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Question

If $$T$$ is the surface tension of a fluid, then the energy needed to break a liquid drop of radius $$R$$ into $$64$$ equal drops is :


A
6πR2T
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B
πR2T
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C
12πR2T
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D
8πR2T
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Solution

The correct option is B $$ 12 \pi R^2 T $$

Let the radius of the small drop is $$r$$.

Since, the volume of the bigger drop will be equal to the $$64$$ smaller drops. therefore,

$$\frac{4}{3}\pi {R^3} = 64 \times \frac{4}{3}\pi {r^3}$$

$$R = 4r$$

The surface energy of the larger drop is given as,

$${E_1} = \left( {4\pi {R^2}} \right)T$$

The surface energy of the small drops is given as,

$${E_2} = 64 \times \left( {4\pi {r^2}} \right)T$$

$$ = 64 \times \left( {4\pi {{\left( {\frac{R}{4}} \right)}^2}} \right)T$$

$$ = \left( {16\pi {R^2}} \right)T$$

The required energy is given as,

$$\Delta E = {E_2} - {E_1}$$

$$ = \left( {16\pi {R^2}} \right)T - \left( {4\pi {R^2}} \right)T$$

$$ = \left( {12\pi {R^2}} \right)T$$

Thus, the required energy to create the $$64$$ small drops is $$\left( {12\pi {R^2}} \right)T$$.


Physics

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