1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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



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

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

Questions Asked from Impulse & Momentum

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