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

Concrete mixture is made by mixing cement, stone and sand in a rotating cylindrical drum. If the drum rotates too fast, the ingredients remain stuck to the wall of the drum and proper mixing of ingredients does not take place. The maximum rotational speed of the drum in revolutions per minute(rpm) to ensure proper mixing is close to :

(Take the radius of the drum to be 1.25 m and its axle to be horizontal) :
A
0.4
B
1.3
C
8.0
D
27.0
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### 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
3

### 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 ${V = \pi {R^2}l}$ ]

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

$\Rightarrow {l \over R} = \sqrt {{3 \over 2}}$
4

### 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, $I$ = ${{M{l^2}} \over 3}$

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

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