1

### 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

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

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

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

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

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

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

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

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

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

$\Rightarrow$  ${M \over m} = {{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}} \Rightarrow M = {{{\theta _0} - \theta 1} \over {{\theta _0} + {\theta _1}}}$
4

### JEE Main 2019 (Online) 8th April Morning Slot

If 1022 gas molecules each of mass 10–26 kg collide with a surface (perpendicular to it) elastically per second over an area 1 m2 with a speed 104 m/s, the pressure exerted by the gas molecules will be of the order of :
A
108 N/m2
B
1016 N/m2
C
104 N/m2
D
2 N/m2