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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2008

MCQ (Single Correct Answer)
A block of mass $$0.50$$ $$kg$$ is moving with a speed of $$2.00$$ $$m{s^{ - 1}}$$ on a smooth surface. It strike another mass of $$1.0$$ $$kg$$ and then they move together as a simple body. The energy loss during the collision is
A
$$0.16J$$
B
$$1.00J$$
C
$$0.67J$$
D
$$0.34$$ $$J$$

Explanation

Let $$m$$ = 0.50 kg and $$M$$ = 1.0 kg

Initial kinetic energy of the system when 1 kg mass is at rest,

$$K.{E_i} = {1 \over 2}m{u^2} + {1 \over 2}M{\left( 0 \right)^2}$$

$$ = {1 \over 2} \times 0.5 \times 2 \times 2 + 0 = 1J$$

For collision, applying conservation of linear momentum

$$\,\,\,\,\,\,\,\,\,\,\,\,m \times u = \left( {m + M} \right) \times v$$

$$\therefore$$ $$0.5 \times 2 = \left( {0.5 + 1} \right) \times v \Rightarrow v = {2 \over 3}m/s$$

Final kinetic energy of the system is

$$K.{E_f} = {1 \over 2}\left( {m + M} \right){v^2}$$

$$ = {1 \over 2}\left( {0.5 + 1} \right) \times {2 \over 3} \times {2 \over 3} = {1 \over 3}J$$

$$\therefore$$ Energy loss during collision

$$ = \left( {1 - {1 \over 3}} \right)J = 0.67J$$
2

AIEEE 2007

MCQ (Single Correct Answer)
A $$2$$ $$kg$$ block slides on a horizontal floor with a speed of $$4m/s.$$ It strikes a uncompressed spring, and compress it till the block is motionless. The kinetic friction force is $$15N$$ and spring constant is $$10, 000$$ $$N/m.$$ The spring compresses by
A
$$8.5cm$$
B
$$5.5cm$$
C
$$2.5cm$$
D
$$11.0cm$$

Explanation

Let the block compress the spring by $$x$$ before coming to rest.

Initial kinetic energy of the block $$=$$ (potential energy of compressed spring) $$+$$ work done due to friction.

$${1 \over 2} \times 2 \times {\left( 4 \right)^2} = {1 \over 2} \times 10000 \times {x^2} + 15 \times x$$

$$10,000{x^2} + 30x - 32 = 0$$

$$ \Rightarrow 5000{x^2} + 15x - 16 = 0$$

$$\therefore$$ $$x = {{ - 15 \pm \sqrt {{{\left( {15} \right)}^2} - 4 \times \left( {5000} \right)\left( { - 16} \right)} } \over {2 \times 5000}}$$

$$\,\,\,\,\, = 0.055m = 5.5cm.$$
3

AIEEE 2006

MCQ (Single Correct Answer)
The potential energy of a $$1$$ $$kg$$ particle free to move along the $$x$$-axis is given by $$V\left( x \right) = \left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right)J$$.

The total mechanical energy of the particle is $$2J.$$ Then, the maximum speed (in $$m/s$$) is

A
$${3 \over {\sqrt 2 }}$$
B
$${\sqrt 2 }$$
C
$${1 \over {\sqrt 2 }}$$
D
$$2$$

Explanation

Velocity is maximum when kinetic energy is maximum and when kinetic energy is maximum then potential energy should be minimum

For minimum potential energy,

$${{dV} \over {dx}} = 0 $$

$$\Rightarrow {x^3} - x = 0 $$

$$\Rightarrow x = \pm 1$$

$$ \Rightarrow$$ Min. Potential energy (P.E.) =$$ {1 \over 4} - {1 \over 2} = - {1 \over 4}J$$

$$K.E{._{\left( {\max .} \right)}} + P.E{._{\left( {\min .} \right)}} = 2\,$$ (Given)

$$\therefore$$ $$K.E{._{\left( {\max .} \right)}} = 2 + {1 \over 4} = {9 \over 4}$$

$$\therefore$$ $${1 \over 2}mv_{\max }^2$$ = $${9 \over 4}$$

$$ \Rightarrow {1 \over 2} \times 1 \times {v^2}_{\max .} = {9 \over 4}$$

$$ \Rightarrow {v_{\max }} = {3 \over {\sqrt 2 }}$$ m/s
4

AIEEE 2006

MCQ (Single Correct Answer)
A particle of mass $$100g$$ is thrown vertically upwards with a speed of $$5$$ $$m/s$$. The work done by the force of gravity during the time the particle goes up is
A
$$-0.5J$$
B
$$-1.25J$$
C
$$1.25J$$
D
$$0.5J$$

Explanation

Kinetic energy at point of throwing is converted into potential energy of the particle during rise.

$$K.E = {1 \over 2}m{v^2} = {1 \over 2} \times 0.1 \times 25 = 1.25\,J$$

$$W = - mgh = - \left( {{1 \over 2}m{v^2}} \right) = - 1.25\,J$$

$$\left[ \, \right.$$ As we know, $$mgh = {1 \over 2}m{v^2}$$ by energy conservation $$\left. \, \right]$$

Questions Asked from Work Power & Energy

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