NEET UG 2014 — Physics Mechanics

2014mcqhard

A certain number of spherical drops of a liquid of radius $r$ coalesce to form a single drop of radius $R$ and volume $V.$ If $T$ is the surface tension of the liquid then:

Official previous-year question

Held on 30 Apr 2014 · Verified 9 Jul 2026.

Options

A
Energy $4VT(\frac{1}{r}-\frac{1}{R})$ is released
B
Energy $=3VT(\frac{1}{r}+\frac{1}{R})$ is released
C
Energy $=3VT(\frac{1}{r}-\frac{1}{R})$ is released
D
Energy is neither released nor absorbed

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Solution

$\Delta U=(ST)(\Delta A)$

$A(initial)=(4\pi {r}^{2})n$

$A(final)=4\pi {R}^{2}$

$\Delta A=(4\pi {r}^{2})n-4\pi {R}^{2}$

$(\frac{4}{3}\pi {r}^{3})n=\frac{4}{3}\pi {R}^{3}$

$n=\frac{{R}^{3}}{{r}^{3}}$

$\Delta A=4\pi [\frac{{R}^{3}}{{r}^{3}}.{r}^{2}-{R}^{2}]=4\pi [\frac{{R}^{3}}{r}-\frac{{R}^{3}}{R}]=(\frac{4\pi {R}^{3}}{3})3[\frac{1}{r}-\frac{1}{R}]$

$\Delta A=3V[\frac{1}{r}-\frac{1}{R}]$

$\Delta U=3VT[\frac{1}{r}-\frac{1}{R}]$

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Solution

$\Delta U=(ST)(\Delta A)$

$A(initial)=(4\pi {r}^{2})n$

$A(final)=4\pi {R}^{2}$

$\Delta A=(4\pi {r}^{2})n-4\pi {R}^{2}$

$(\frac{4}{3}\pi {r}^{3})n=\frac{4}{3}\pi {R}^{3}$

$n=\frac{{R}^{3}}{{r}^{3}}$

$\Delta A=4\pi [\frac{{R}^{3}}{{r}^{3}}.{r}^{2}-{R}^{2}]=4\pi [\frac{{R}^{3}}{r}-\frac{{R}^{3}}{R}]=(\frac{4\pi {R}^{3}}{3})3[\frac{1}{r}-\frac{1}{R}]$

$\Delta A=3V[\frac{1}{r}-\frac{1}{R}]$

$\Delta U=3VT[\frac{1}{r}-\frac{1}{R}]$

NEET UG 2014 — Physics Mechanics

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NEET UG 2014 — Physics Mechanics

Options

Did you get this right?

Solution

Your note

NEET UG Mechanics in other years

More NEET UG Mechanics questions

Other Physics topics for NEET UG

Electromagnetism

Modern Physics

Thermodynamics

Optics

Waves & Oscillations

Keep practicing Mechanics