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### 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.

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

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

$\tau = {{2\mu FR} \over 3}$
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### 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$\alpha$

$\alpha$ = ${{2F} \over {3MR}}$
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### JEE Main 2019 (Online) 10th January Evening Slot

Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is : A
${{17} \over {15}}$ MR2
B
${{137} \over {15}}$ MR2
C
${{209} \over {15}}$ MR2
D
${{152} \over {15}}$ MR2

## Explanation

For Ball

using parallel axis theorem.

Iball = ${2 \over 5}$MR2 + M(2R)2

= ${{22} \over 5}$ MR2

2 Balls   so  ${{44} \over 5}$MR2

Irod = for rod ${{M{{(2R)}^2}} \over R}$ = ${{M{R^2}} \over 3}$

Isystem = IBall + Irod

= ${{44} \over 5}M{R^2} + {{M{R^2}} \over 3}$

= ${{137} \over {15}}$ MR2