1

JEE Main 2016 (Online) 9th April Morning Slot

A cubical block of side 30 cm is moving with velocity 2 ms−1 on a smooth horizontal surface. The surface has a bump at a point O as shown in figure. The angular velocity (in rad/s) of the block immediately after it hits the bump, is :

A
5.0
B
6.7
C
9.4
D
13.3

Explanation

Before hitting point 0,

angular moment = mv $\times$ ${a \over 2}$

After hitting point 0,

Angular momentum = ${\rm I}\omega$

$\therefore$    ${\rm I}\omega$ = ${{mva} \over 2}$

$\Rightarrow$   $\omega$ = ${{mva} \over {2{\rm I}}}$

${\rm I}$ = moment of inertia about edge,

=   ${{m{a^2}} \over 6} + m{\left( {{a \over {\sqrt 2 }}} \right)^2}$

=   ${{m{a^2}} \over 6} + {{m{a^2}} \over 2}$

=   ${{2m{a^2}} \over 3}$

$\therefore$   $\omega$ = ${{mva} \over {2 \times {{2m{a^2}} \over 3}}}$ = ${{3v} \over {4a}}$ = ${{3 \times 2} \over {4 \times 0.3}}$ = 5 rad/s
2

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
3

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
4

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

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

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