1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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,

$$T \propto {R^{{{n + 1} \over 2}}}$$
3
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

= $${E_f} - {E_i}$$

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

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

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