JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Fluid Mechanics

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The amount of work done to break a big water drop of radius 'R' into 27 small drops of equal radius is 10 J. The work done required to break the same big drop into 64 small drops of equal radius will be :-

(A)15 J✓ Correct
(B)10 J
(C)20 J
(D)5 J

Answer: (A)

Solution

When a drop of radius

R

breaks into

n

identical drops of radius

r

, volume is conserved:

\frac{4}{3} π R^{3} = n \cdot \frac{4}{3} π r^{3} \Rightarrow r = \frac{R}{n ^{1/3}}

The work done equals the increase in surface energy:

W = S (n \cdot 4 π r^{2} - 4 π R^{2}) = 4 π S R^{2} (n^{1/3} - 1)

For 27 drops:

W_{27} = 4 π S R^{2} (2 7^{1/3} - 1) = 4 π S R^{2} (3 - 1) = 10

J ... (i)

For 64 drops:

W_{64} = 4 π S R^{2} (6 4^{1/3} - 1) = 4 π S R^{2} (4 - 1)

... (ii)

Dividing (ii) by (i):

\frac{W _{64}}{W _{27}} = \frac{4 - 1}{3 - 1} = \frac{3}{2}

W_{64} = \frac{3}{2} \times 10 = 15

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Fluid Mechanics

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