1

### JEE Main 2016 (Online) 10th April Morning Slot

In the figure shown ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then ${{BC} \over {AB}}$ is close to : A
1.85
B
1.37
C
1.5
D
3

## Explanation Here   AB = x

and   BC = y

and   $\lambda$ = linear mass density.

As centre of mass is below point A, so horizontal distance of the centre of mass from B is = xcos60o = ${x \over 2}$

$\therefore$   XCM = ${x \over 2}$ = ${{{m_1}{x_1} + {m_2}{x_2}} \over {{m_1} + {m_2}}}$

$\Rightarrow$   ${x \over 2}$ = ${{\left( {\lambda x} \right)\left( {{x \over 2}} \right)\cos {{60}^o} + \left( {\lambda y} \right)\left( {{y \over 2}} \right)} \over {\lambda \left( {x + y} \right)}}$

$\Rightarrow$   ${x \over 2}$ = ${{{{{x^2}} \over 4} + {{{y^2}} \over 2}} \over {x + y}}$

$\Rightarrow$   x2 + xy = ${{{x^2}} \over 2} + {y^2}$

$\Rightarrow$   x2 + 2xy $-$ 2y2 = 0

$\therefore$   x = ${{ - 2y \pm \sqrt {{{\left( {2y} \right)}^2} - 4.1\left( { - 2{y^2}} \right)} } \over {2.1}}$

=    ${{ - 2y \pm \sqrt {12{y^2}} } \over 2}$

=   $-$ y $\pm$ $\sqrt 3$y

${x \over y} \ne - \sqrt 3 - 1$  as  ${x \over y}$ = positive.

$\therefore$   ${x \over y}$ = $\sqrt 3 - 1$

$\Rightarrow$   ${y \over x}$ = ${1 \over {\sqrt 3 - 1}} \times {{\sqrt 3 + 1} \over {\sqrt 3 + 1}}$

=   ${{\sqrt 3 + 1} \over 2}$

=   ${{2.732} \over 2}$

=   1.366 $\simeq$ 1.37
2

### JEE Main 2017 (Offline)

The moment of inertia of a uniform cylinder of length $l$ and radius R about its perpendicular bisector is $I$. What is the ratio ${l \over R}$ such that the moment of inertia is minimum?
A
${3 \over {\sqrt 2 }}$
B
$\sqrt {{3 \over 2}}$
C
${{\sqrt 3 } \over 2}$
D
1

## Explanation

The volume of the cylinder V = $\pi {R^2}l$

$\therefore$ ${R^2} = {V \over {\pi l}}$

We know, moment of inertia of a uniform cylinder of length $l$ and radius R about its perpendicular bisector is,

$I = {{M{l^2}} \over {12}} + {{M{R^2}} \over 4}$

[ Putting ${R^2} = {V \over {\pi l}}$ in this equation]

$\Rightarrow$ $I = {{M{l^2}} \over {12}} + {{MV} \over {4\pi l}}$

Here $I$ is a function of $l$ as M and V are constant.

$I$ will be maximum or minimum when ${{{dI} \over {dl}}}$ = 0.

$\Rightarrow {{Ml} \over 6} - {{MV} \over {4\pi {l^2}}} = 0$

$\Rightarrow {{Ml} \over 6} = {{MV} \over {4\pi {l^2}}}$

$\Rightarrow {l \over 6} = {{\pi {R^2}l} \over {4\pi {l^2}}}$ [ as ${V = \pi {R^2}l}$ ]

$\Rightarrow {{{R^2}} \over {{l^2}}} = {4 \over 6}$

$\Rightarrow {l \over R} = \sqrt {{3 \over 2}}$
3

### JEE Main 2017 (Offline)

A slender uniform rod of mass M and length $l$ is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle $\theta$ with the vertical is A
${{2g} \over {3l}}\cos \theta$
B
${{3g} \over {2l}}\sin \theta$
C
${{2g} \over {3l}}\sin \theta$
D
${{3g} \over {3l}}\sin \theta$

## Explanation

Forces acting on the rod are shown below. Torque about pivot point O due to force Nx and Ny are zero.

Mgcos$\theta$ is passing through O, so torque will be zero due to this force.

So torque about the point O is

$\tau = Mg\sin \theta \times {l \over 2}$

We know, $\tau = I\alpha$

$\therefore$ $I\alpha = Mg\sin \theta \times {l \over 2}$

Moment of inertia of rod about point O, $I$ = ${{M{l^2}} \over 3}$

$\therefore$ ${{M{l^2}} \over 3} \times \alpha = Mg\sin \theta \times {l \over 2}$

$\Rightarrow \alpha = {3 \over 2}{g \over l}\sin \theta$
4

### JEE Main 2017 (Online) 8th April Morning Slot

Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre O and perpendicular to its plane is Io as shown in the figure. A cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is : A
${7 \over 8}$ Io
B
${15 \over 16}$ Io
C
${{3\,{{\rm I}_o}} \over 4}$
D
${{31\,{{\rm I}_o}} \over 32}$

## Explanation

Let, side of triangle ABC = $\ell$

According to perpendicular axes theorem, moment of inertia of triangle about it center and perpendicular to its plane,

IO = ${1 \over {12}}$ m$\ell$2

In, triangle DEF,

DE = DF = EF = ${1 \over 2}$ AB = ${1 \over 2}$ $\ell$

$\therefore\,\,\,$ moment of inertia of triangle DEF,

IDEF = ${1 \over {12}} \times {m \over 4} \times {\left( {{\ell \over 2}} \right)^2}$

= ${1 \over {12}} \times {{m{\ell ^2}} \over {16}}$

= ${{{I_O}} \over {16}}$

$\therefore\,\,\,$ Moment of inertia of the remaining part,

Iremain = IO $-$ ${{{{\rm I}_O}} \over {16}}$ = ${{15\,{{\rm I}_O}} \over {16}}$