1

### 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
2

### JEE Main 2019 (Online) 11th January Morning Slot

A particle is moving along a circular path with a constant speed of 10 ms–1. What is the magnitude of the change in velocity of the particle, when it moves through an angle of 60o around the centre of the circle?
A
zero
B
10 m/s
C
$10\sqrt 2 m/s$
D
$10\sqrt 3 m/s$

## Explanation $\left| {\Delta \overrightarrow v } \right| = \sqrt {v_1^2 + v_2^2 + 2{v_1}{v_2}\cos \left( {\pi - \theta } \right)}$

$= 2v\sin {\theta \over 2}$        since $\left[ {\left| {\overline v {}_1} \right| = \left| {{{\overline v }_2}} \right|} \right]$

$= \left( {2 \times 10} \right) \times \sin \left( {{{30}^o}} \right)$

= 10 m/s
3

### JEE Main 2019 (Online) 11th January Morning Slot

A slab is subjected to two forces $\overrightarrow {{F_1}}$ and $\overrightarrow {{F_2}}$ of same magnitude F as shown in the figure. Force $\overrightarrow {{F_2}}$ is in XY-plane while force $\overrightarrow {{F_1}}$ acts along z = axis at the point $\left( {2\overrightarrow i + 3\overrightarrow j } \right).$. The moment of these forces about point O will be : A
$\left( {3\widehat i - 2\widehat j - 3\widehat k} \right)F$
B
$\left( {3\widehat i + 2\widehat j - 3\widehat k} \right)F$
C
$\left( {3\widehat i + 2\widehat j + 3\widehat k} \right)F$
D
$\left( {3\widehat i - 2\widehat j + 3\widehat k} \right)F$

## Explanation

Torque for F1 force

${\overrightarrow F _1}$ = ${F \over 2}\left( { - \widehat i} \right) + {{F\sqrt 3 } \over 2}\left( { - \widehat j} \right)$

${\overrightarrow r _1} = 0\widehat i + 6\widehat j$

${\overrightarrow \tau _{_{{F_1}}}} = \overrightarrow {{r_1}} \times \overrightarrow {{F_1}} = 3F\widehat k$

Torque for F2 force

$\overrightarrow {{F_2}} = F\widehat k$

$\overrightarrow {{r_2}} = 2\widehat i + 3\widehat j$

${\overrightarrow \tau _{_{{F_2}}}} = \overrightarrow {{r_2}} \times \overrightarrow {{F_2}} = 3F\widehat i + 2F\left( { - \widehat j} \right)$

${\overrightarrow \tau _{net}} = {\overrightarrow \tau _{{F_1}}} + {\overrightarrow \tau _{{f_2}}}$

$= 3F\widehat i + 2F\left( { - \widehat j} \right) + 3F\left( {\widehat k} \right)$
4

### JEE Main 2019 (Online) 11th January Morning Slot

An equilateral triangle ABC is cut from a thin solid sheet of wood. (see figure) D, E and F are the mid-points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is I0. If the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I. then : A
${\rm I} = {{{{\rm I}_0}} \over 4}$
B
${\rm I} = {{15} \over {16}}{{\rm I}_0}$
C
${\rm I} = {9 \over {16}}{{\rm I}_0}$
D
${\rm I} = {3 \over 4}{{\rm I}_0}$

## Explanation

Suppose M is mass and a is side of larger triangle, then ${M \over 4}$ and ${a \over 2}$ will be mass and side length of smaller triangle.

${{{{\rm I}_{removed}}} \over {{{\rm I}_{original}}}} = {{{M \over 4}{{\left( {{a \over 2}} \right)}^2}} \over {M{{\left( a \right)}^2}}}$

${{\rm I}_{removed}} = {{{{\rm I}_0}} \over {16}}$

So, ${\rm I}$ = ${\rm I}$ 0 $-$ ${{{{\rm I}_0}} \over {16}}$ = ${{15{{\rm I}_0}} \over {16}}$