 JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

AIEEE 2002

Moment of inertia of a circular wire of mass $M$ and radius $R$ about its diameter is
A
${{M{R^2}} \over 2}$
B
$M{R^2}$
C
$2M{R^2}$
D
${{M{R^2}} \over 4}$

Explanation

Moment of Inertia of a circular wire about an axis $nn'$ passing through the centre of the circle and perpendicular to the plane of the circle $= M{R^2}$ As shown in the figure, $X$-axis and $Y$-axis lies in the plane of the ring. Then by perpendicular axis theorem

${I_X} + {I_Y} = {I_Z}$

$\Rightarrow 2{I_X} = M{R^2}\,$ $\left[ \, \right.$ as ${I_X} - {I_Y}$ (by symmetry) and ${I_Z} = M{R^2}$ $\left. \, \right]$

$\therefore$ ${I_X} = {1 \over 2}M{R^2}$
2

AIEEE 2002

Two identical particles move towards each other with velocity $2v$ and $v$ respectively. The velocity of center of mass is
A
$v$
B
$v/3$
C
$v/2$
D
zero

Explanation The velocity of center of mass of two particle system is

${v_c} = {{{m_1}{v_1} + {m_2}{v_2}} \over {{m_1} + {m_2}}}$

$= {{m\left( {2v} \right) + m\left( { - v} \right)} \over {m + m}}$

$= {v \over 2}$
3

AIEEE 2002

A particle of mass $m$ moves along line PC with velocity $v$ as shown. What is the angular momentum of the particle about P? A
$mvL$
B
$mvl$
C
$mvr$
D
zero

Explanation

Angular momentum $(L)$

$=$ ( linear momentum ) $\times$ ( perpendicular distance of the line of action of momentum from the axis of rotation)

$= mv \times r$

$= mv \times 0$

$=0$

[ Here $r=0$ because the particle is moving through the line PQ and r is the perpendicular distance from line PQ of the particle ]
4

AIEEE 2002

A solid sphere, a hollow sphere and a ring are released from top of an inclined plane (frictionless) so that they slide down the plane. Then maximum acceleration down the plane is for (no rolling)
A
solid sphere
B
hollow sphere
C
ring
D
all same

Explanation

Each bodies is sliding along the frictionless inclined plane and there is no rolling, therefore the acceleration of all the bodies is same $\left( {g\,\sin \,\theta } \right).$