  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  B
πR2T  C
12πR2T  D
8πR2T  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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