JEE Main 2026 — Physics Mechanics

2026mcqmedium

A liquid drop of diameter $2$ mm breaks into $512$ droplets. The change in surface energy is $\alpha \times 10^{-6}$ J. The value of $\alpha$ is _______. (Take surface tension of liquid $= 0.08$ N/m)

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Options

A
$10$
B
$7$
C
$8$
D
$11$

Did you get this right?

Solution

Let the radius of the initial drop be $R$ and the radius of each small droplet be $r$ .

Given diameter $D = 2$ mm, so $R = 1$ mm $= 10^{-3}$ m.

By conservation of volume:

$\dfrac{4}{3}\pi R^3 = 512 \times \dfrac{4}{3}\pi r^3$

$R^3 = 512 r^3 \Rightarrow R = 8r \Rightarrow r = \dfrac{R}{8}$

Initial surface area, $A_i = 4\pi R^2$

Final surface area, $A_f = 512 \times 4\pi r^2 = 512 \times 4\pi \left(\dfrac{R}{8}\right)^2 = 32\pi R^2$

Change in surface area, $\Delta A = A_f - A_i = 32\pi R^2 - 4\pi R^2 = 28\pi R^2$

Change in surface energy, $\Delta U = T \Delta A$

$\Delta U = 0.08 \times 28\pi R^2$

Substituting $R = 10^{-3}$ m and $\pi \approx \dfrac{22}{7}$ :

$\Delta U = 0.08 \times 28 \times \dfrac{22}{7} \times (10^{-3})^2$

$\Delta U = 0.08 \times 4 \times 22 \times 10^{-6}$

$\Delta U = 7.04 \times 10^{-6}$ J

Comparing with $\alpha \times 10^{-6}$ J, we get $\alpha \approx 7$ .

Answer: $7$

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Solution

Let the radius of the initial drop be $R$ and the radius of each small droplet be $r$ .

Given diameter $D = 2$ mm, so $R = 1$ mm $= 10^{-3}$ m.

By conservation of volume:

$\dfrac{4}{3}\pi R^3 = 512 \times \dfrac{4}{3}\pi r^3$

$R^3 = 512 r^3 \Rightarrow R = 8r \Rightarrow r = \dfrac{R}{8}$

Initial surface area, $A_i = 4\pi R^2$

Final surface area, $A_f = 512 \times 4\pi r^2 = 512 \times 4\pi \left(\dfrac{R}{8}\right)^2 = 32\pi R^2$

Change in surface area, $\Delta A = A_f - A_i = 32\pi R^2 - 4\pi R^2 = 28\pi R^2$

Change in surface energy, $\Delta U = T \Delta A$

$\Delta U = 0.08 \times 28\pi R^2$

Substituting $R = 10^{-3}$ m and $\pi \approx \dfrac{22}{7}$ :

$\Delta U = 0.08 \times 28 \times \dfrac{22}{7} \times (10^{-3})^2$

$\Delta U = 0.08 \times 4 \times 22 \times 10^{-6}$

$\Delta U = 7.04 \times 10^{-6}$ J

Comparing with $\alpha \times 10^{-6}$ J, we get $\alpha \approx 7$ .

Answer: $7$

JEE Main 2026 — Physics Mechanics

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JEE Main 2026 — Physics Mechanics

Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics