1

### JEE Main 2019 (Online) 9th January Morning Slot

Two masses m and ${m \over 2}$ are connected at the two ends of a massless rigid rod of length l. The rod is suspended by a thin wire of torsional constant k, at the centre of mass of the rod-mass system(see figure). Because of torsional constant k, the restoring torque is $\tau$ = k$\theta$ for angular displacement $\theta$. If the rod is rotated by $\theta$0 and released, the tension in it when it passes through its mean position will be :

A
${{3k{\theta _0}^2} \over l}$
B
${{2k{\theta _0}^2} \over l}$
C
${{k{\theta _0}^2} \over l}$
D
${{k{\theta _0}^2} \over {2l}}$

## Explanation

$\therefore$   ${r_1} = \left( {{{{n \over 2}} \over {m + {m \over 2}}}} \right)l$

${r_1} = {l \over 3}$

When rod is rotated by $\theta$0 and released, then it look this.

After roating $\theta$0 and released, when the rod go through mean position it will have maximum speed.

Vmax $=$ A$\omega$

$=$ r1$\theta$0$\omega$

$=$ $\left( {{l \over 3}{\theta _0}} \right)\omega$

We know,

$\omega$ $= \sqrt {{K \over {\rm I}}}$

and    ${\rm I} = m{\left( {{l \over 3}} \right)^2} + {m \over 2}{\left( {{{2l} \over 3}} \right)^2}$

$= {{m{l^2}} \over 9} + {{4m{l^2}} \over {18}}$

$= {{2m{l^2} + 4m{l^2}} \over {18}}$

$= {{m{l^2}} \over 3}$

$\therefore$   $\omega = \sqrt {{{3K} \over {m{l^2}}}}$

$\therefore$   Tension in the rod when it passes through the mean position is,

T = ${{mv_{\max }^2} \over {{r_1}}}$

$= {{m{{\left( {{l \over 3}{\theta _0}\omega } \right)}^2}} \over {{l \over 3}}}$

$= {{m{{\left( {{l \over 3}} \right)}^2}\theta _0^2{\omega ^2}} \over {{\ell \over 3}}}$

$= m\left( {{l \over 3}} \right)\theta _0^2\left( {{{3K} \over {m{l^2}}}} \right)$

$= {{K\theta _0^2} \over l}$
2

### JEE Main 2019 (Online) 10th January Evening Slot

A rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be -

A
${g \over {13l}}$
B
${g \over {2l}}$
C
${g \over {3l}}$
D
${7g \over {3l}}$

## Explanation

Applying torque equation about point P.

2M0 (2l) $-$ 5 M0 gl = I$\alpha$

I = 2M0 (2l)2 + 5M0 l2 = 13 M0l2d

$\therefore$  $\alpha$ = $-$ ${{{M_0}gl} \over {13{M_0}{\ell ^2}}}$ $\Rightarrow$ $\alpha$ = $-$ ${g \over {13\ell }}$

$\therefore$  $\alpha$ = ${g \over {13\ell }}$ anticlockwise
3

### JEE Main 2019 (Online) 11th January Evening Slot

A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here k is a constant. The value of T is :
A
$2\sqrt {{k \over p}}$
B
$2\sqrt {{p \over k}}$
C
$\sqrt {{{2p} \over 2}}$
D
$\sqrt {{{2k} \over p}}$

## Explanation

${{dp} \over {dt}} = F = kt$

$\int_P^{3P} {dP} = \int_0^T {kt\,dt}$

$2p = {{K{T^2}} \over 2}$

$T = 2\sqrt {{P \over K}}$
4

### JEE Main 2019 (Online) 12th January Morning Slot

The position vector of the centre of mass $\overrightarrow r {\,_{cm}}\,$ of an asymmetric uniform bar of negligible area of crosssection as shown in figure is :

A
${\overrightarrow r _{cm}}\, = {{11} \over 8}L\,\,\widehat x + {8 \over 8}L\widehat y$
B
${\overrightarrow r _{cm}}\, = {5 \over 8}L\,\,\widehat x + {{13} \over 8}L\widehat y$
C
${\overrightarrow r _{cm}}\, = {{13} \over 8}L\,\,\widehat x + {5 \over 8}L\widehat y$
D
${\overrightarrow r _{cm}}\, = {3 \over 8}L\,\,\widehat x + {{11} \over 8}L\widehat y$

## Explanation

${X_{cm}} = {{2mL + 2mL + {{5mL} \over 2}} \over {4m}} = {{13} \over 8}L$

${Y_{cm}} = {{2m \times L + m \times \left( {{L \over 2}} \right) + m \times 0} \over {4m}} = {{5L} \over 8}$