Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A circular disc D_{1} of mass M and radius R has two identical discs D_{2} and D_{3} 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 D_{1} as shown in the figure, will be :

A

3MR^{2}

B

MR^{2}

C

$${2 \over 3}$$ MR^{2}

D

$${4 \over 5}$$ MR^{2}

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

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

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

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

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

2

MCQ (Single Correct Answer)

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

$${\pi \over 3}$$

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

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

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

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

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

3

MCQ (Single Correct Answer)

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

$${\rm I} = {3 \over 4}{{\rm I}_0}$$

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

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

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

So, $${\rm I}$$ = $${\rm I}$$

4

MCQ (Single Correct Answer)

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

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

Torque for F_{1} 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 F_{2} force

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

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

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

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

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

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

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

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

Torque for F

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

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

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

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

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

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