### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2006

A force of $- F\widehat k$ acts on $O,$ the origin of the coordinate system. The torque about the point $(1, -1)$ is
A
$F\left( {\widehat i - \widehat j} \right)$
B
$- F\left( {\widehat i + \widehat j} \right)$
C
$F\left( {\widehat i + \widehat j} \right)$
D
$- F\left( {\widehat i - \widehat j} \right)$

## Explanation

We know, Torque $\overrightarrow \tau = \overrightarrow r \times \overrightarrow F$

$= \left( {\widehat i - \widehat j} \right) \times \left( { - F\widehat k} \right)$

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

### AIEEE 2006

Four point masses, each of value $m,$ are placed at the corners of a square $ABCD$ of side $l$. The moment of inertia of this system about an axis passing through $A$ and parallel to $BD$ is
A
$2m{l^2}$
B
$\sqrt 3 m{l^2}$
C
$3m{l^2}$
D
$m{l^2}$

## Explanation

Let ${I_{A}}$ is the moment of inertia about an axis passing through A and parallel to BD.

${I_{A}} = M.I\,$ due to the point mass at $B+$

$M.I$ due to the point mass at $D+$

$M.I$ due to the point mass at $C.$

${I_{A}} = 2 \times m{\left( {{\textstyle{\ell \over {\sqrt 2 }}}} \right)^2} + m{\left( {\sqrt 2 \ell } \right)^2}$

$= m{\ell ^2} + 2m{\ell ^2} = 3m{\ell ^2}$

3

### AIEEE 2006

A thin circular ring of mass $m$ and radius $R$ is rotating about its axis with a constant angular velocity $\omega$. Two objects each of mass $M$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity $\omega ' =$
A
${{\omega \left( {m + 2M} \right)} \over m}$
B
${{\omega \left( {m - 2M} \right)} \over {\left( {m + 2M} \right)}}$
C
${{\omega m} \over {\left( {m + M} \right)}}$
D
${{\omega m} \over {\left( {m + 2M} \right)}}$

## Explanation

Here angular momentum is conserved.

Applying conservation of angular momentum $I'\omega ' = I\omega \,\,$

$\left( {m{R^2} + 2M{R^2}} \right)\omega \,' = m{R^2}\omega$

$\Rightarrow \omega \,' = \omega \left[ {{m \over {m + 2M}}} \right]$
4

### AIEEE 2006

Consider a two particle system with particles having masses ${m_1}$ and ${m_2}$. If the first particle is pushed towards the center of mass through a distance $d,$ by what distance should the second particle is moved, so as to keep the center of mass at the same position?
A
${{{m_2}} \over {{m_1}}}\,\,d$
B
${{{m_1}} \over {{m_1} + {m_2}}}d$
C
${{{m_1}} \over {{m_2}}}d$
D
$d$

## Explanation

Initially,

$0 = {{{m_1}\left( { - {x_1}} \right) + {m_2}{x_2}} \over {{m_1} + {m_2}}} \Rightarrow {m_1}{x_1} = {m_2}{x_2}$

Finally,

mass m1 moved towards the center a distance d so the distance of mass m1 from the origin is x1 - d, and now let mass m2 need to move d' to keep the center at the origin.

$\therefore$ $0 = {{{m_1}\left( {d - {x_1}} \right) + {m_2}\left( {{x_2} - d'} \right)} \over {{m_1} + {m_2}}}$

$\Rightarrow 0 = {m_1}d - {m_1}{x_1} + {m_2}{x_2} - {m_2}d'$

$\Rightarrow d' = {{{m_1}} \over {{m_2}}}d$