Q123 / 150 · MHT CET 2009 · Physics

The change in energy when a big dropis split in sma $\| n$ droplets is

  • A
    $4 R^{2}\left(n^{2 / 3}-1\right) I$
  • Correct answer
    B
    $4 \pi R^{2}\left(n^{1 / 3}-1\right) T$
  • C
    $4 \pi R^{2}\left(n^{-1 / 3}-1\right) T$
  • D
    $4 \pi R^{2}\left[n^{-2 / 3}-1\right] T$

Solution

When a drop of radius $R$ split into $n$ smaller drops (each of radius $r$ ), then surface area of liquid increases. Hence, the work is to be done against surface tension. Since, the volume of liquid remains constant $$ \begin{array}{l} \therefore \quad \frac{4}{3} \pi R^{3}=n \frac{4}{3} \pi r^{3} \quad \therefore R^{3}=n r^{3} \\ \text { Work done }=T \times \Delta A \\ =T\left[n 4 \pi r^{2}-4 \pi R^{2}\right] \\ =4 \pi R^{2} T\left(n^{1 / 3}-1\right) \end{array} $$

Hence, the correct answer is option: 'B'.

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