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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2006

MCQ (Single Correct Answer)
A force of $$ - F\widehat k$$ acts on $$O,$$ the origin of the coordinate system. The torque about the point $$(1, -1)$$ is
A
$$F\left( {\widehat i - \widehat j} \right)$$
B
$$ - F\left( {\widehat i + \widehat j} \right)$$
C
$$F\left( {\widehat i + \widehat j} \right)$$
D
$$ - F\left( {\widehat i - \widehat j} \right)$$

Explanation

We know, Torque $$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F $$

$$= \left( {\widehat i - \widehat j} \right) \times \left( { - F\widehat k} \right) $$

$$= F\left( {\widehat i + \widehat j} \right)$$
2

AIEEE 2006

MCQ (Single Correct Answer)
Four point masses, each of value $$m,$$ are placed at the corners of a square $$ABCD$$ of side $$l$$. The moment of inertia of this system about an axis passing through $$A$$ and parallel to $$BD$$ is
A
$$2m{l^2}$$
B
$$\sqrt 3 m{l^2}$$
C
$$3m{l^2}$$
D
$$m{l^2}$$

Explanation

Let $${I_{A}}$$ is the moment of inertia about an axis passing through A and parallel to BD.

$${I_{A}} = M.I\,$$ due to the point mass at $$B+$$

$$M.I$$ due to the point mass at $$D+$$

$$M.I$$ due to the point mass at $$C.$$

$${I_{A}} = 2 \times m{\left( {{\textstyle{\ell \over {\sqrt 2 }}}} \right)^2} + m{\left( {\sqrt 2 \ell } \right)^2}$$

$$ = m{\ell ^2} + 2m{\ell ^2} = 3m{\ell ^2}$$

3

AIEEE 2006

MCQ (Single Correct Answer)
A thin circular ring of mass $$m$$ and radius $$R$$ is rotating about its axis with a constant angular velocity $$\omega $$. Two objects each of mass $$M$$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity $$\omega ' = $$
A
$${{\omega \left( {m + 2M} \right)} \over m}$$
B
$${{\omega \left( {m - 2M} \right)} \over {\left( {m + 2M} \right)}}$$
C
$${{\omega m} \over {\left( {m + M} \right)}}$$
D
$${{\omega m} \over {\left( {m + 2M} \right)}}$$

Explanation

Here angular momentum is conserved.

Applying conservation of angular momentum $$I'\omega ' = I\omega \,\,$$

$$\left( {m{R^2} + 2M{R^2}} \right)\omega \,' = m{R^2}\omega $$

$$ \Rightarrow \omega \,' = \omega \left[ {{m \over {m + 2M}}} \right]$$
4

AIEEE 2006

MCQ (Single Correct Answer)
Consider a two particle system with particles having masses $${m_1}$$ and $${m_2}$$. If the first particle is pushed towards the center of mass through a distance $$d,$$ by what distance should the second particle is moved, so as to keep the center of mass at the same position?
A
$${{{m_2}} \over {{m_1}}}\,\,d$$
B
$${{{m_1}} \over {{m_1} + {m_2}}}d$$
C
$${{{m_1}} \over {{m_2}}}d$$
D
$$d$$

Explanation

Initially,

$$0 = {{{m_1}\left( { - {x_1}} \right) + {m_2}{x_2}} \over {{m_1} + {m_2}}} \Rightarrow {m_1}{x_1} = {m_2}{x_2}$$

Finally,

mass m1 moved towards the center a distance d so the distance of mass m1 from the origin is x1 - d, and now let mass m2 need to move d' to keep the center at the origin.

$$\therefore$$ $$0 = {{{m_1}\left( {d - {x_1}} \right) + {m_2}\left( {{x_2} - d'} \right)} \over {{m_1} + {m_2}}}$$

$$ \Rightarrow 0 = {m_1}d - {m_1}{x_1} + {m_2}{x_2} - {m_2}d'$$

$$ \Rightarrow d' = {{{m_1}} \over {{m_2}}}d$$

Questions Asked from Rotational Motion

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