 JEE Mains Previous Years Questions with Solutions

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1

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

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

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

$= {{4\ell } \over 3}$
2

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.
3

AIEEE 2005

The moment of inertia of a uniform semicircular disc of mass $M$ and radius $r$ about a line perpendicular to the plane of the disc through the center is
A
${2 \over 5}M{r^2}$
B
${1 \over 4}Mr$
C
${1 \over 2}M{r^2}$
D
$M{r^2}$

Explanation

Let mass of the semi circular disc = M

Now assume a disc which is combination of two semi circular parts. Let $I$ be the moment of inertia of the uniform semicircular disc. So $2I$ will be the moment of inertia of the full circular disc and 2M will be the mass.

$\Rightarrow 2I = {{2M{r^2}} \over 2}$

$\Rightarrow I = {{M{r^2}} \over 2}$
4

AIEEE 2005

An annular ring with inner and outer radii ${R_1}$ and ${R_2}$ is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, ${{{F_1}} \over {{F_2}}}\,$ is
A
${\left( {{{{R_1}} \over {{R_2}}}} \right)^2}$
B
${{{{R_2}} \over {{R_1}}}}$
C
${{{{R_1}} \over {{R_2}}}}$
D
$1$

Explanation

Let the mass of each particle is m.

Then force experienced by each particle, $F = m{\omega ^2}R$

$\therefore$ ${{{F_1}} \over {{F_2}}} = {{m{\omega ^2}{R_1}} \over {m{\omega ^2}{R_2}}}$

$\Rightarrow$ ${{{F_1}} \over {{F_2}}} = {{{R_1}} \over {{R_2}}}$