1
MCQ (Single Correct Answer)

### JEE Main 2017 (Online) 8th April Morning Slot

If the Earth has no rotational motion, the weight of a person on the equator is W. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh ${3 \over 4}$ W. Radius of the Earth is 6400 km and g=10 m/s2.
A
1.1 $\times$ 10−3 rad/s
B
0.83 $\times$ 10−3 rad/s
C
0.63 $\times$ 10−3 rad/s
D
0.28 $\times$ 10−3 rad/s

## Explanation

Initially when earth is not rotating then weight of the person is $w$.

When earth rotares about it's axis then weight = ${{3\omega } \over 4}$

Then,

g' = g $-$ $\omega$2R cos2$\theta$

$\Rightarrow$ $\,\,\,$ ${{3g} \over 4}$ = g $-$ $\omega$2R cos2 0o

$\Rightarrow$ $\,\,\,$ $\omega$2R = ${g \over 4}$

$\,\,\,$ $\omega$ = $\sqrt {{g \over {4R}}}$

$\Rightarrow$ $\omega = \sqrt {{{10} \over {4 \times 6400 \times {{10}^3}}}}$

= 0.63 $\times$ 10$-$3 rad/s
2
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
3
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,

$T \propto {R^{{{n + 1} \over 2}}}$
4
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

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