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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
A block of mass $$m$$ is placed on a surface with a vertical cross section given by $$y = {{{x^3}} \over 6}.$$ If the coefficient of friction is $$0.5,$$ the maximum height above the ground at which the block can be placed without slipping is:
A
$${1 \over 6}m$$
B
$${2 \over 3}m$$
C
$${1 \over 3}m$$
D
$${1 \over 2}m$$

Explanation

At limiting equilibrium, $$\mu = \tan \theta $$

Equation of the surface,

$$y = {{{x^3}} \over 6}$$

Slope, $$\tan \theta = \mu = {{dy} \over {dx}} = {{{x^2}} \over 2}$$


Given that, Coefficient of friction $$\mu = 0.5$$

$$\therefore$$ $$\,\,\,\,0.5 = {{{x^2}} \over 2}$$

$$ \Rightarrow \,\,\,x = \pm \,1$$

Now, $$y = {{{x^3}} \over 6} = {1 \over 6}m$$
2

AIEEE 2010

MCQ (Single Correct Answer)
Two fixed frictionless inclined planes making an angle $${30^ \circ }$$ and $${60^ \circ }$$ with the vertical are shown in the figure. Two blocks $$A$$ and $$B$$ are placed on the two planes. What is the relative vertical acceleration of $$A$$ with respect to $$B$$ ?
A
$$4.9m{s^{ - 2}}$$ in horizontal direction
B
$$9.8m{s^{ - 2}}$$ in vertical direction
C
Zero
D
$$4.9m{s^{ - 2}}$$ in vertical direction

Explanation

Along inclined plane the equation of motion of the body

$$mg\,\sin \,\theta = ma$$ $$\,\,\,\,\,\,\,\,$$ $$\therefore$$ $$a = g\,\sin \,\theta $$

where $$a$$ is along the inclined plane.

$$\therefore$$ vertical component of acceleration is $$\left( {g\sin \theta } \right)\sin \theta $$ = $$g\,{\sin ^2}\theta $$ (Along vertical)

For block A,
$${a_{A\left( {along\,vertical} \right)}} = g{\sin ^2}60^\circ $$

For block B,
$${a_{B\left( {along\,vertical} \right)}} = g{\sin ^2}30^\circ $$

$$\therefore$$ relative vertical acceleration of $$A$$ with respect to $$B$$ is

$$g{\sin ^2}60^\circ $$ - $$g{\sin ^2}30^\circ $$

=$$g\left( {{{\sin }^2}60 - {{\sin }^2}\left. {30} \right]} \right.$$

$$= 4.9$$ $$\,\,m/{s^2}$$ in vertical direction
3

AIEEE 2007

MCQ (Single Correct Answer)
A block of mass $$m$$ is connected to another block of $$mass$$ $$M$$ by a spring (massless) of spring constant $$k.$$ The block are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then a constant force $$F$$ starts acting on the block of mass $$M$$ to pull it. Find the force of the block of mass $$m.$$
A
$${{MF} \over {\left( {m + M} \right)}}$$
B
$${{mF} \over M}$$
C
$${{\left( {M + m} \right)F} \over m}$$
D
$${{mF} \over {\left( {m + M} \right)}}$$

Explanation


From free body-diagram of $$m$$

we get $$T = ma$$

From free body-diagram of $$M$$

we get $$F-T=Ma$$

where $$T$$ is force due to spring

$$ \Rightarrow F - ma = Ma$$

$$ \Rightarrow$$ $$F=Ma+ma$$

$$\therefore$$ $$a = {F \over {M + m}}$$

Now, force acting on the block of mass $$m$$ is

$$ma = m\left( {{F \over {M + m}}} \right) = {{mF} \over {m + M}}.$$
4

AIEEE 2006

MCQ (Single Correct Answer)
A player caught a cricket ball of mass $$150$$ $$g$$ moving at a rate of $$20$$ $$m/s.$$ If the catching process is completed in $$0.1s,$$ the force of the blow exerted by the ball on the hand of the player is equal to
A
$$150$$ $$N$$
B
$$3$$ $$N$$
C
$$30$$ $$N$$
D
$$300$$ $$N$$

Explanation

We know, Force$$ \times $$ time = Impulse = Change in momentum

$$\therefore$$ $$F \times t = m\left( {v - u} \right)$$

$$ \Rightarrow $$ $$F = {{m\left( {v - u} \right)} \over t} = {{0.15\left( {0 - 20} \right)} \over {0.1}} = 30N$$

Questions Asked from Laws of Motion

On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
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