 JEE Mains Previous Years Questions with Solutions

4.5
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1

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 $I'\omega ' = I\omega \,\,$

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

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

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

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

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

3

AIEEE 2005

A T shaped object with dimensions shown in the figure, is lying on a smooth floor. A force $'\,\,\overrightarrow F \,\,'$ is applied at the point $P$ parallel to $AB,$ such that the object has only the translational motion without rotation. Find the location of $P$ with respect to $C.$ A
${3 \over 2}L$
B
${2 \over 3}L$
C
$L$
D
${4 \over 3}L$

Explanation This is a case of translation motion without rotation, the force $\overrightarrow F$ has to be applied at center of mass.

i.e. the point $'P'$ has to be at the centre of mass

$y = {{{m_1}{y_1} + {m_2}{y_2}} \over {{m_1} + {m_2}}}$

$= {{m \times 2\ell + 2m \times \ell } \over {3m}}$

$= {{4\ell } \over 3}$
4

AIEEE 2005

A body $A$ of mass $M$ while falling vertically downloads under gravity breaks into two-parts; a body $B$ of mass ${1 \over 3}$ $M$ and a body $C$ of mass ${2 \over 3}$ $M.$ The center of mass of bodies $B$ and $C$ taken together shifts compared to that of bodies $B$ and $C$ taken together shifts compared to that of body $A$ towards
A
does not shift
B
depends on height of breaking
C
body $B$
D
body $C$

Explanation

The center of mass does not shift as no external force is applied horizontally. So the center of mass of the system continues its original path. It is only the internal forces which comes into play while breaking.