1

### JEE Main 2016 (Online) 10th April Morning Slot

A particle of mass m is acted upon by a force F given by the empirical law F =${R \over {{t^2}}}\,v\left( t \right).$ If this law is to be tested experimentally by observing the motion starting from rest, the best way is to plot :
A
$\upsilon$(t) against t2
B
log $\upsilon$(t) against ${1 \over {{t^2}}}$
C
log $\upsilon$(t) against t
D
log $\upsilon$(t) against ${1 \over {{t}}}$

## Explanation

Given,

F = ${R \over {{t^2}}}$ v(t)

$\Rightarrow$   m ${{dv} \over {dt}}$ = ${R \over {{t^2}}}$ (v)

$\Rightarrow$   ${{dv} \over v}$ = ${R \over m}$ ${{dt} \over {{t^2}}}$

Intergrating both sides,

$\int {{{dv} \over v} = {R \over m}\int {{{dt} \over {{t^2}}}} }$

$\Rightarrow$   lnv = ${{R \over m}}$ $\times$ $\left( { - {1 \over t}} \right)$ + C

$\Rightarrow$   lnv = $-$ ${{R \over m}}$ $\left( {{1 \over t}} \right)$ + C

Graph between lnv and ${{1 \over t}}$ will be straight line curve.
2

### JEE Main 2017 (Online) 9th April Morning Slot

A conical pendulum of length 1 m makes an angle $\theta$ = 45o w.r.t. Z-axis and moves in a circle in the XY plane. The radius of the circle is 0.4 m and its center is vertically below O. The speed of the pendulum, in its circular path, will be: (Take g = 10 ms−2 ) A
0.4 m/s
B
4 m/s
C
0.2 m/s
D
2 m/s

## Explanation

FBD of pendulum is : $\therefore\,\,\,$ T sin $\theta$ = ${{m{v^2}} \over r}$

T cos $\theta$ = mg

$\therefore\,\,\,$ tan $\theta$ = ${{{v^2}} \over {rg}}$

$\Rightarrow $$\,\,\, tan45o = {{{v^2}} \over {rg}} \Rightarrow$$\,\,\,$ v2 = rg

$\Rightarrow $$\,\,\, v = \sqrt {0.4 \times 10} = 2 m/s 3 MCQ (Single Correct Answer) ### JEE Main 2018 (Offline) Two masses m1 = 5 kg and m2 = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is : A 10.3 kg B 18.3 kg C 27.3 kg D 43.3 kg ## Explanation Moving block will stop when the friction force between m2 and surface is \ge tension force. So condition for stopping the moving block, f \ge T \Rightarrow \mu N \ge T \Rightarrow \mu \left( {m + {m_2}} \right)g \ge {m_1}g When m is minimum then, \mu \left( {m + {m_2}} \right)g = {m_1}g \Rightarrow m = {{{m_1} - \mu {m_2}} \over \mu } \Rightarrow m = {{5 - 0.15 \times 10} \over {0.15}} = 23.33 kg So if m \ge 23.33 kg then the motion will stop. From option the minimum possible m is 27.3 kg. 4 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 15th April Morning Slot A given object takes n times more time to slide down a {45^ \circ } rough inclined plane as it takes to slide down a perfectly smooth {45^ \circ } incline. The coefficient of kinetic friction between the object and the incline is : A {1 \over {2 - {n^2}}} B 1 - {1 \over {{n^2}}} C \sqrt {1 - {1 \over {{n^2}}}} D \sqrt {{1 \over {1 - {n^2}}}} ## Explanation Let, t1 and t2 are time taken to move on the smooth and rough surface for smooth surface, S = {1 \over 2} g sin45o t_1^2 \Rightarrow$$\,\,\,\,$ t1 = $\sqrt {{{2\sqrt 2 S} \over g}}$

For rough surface,

S = ${1 \over 2}$ g (sin45o $-$ $\mu$k cos45o) $t_2^2$

$\Rightarrow $$\,\,\,\, t2 = \sqrt {{{2\sqrt 2 S} \over {g\left( {1 - {\mu _k}} \right)}}} Here {\mu _k} = Kinetic friction. According to question, t2 = n t1 \Rightarrow$$\,\,\,\,$ ${{2\sqrt 2 \,S} \over {g\left( {1 - {\mu _k}} \right)}}$ = n2 $\times$ ${{2\sqrt 2 \,S} \over g}$

$\Rightarrow $$\,\,\,\, 1 - \mu k = {1 \over {{n^2}}} \Rightarrow$$\,\,\,\,$ ${\mu _k}$ = 1 $-$ ${1 \over {{n^2}}}$