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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2016 (Offline)

MCQ (Single Correct Answer)
A roller is made by joining together two cones at their vertices $$0$$. It is kept on two rails $$AB$$ and $$CD$$, which are placed asymmetrically (see figure), with its axis perpendicular to $$CD$$ and its center $$O$$ at the center of line joining $$AB$$ and $$CD$$ (see figure). It is given a light push so that it starts rolling with its center $$O$$ moving parallel to $$CD$$ in the direction shown. As it moves, the roller will tend to :
A
go straight
B
turn left and right alternately
C
turn left
D
turn right

Explanation


As shown in the diagram, the normal reaction of $$AB$$ on roller will shift towards $$O.$$ This will lead to tending of the system of cones to turn left.
2

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
From a solid sphere of mass $$M$$ and radius $$R$$ a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its face is:
A
$${{4M{R^2}} \over {9\sqrt {3\pi } }}$$
B
$${{4M{R^2}} \over {3\sqrt {3\pi } }}$$
C
$${{M{R^2}} \over {32\sqrt {2\pi } }}$$
D
$${{M{R^2}} \over {16\sqrt {2\pi } }}$$

Explanation

From the diagram you can see diagonal of the cube is equal to the diameter of the sphere.

$$\therefore$$ $$\sqrt 3 a = 2R$$

$$ \Rightarrow $$ $$a = {2 \over {\sqrt 3 }}R$$

Let M is the mass of sphere and M' is the mass of the cube.

So $${M \over {M'}} = {{{4 \over 3}\pi {R^3}} \over {{a^3}}}$$

$$ = {{{4 \over 3}\pi {R^3}} \over {{{\left( {{2 \over {\sqrt 3 }}R} \right)}^3}}} = {{\sqrt 3 } \over 2}\pi $$

$$\therefore$$ $$M' = {{2M} \over {\sqrt 3 \pi }}$$

Moment of inertia of the cube about the axis passing through its center and perpendicular to one of its face,

$$I = {{M'{a^2}} \over 6} = {{{{2M} \over {\sqrt 3 \pi }} \times \left( {{2 \over {\sqrt 3 }}R} \right)} \over 6}$$

$$ = {{4M{R^2}} \over {9\sqrt 3 \pi }}$$
3

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
Distance of the center of mass of a solid uniform cone from its vertex is $$z{}_0$$. If the radius of its base is $$R$$ and its height is $$h$$ then $$z{}_0$$ is equal to :
A
$${{5h} \over 8}$$
B
$${{3{h^2}} \over {8R}}$$
C
$${{{h^2}} \over {4R}}$$
D
$${{3h} \over 4}$$

Explanation

Let the density of solid cone $$\rho $$.

$$dm = \rho \pi {r^2}dy$$

$${y_{cm}} = {{\int {ydm} } \over {\int {dm} }}$$
$$ = {{\int\limits_0^h {\pi {r^2}} dy\rho \times y} \over {{1 \over 3}\pi {R^2}h\rho }}$$
$$ = {{3h} \over 4}$$
4

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
A mass $$'m'$$ is supported by a massless string wound around a uniform hollow cylinder of mass $$m$$ and radius $$R.$$ If the string does not slip on the cylinder, with what acceleration will the mass fall or release?
A
$${{2g} \over 3}$$
B
$${{g} \over 2}$$
C
$${{5g} \over 6}$$
D
$$g$$

Explanation


Here string is not slipping over pulley.

From figure,

Acceleration $$a = R\alpha \,\,\,\,...(i)$$

Applying Newton's second law on the block we get,

$$mg-T=ma$$ $$...(ii)$$

Torque on cylinder due to tension in string about the fixed point,

$$T \times R = I\alpha$$

$$T \times R = m{R^2}\alpha = m{R^2}\left( {{a \over R}} \right)$$

[ as $$I$$ = MR2 for hollow cylinder]

or $$T=ma$$

$$ \Rightarrow \,\,mg - ma = ma$$

$$ \Rightarrow \,\,\,a = {g \over 2}$$

Questions Asked from Rotational Motion

On those following papers in MCQ (Single Correct Answer)
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