Question

A. liquid drop of radius ‘R’ breaks into 64 tiny droplets each of radius ‘r’. If the surface tension of the liquid is ‘T’ then gain in energy is:

A
48πR2T
B
12πr2T
C
96πr2T
D
192πr2T

Solution

The correct option is C $$192\pi r^{2}T$$Let the density of the liquid be  $$\rho$$.Mass of the liquid drops must be equal to the total mass of the droplets.$$\therefore$$   $$\rho \times \dfrac{4\pi}{3}R^3 = 64 \times \rho \times \dfrac{4\pi}{3} r^3$$            $$\implies R =4r$$Change in surface area   $$\Delta A = 64 \times 4\pi r^2 - 4\pi R^2$$$$\therefore$$    $$\Delta A = 64 \times 4\pi \times r^2 - 4\pi (4r)^2 =192 \pi r^2$$$$\therefore$$ Gain in energy      $$E = T \Delta A = 192 \pi r^2 T$$Physics

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