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1

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)
A simple pendulum, made of a string of length $$\ell $$ and a bob of mass m, is released from a small angle $${{\theta _0}}$$. It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle $${{\theta _1}}$$. Then M is given by :
A
$${m \over 2}\left( {{{{\theta _0} + {\theta _1}} \over {{\theta _0} - {\theta _1}}}} \right)$$
B
$${m \over 2}\left( {{{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}}} \right)$$
C
$$m\left( {{{{\theta _0} + {\theta _1}} \over {{\theta _0} - {\theta _1}}}} \right)$$
D
$$m\left( {{{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}}} \right)$$

Explanation



v = $$\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)} $$

v1 = $$\sqrt {2g\ell \left( {1 - \cos {\theta _1}} \right)} $$

By momentum conservation

m$$\sqrt {2gl\left( {1 - \cos {\theta _0}} \right)} $$
$$= M{V_m} - m\sqrt {2g\left( {1 - \cos \theta } \right)} $$

$$ \Rightarrow $$$$m\sqrt {2g\ell } \left\{ {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right\}$$

$$ = $$ MVm

and  e = 1 = $${{{V_m} + \sqrt {2g\ell \left( {1 - \cos {\theta _1}} \right)} } \over {\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)} }}$$

$$\sqrt {2g\ell } $$ $$\left( {\sqrt {1 - \cos {\theta _0}} - \sqrt {1 - \cos {\theta _1}} } \right) $$
$$= $$ Vm     . . .(I)

m$$\sqrt {2g\ell } \left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)$$
$$ = $$ MVM     . . .(II)

Dividing

$${{\left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)} \over {\left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)}} = {M \over m}$$

By componendo divided

$${{m - M} \over {m + M}}$$ = $${{\sqrt {1 - \cos {\theta _1}} } \over {\sqrt {1 - \cos {\theta _0}} }} = {{\sin \left( {{{{\theta _1}} \over 2}} \right)} \over {\sin \left( {{{{\theta _0}} \over 2}} \right)}}$$

$$ \Rightarrow $$  $${M \over m} = {{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}} \Rightarrow M = {{{\theta _0} - \theta 1} \over {{\theta _0} + {\theta _1}}}$$
2

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)
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



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

$${Y_{cm}} = {{2m \times L + m \times \left( {{L \over 2}} \right) + m \times 0} \over {4m}} = {{5L} \over 8}$$
3

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)
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) 10th January Evening Slot

MCQ (Single Correct Answer)
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

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