A particle is moving in a… - JEE Main 2015 Physics PYQ

JEE Main 2015 — Physics Mechanics

2015mcqmedium

A particle is moving in a circle of radius $r$ under the action of a force $F=\alpha {r}^{2}$ which is directed towards centre of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy $=0$ for $r=0$ ):

Official previous-year question

Held on 11 Apr 2015 · Verified 6 Jul 2026.

Options

A
$\frac{5}{6}\alpha {r}^{3}$
B
$\alpha {r}^{3}$
C
$\frac{1}{2}\alpha {r}^{3}$
D
$\frac{4}{3}\alpha {r}^{3}$

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Solution

Force $F=\alpha {r}^{2}$ will act as centripetal force

$\Rightarrow \alpha {r}^{2}=\frac{m{v}^{2}}{r}\Rightarrow m{v}^{2}=\alpha {r}^{3}$

$\therefore$ Kinetic energy $K.E. =\frac{1}{2}m{v}^{2}$

$\Rightarrow K.E.=\frac{\alpha {r}^{3}}{2}$

Also $dU=-\vec{F}.d\vec{r}$

$\Delta U=-\int \vec{F}.d\vec{r}$

$=-\int \alpha {r}^{2}(-dr)$

$=\alpha \int _{0}^{r}{r}^{2}dr$

$U-0=\frac{\alpha {r}^{3}}{3}$

$\Rightarrow U=\frac{\alpha {r}^{3}}{3}$

So total mechanical energy

$E=U+K$

$=\frac{\alpha {r}^{3}}{3}+\frac{\alpha {r}^{3}}{2}$

$E=\frac{5\alpha {r}^{3}}{6}$

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Solution

Force $F=\alpha {r}^{2}$ will act as centripetal force

$\Rightarrow \alpha {r}^{2}=\frac{m{v}^{2}}{r}\Rightarrow m{v}^{2}=\alpha {r}^{3}$

$\therefore$ Kinetic energy $K.E. =\frac{1}{2}m{v}^{2}$

$\Rightarrow K.E.=\frac{\alpha {r}^{3}}{2}$

Also $dU=-\vec{F}.d\vec{r}$

$\Delta U=-\int \vec{F}.d\vec{r}$

$=-\int \alpha {r}^{2}(-dr)$

$=\alpha \int _{0}^{r}{r}^{2}dr$

$U-0=\frac{\alpha {r}^{3}}{3}$

$\Rightarrow U=\frac{\alpha {r}^{3}}{3}$

So total mechanical energy

$E=U+K$

$=\frac{\alpha {r}^{3}}{3}+\frac{\alpha {r}^{3}}{2}$

$E=\frac{5\alpha {r}^{3}}{6}$

Options

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Solution

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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