Q158 / 200 · MHT CET - Medical 2026 (24 April Shift 1) · Physics
A certain number of spherical drops of a liquid of radius 'r' coalesce to form a single drop of the radius 'R' and volume V. If T is the surface tension of the liquid, then the energy released in this process is
- A$VT\left(\dfrac{1}{r}-\dfrac{1}{R}\right)$
- Correct answerB$3VT\left(\dfrac{1}{r}-\dfrac{1}{R}\right)$
- C$2VT\left(\dfrac{1}{r}-\dfrac{1}{R}\right)$
- D$4VT\left(\dfrac{1}{r}-\dfrac{1}{R}\right)$
Solution
Let $n$ be the number of small drops of radius $r$.
Equating the initial and final volumes:
$n \left(\dfrac{4}{3}\pi r^3\right) = \dfrac{4}{3}\pi R^3 = V$
The initial surface area is $A_i = n (4\pi r^2)$.
Using $n = \dfrac{V}{\dfrac{4}{3}\pi r^3}$, we get $A_i = \dfrac{3V}{r}$.
The final surface area is $A_f = 4\pi R^2$.
Using $V = \dfrac{4}{3}\pi R^3$, we get $A_f = \dfrac{3V}{R}$.
The decrease in surface area is $\Delta A = A_i - A_f = 3V\left(\dfrac{1}{r} - \dfrac{1}{R}\right)$.
The energy released is given by $E = T \Delta A = 3VT\left(\dfrac{1}{r} - \dfrac{1}{R}\right)$.
Answer: $3VT\left(\dfrac{1}{r}-\dfrac{1}{R}\right)$
Equating the initial and final volumes:
$n \left(\dfrac{4}{3}\pi r^3\right) = \dfrac{4}{3}\pi R^3 = V$
The initial surface area is $A_i = n (4\pi r^2)$.
Using $n = \dfrac{V}{\dfrac{4}{3}\pi r^3}$, we get $A_i = \dfrac{3V}{r}$.
The final surface area is $A_f = 4\pi R^2$.
Using $V = \dfrac{4}{3}\pi R^3$, we get $A_f = \dfrac{3V}{R}$.
The decrease in surface area is $\Delta A = A_i - A_f = 3V\left(\dfrac{1}{r} - \dfrac{1}{R}\right)$.
The energy released is given by $E = T \Delta A = 3VT\left(\dfrac{1}{r} - \dfrac{1}{R}\right)$.
Answer: $3VT\left(\dfrac{1}{r}-\dfrac{1}{R}\right)$
Hence, the correct answer is option: 'B'.