of the total surface energies before and after the change is