1

### JEE Main 2016 (Online) 9th April Morning Slot

A rocket is fired vertically from the earth with an acceleration of 2g, where g is the gravitational acceleration. On an inclined plane inside the rocket, making an angle $\theta$ with the horizontal, a point object of mass m is kept. The minimum coefficient of friction $\mu$min between the mass and the inclined surface such that the mass does not move is :
A
tan$\theta$
B
2tan$\theta$
C
3tan$\theta$
D
tan2$\theta$

## Explanation

Rocket is moving upward with acceleration 2g and gravitation acceleration is g downward direction.

So, acceleration experienced by the point object,

= 2g $-$ ($-$ g) = 3g At equilibrium,

N = 3mgcos$\theta$

$\mu$N = 3mgsin$\theta$

$\Rightarrow$   $\mu$ (3mgcos$\theta$) = 3mg sin$\theta$

$\Rightarrow$   $\mu$ = tan$\theta$
2

### 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.
3

### 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
4

### 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.