1

### JEE Main 2019 (Online) 9th January Morning Slot

If the angular momentum of a planet of mass m, moving around the Sun in a circular orbit is L, about the center of the Sun, its areal velocity is :
A
${L \over m}$
B
${4L \over m}$
C
${L \over 2m}$
D
${2L \over m}$

## Explanation

dA = ${1 \over 2}$ r2d$\theta$

$\therefore$   ${{dA} \over {dt}} = {1 \over 2}{r^2}{{d\theta } \over {dt}}$

$\Rightarrow$   ${{dA} \over {dt}} = {1 \over 2}{r^2}\omega$    . . . . . (1)

We know,

angular momentum,

L = $mvr$

= $m\left( {\omega r} \right)r$

= mr2$\omega$

$\therefore$   $\omega$ = ${L \over {m{r^2}}}$     . . . . . (2)

Put value of $\omega$ in equation(1),

${{dA} \over {dt}}$ = ${1 \over 2}{r^2}$ (${L \over {m{r^2}}}$)

= ${L \over {2m}}$
2

### JEE Main 2019 (Online) 9th January Evening Slot

A rod of length 50 cm is pivoted at one end. It is raised such that if makes an angle of 30o from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s$-$1) will be (g = 10 ms$-$2)

A
$\sqrt {{{30} \over 2}}$
B
$\sqrt {30}$
C
${{\sqrt {20} } \over 3}$
D
${{\sqrt {30} } \over 2}$

## Explanation

When this rod move from initial position to final position then,

Gain in kinetic energy = loss in potential energy

## Explanation

Consider a strip of radius x & thickness dx,

Torque due to friction on this strip.

$\int {d\tau = \int\limits_0^R {{{x\mu F.2\pi xdx} \over {\pi {R^2}}}} }$

$\tau = {{2\mu F} \over {{R^2}}}.{{{R^3}} \over 3}$

$\tau = {{2\mu FR} \over 3}$
4

### JEE Main 2019 (Online) 10th January Morning Slot

A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is -
A
${F \over {2mR}}$
B
${2F \over {3mR}}$
C
${3F \over {2mR}}$
D
${F \over {3mR}}$

## Explanation

FR = ${3 \over 2}$ MR2$\alpha$

$\alpha$ = ${{2F} \over {3MR}}$