Q125 / 150 · MHT CET 2025 (22 Apr Shift 2) · Physics

The energy needed for breaking a liquid drop of radius ' R ' into ' n ' droplets each of radius ' $r$ ' is $\text { [ } \mathrm{T}=\text { surface tension of the liquid] }$

  • A
    $4 \pi \mathrm{~T} \mathrm{R}^2\left[\frac{\mathrm{R}}{\mathrm{r}}+1\right]$
  • Correct answer
    B
    $4 \pi \mathrm{~T} \mathrm{R}^2\left[\frac{\mathrm{R}}{\mathrm{r}}-1\right]$
  • C
    $4 \pi \mathrm{~T} \mathrm{R}^2\left[\frac{\mathrm{r}}{\mathrm{R}}+1\right]$
  • D
    $4 \pi \mathrm{Tr}^2\left[\frac{\mathrm{R}}{\mathrm{r}}-1\right]$

Solution

Energy for breaking a liquid droplet: The energy required arises from the increase in surface area, given as the product of surface tension and the change in surface area.

Let the radius of the large drop be $R$ and its surface tension be $T$. Upon breaking into $n$ droplets of radius $r$, volume conservation requires $R^3 = n r^3$, so $n = R^3 / r^3$.

Initial surface area is $4\pi R^2$; final surface area is $n \cdot 4\pi r^2 = 4\pi R^3 / r$. The change is $\Delta A = 4\pi R^2 (R / r - 1)$.

Energy absorbed is $E = T \cdot \Delta A = 4\pi T R^2 (R / r - 1)$.

This corresponds to option B.

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

Relevant Questions