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1

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

A circular disc D1 of mass M and radius R has two identical discs D2 and D3 of the same mass M and radius R attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis OO' ,passing through the centre of D1 as shown in the figure, will be :

A
3MR2
B
MR2
C
$${2 \over 3}$$ MR2
D
$${4 \over 5}$$ MR2

## Explanation

I $$=$$ $${{M{R^2}} \over 2} + 2\left( {{{M{R^2}} \over 4} + M{R^2}} \right)$$

$$= {{M{R^2}} \over 2} + {{M{R^2}} \over 2} + 2M{R^2}$$

$$=$$ $$3M{R^2}$$
2

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

The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the origin. If the force acting on it is 1 N, and the distance of the particle from the origin is 5m, the angle between the force and the position vector is (in radians) :
A
$${\pi \over 8}$$
B
$${\pi \over 6}$$
C
$${\pi \over 4}$$
D
$${\pi \over 3}$$

## Explanation

2.5 = 1 $$\times$$ 5 sin $$\theta$$

sin$$\theta$$ = 0.5 = $${1 \over 2}$$

$$\theta$$ = $${\pi \over 6}$$
3

### 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 $$-$$ $${{{{\rm I}_0}} \over {16}}$$ = $${{15{{\rm I}_0}} \over {16}}$$
4

### 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

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

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

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

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

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

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