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JEE PHYSICS MCQ: MECHANICAL PROPERTIES OF FLUIDS

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Application +4 | -1

If a drop of mercury having radius

1 \mathrm{~cm}

is broken into

10^6

identical small droplets. If surface of mercury is

35 \times 10^{-3} \mathrm{~N} / \mathrm{m}^2

, then find out work done in this process.

View answer and explanation

Correct option: B

4.35 \times 10^{-3} \mathrm{~J}

Option B (

4.35 \times 10^{-3} \mathrm{~J}

) is correct because it is the work done in breaking the mercury drop into smaller droplets, calculated using the formula

W = 4 \pi R^2 T [n^{1/3} - 1]

.

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