1
MCQ (Single Correct Answer)

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
2
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.
3
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 $$-$$ $${1 \over {{n^2}}}$$
4
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Evening Slot

A body of mass 2 kg slides down with an acceleration of 3 m/s2 on a rough inclined plane having a slope of $${30^o}$$. The external force required to take the same body up the plane with the same acceleration will be : (g = 10 m/s2)
A
14 N
B
20 N
C
6 N
D
4 N

Explanation

When mass slide down then,

Mgsin$$\theta $$ $$-$$ $$\mu $$Mg cos$$\theta $$ = Ma

$$ \Rightarrow $$ $$\,\,\,$$ a = g(sin30o $$-$$ $$\mu $$ cos30o)

When mass pushed upward with force F,

F $$-$$ Mgsin$$\theta $$ $$-$$ $$\mu $$ Mgcos$$\theta $$ = Ma

$$ \Rightarrow $$ $$\,\,\,$$ F = Mg(sin30o + $$\mu $$ cos 30o) + Mg(sin30o $$-$$ $$\mu $$ cos30o)

$$ \Rightarrow $$ $$\,\,\,$$ F = 2Mg sin30o

= 2 $$ \times $$ 2 $$ \times $$ 10 $$ \times $$ $${1 \over 2}$$

= 20 N

Questions Asked from Laws of Motion

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