 JEE Mains Previous Years Questions with Solutions

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1

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 $C.$

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

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

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.