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### JEE Main 2017 (Online) 9th April Morning Slot

The mass density of a spherical body is given by
$\rho$ (r) = ${k \over r}$ for r $\le$ R and $\rho$ (r) = 0 for r > R,

where r is the distance from the centre.

The correct graph that describes qualitatively the acceleration, a, of a test particle as a function of r is :
A B C D 2

### JEE Main 2018 (Offline)

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then :
A
T $\propto$ Rn/2
B
T $\propto$ R3/2 for any n
C
T $\propto$ Rn/2 +1
D
T $\propto$ R(n+1)/2

## Explanation

We know, Central force in circular motion, F = $m{\omega ^2}R$

According to the question,

$F \propto {1 \over {{R^n}}}$

$\therefore$ $m{\omega ^2}R$ $\propto {1 \over {{R^n}}}$

$\Rightarrow m{\omega ^2}R = {k \over {{R^n}}}$

$\Rightarrow {\omega ^2} = {k \over {m{R^{n + 1}}}}$

$\therefore$ $\omega \propto {1 \over {{R^{{{n + 1} \over 2}}}}}$ .......(1)

And we know, $T = {{2\pi } \over \omega }$

$\therefore$ $T \propto {1 \over \omega }$ ...... (2)

From (1) and (2) we can conclude that,

$T \propto {R^{{{n + 1} \over 2}}}$
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### JEE Main 2018 (Offline)

A particle is moving in a circular path of radius $a$ under the action of an attractive potential $U = - {k \over {2{r^2}}}$ Its total energy is:
A
$- {3 \over 2}{k \over {{a^2}}}$
B
Zero
C
$- {k \over {4{a^2}}}$
D
${k \over {2{a^2}}}$

## Explanation

We know, Total energy = Kinetic energy + Potential energy

Potential energy given as $U = - {k \over {2{r^2}}}$

We need to find Kinetic Energy.

As Force acting on the particle (F) = $- {{dU} \over {dr}}$

$\Rightarrow F = - {d \over {dr}}\left( {{{ - k} \over {2{r^2}}}} \right)$

$= {k \over 2} \times \left( { - 2} \right) \times {r^{ - 3}}$

$= - {k \over {{r^3}}}$

Because of this force particle is having circular motion so it will provide possible centripetal force.

$\left| F \right| = {{m{v^2}} \over r}$

$\Rightarrow {{m{v^2}} \over r} = {k \over {{r^3}}}$

$\Rightarrow$ $m{v^2} = {k \over {{r^2}}}$

We know kinetic energy of particle, K = ${1 \over 2}m{v^2}$ = ${k \over {2{r^2}}}$

As Total energy = Kinetic energy + Potential energy

So Total energy = ${k \over {2{r^2}}}$ $- {k \over {2{r^2}}}$ = 0
4

### JEE Main 2018 (Online) 15th April Morning Slot

A body of mass m is moving in a circular orbit of radius R about a planet of mass M. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius ${R \over 2},$ and the other mass, in a circular orbit of radius ${3R \over 2}$. The difference between the final and initial total energies is :
A
$- {{GMm} \over {2R}}$
B
$+ {{GMm} \over {6R}}$
C
${{GMm} \over {2R}}$
D
$- {{GMm} \over {6R}}$

## Explanation

Initially gravitational potenrial energy

Ei = $-$ ${{GMm} \over {2R}}$

Final gravitational potential energy

Ef = $-$ ${{GM\left( {{m \over 2}} \right)} \over {2\left( {{R \over 2}} \right)}}$ $-$ ${{GM\left( {{m \over 2}} \right)} \over {2\left( {{{3R} \over 2}} \right)}}$

= $-$ ${{GMm} \over {2R}} - {{GMm} \over {6R}}$

= $-$ ${{4GMm} \over {6R}}$

= $-$ ${{2GMm} \over {3R}}$

$\therefore\,\,\,\,$ Required difference in energies

= ${E_f} - {E_i}$

= $-$ ${{GMm} \over R}\left( {{2 \over 3} - {1 \over 2}} \right)$

= $-$ ${{GMm} \over {6R}}$