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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2006

MCQ (Single Correct Answer)
A bomb of mass $$16kg$$ at rest explodes into two pieces of masses $$4$$ $$kg$$ and $$12$$ $$kg.$$ The velocity of the $$12$$ $$kg$$ mass is $$4\,\,m{s^{ - 1}}.$$ The kinetic energy of the other mass is
A
$$144$$ $$J$$
B
$$288$$ $$J$$
C
$$192$$ $$J$$
D
$$96$$ $$J$$

Explanation

Here linear momentum is conserved as no external force is acting on the bomb.

Let the velocity and mass of $$4$$ $$kg$$ piece be $${v_1}$$ and $${m_1}$$ and that of $$12$$ $$kg$$ piece be $${v_2}$$ and $${m_2}$$.

Applying conservation of linear momentum

$$0 = {m_2}{v_2} - {m_1}{v_1}$$

$$\Rightarrow {v_1} = {{12 \times 14} \over 4} = 12\,m{s^{ - 1}}$$

$$\therefore$$ $$K.E{_1} = {1 \over 2}{m_1}v_1^2 = {1 \over 2} \times 4 \times 144 = 288\,J$$
2

AIEEE 2005

MCQ (Single Correct Answer)
A mass $$'m'$$ moves with a velocity $$'v'$$ and collides inelastically with another identical mass. After collision the $${1^{st}}$$ mass moves with velocity $${v \over {\sqrt 3 }}$$ in a direction perpendicular to the initial direction of motion. Find the speed of the $${2^{nd}}$$ mass after collision.
A
$${\sqrt 3 v}$$
B
$$v$$
C
$${v \over {\sqrt 3 }}$$
D
$${2 \over {\sqrt 3 }}v$$

Explanation

Assume speed of second mass = $${v_1}$$

As momentum is conserved,

In $$x$$-direction, $$mv = m{v_1}\cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....(1)$$

In $$y$$-direction, $${{mv} \over {\sqrt 3 }} = m{v_1}\,\sin \theta \,\,\,\,\,\,\,\,\,\,...(2)$$

Squaring and adding eqns.$$(1)$$ and $$(2)$$

$${\left( {m{v_1}\cos \theta } \right)^2} + {\left( {m{v_1}\sin \theta } \right)^2}$$$$ = {\left( {mv} \right)^2} + {\left( {{{mv} \over {\sqrt 3 }}} \right)^2}$$

$$ \Rightarrow $$ $$v_1^2 = {v^2} + {{{v^2}} \over {\sqrt 3 }} $$

$$\Rightarrow {v_1} = {2 \over {\sqrt 3 }}v$$

3

AIEEE 2004

MCQ (Single Correct Answer)
A machine gun fires a bullet of mass $$40$$ $$g$$ with a velocity $$1200m{s^{ - 1}}.$$ The man holding it can exert a maximum force of $$144$$ $$N$$ on the gun. How many bullets can he fire per second at the most?
A
Two
B
Four
C
One
D
Three

Explanation

Assume the man can fire $$n$$ bullets in one second.

$$\therefore$$ change in momentum per second $$ = n \times mv = F$$

[ $$m=$$ mass of bullet, $$v=$$ velocity, $$F$$ = force) ]

$$\therefore$$ $$n = {F \over {mv}} = {{144 \times 1000} \over {40 \times 1200}} = 3$$

Questions Asked from Impulse & Momentum

On those following papers in MCQ (Single Correct Answer)
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JEE Main 2019 (Online) 12th April Evening Slot (1)
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JEE Main 2019 (Online) 10th April Evening Slot (2)
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JEE Main 2019 (Online) 9th April Evening Slot (1)
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JEE Main 2019 (Online) 11th January Evening Slot (1)
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JEE Main 2015 (Offline) (1)
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AIEEE 2010 (2)
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AIEEE 2004 (1)
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