The work done in splitting a water drop of radius R into 64 droplets is ( $\mathrm{T}=$ Surface tension of…

The work done in splitting a water drop of radius R into 64 droplets is ( $\mathrm{T}=$ Surface tension of water)
  1. $6 \pi \mathrm{TR}^2$
  2. $12 \pi \mathrm{TR}^2$
  3. $8 \pi \mathrm{TR}^2$
  4. $\quad 24 \pi \mathrm{TR}^2$

Solution

Given the radius of drop is R . Let $r$ be the radius of smaller droplets. As the total volume remains the same, $\begin{array}{ll} & \frac{4}{3} \pi R^3=64 \times \frac{4}{3} \pi r^3 \\ \therefore \quad & r=\frac{R}{(64)^{\frac{1}{3}}}=\frac{R}{4} \end{array}$
Initial surface energy $E_1=4 \pi R^2 T$ Final surface energy $E_2=64 \times 4 \pi \times\left(\frac{R}{4}\right)^2 \times T$ $=16 \pi \mathrm{R}^2 \mathrm{~T}$ $\therefore \quad$ Work done, $\mathrm{W}=\mathrm{E}_2-\mathrm{E}_1$ $\mathrm{W}=12 \pi \mathrm{TR}^2$

Asked in: MHT CET 2024 (09 May Shift 2)

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