 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2005

A body of mass $m$ is accelerated uniformly from rest to a speed $v$ in a time $T.$ The instantaneous power delivered to the body as a function of time is given by
A
${{m{v^2}} \over {{T^2}}}.{t^2}$
B
${{m{v^2}} \over {{T^2}}}.t$
C
${1 \over 2}{{m{v^2}} \over {{T^2}}}.{t^2}$
D
${1 \over 2}{{m{v^2}} \over {{T^2}}}.t$

Explanation

$u = 0;v = u + aT;v = aT$

Instantaneous power $= F \times v = m.\,a.\,at = m.{a^2}.t$

$\therefore$ Instantaneous power $= {{m{v^2}t} \over {{T^2}}}$
2

AIEEE 2005

A spherical ball of mass $20$ $kg$ is stationary at the top of a hill of height $100$ $m$. It rolls down a smooth surface to the ground, then climbs up another hill of height $30$ $m$ and finally rolls down to a horizontal base at a height of $20$ $m$ above the ground. The velocity attained by the ball is
A
$20$ $m/s$
B
$40$ $m/s$
C
$10\sqrt {30} \,\,\,m/s$
D
$10\,\,m/s$

Explanation Loss in potential energy $=$ gain in kinetic energy

$m \times g \times 80 = {1 \over 2}m{v^2}$

$\Rightarrow$ $10 \times 80 = {1 \over 2}{v^2}$

$\Rightarrow $${v^2} = 1600 or v = 40\,m/s 3 AIEEE 2005 MCQ (Single Correct Answer) The upper half of an inclined plane with inclination \phi is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by A 2\,\cos \,\,\phi B 2\,sin\,\,\phi C \,\tan \,\,\phi D 2\,\tan \,\,\phi Explanation Let the length of the inclined plane is = l. So only {l \over 2} part will have friction. According to work-energy theorem, W = \Delta k = 0 (Since initial and final speeds are zero) \therefore Work done by friction + Work done by gravity =0 i.e., - \left( {\mu \,mg\,\cos \,\phi } \right){\ell \over 2} + mg\ell \,\sin \,\phi = 0 or {\mu \over 2}\cos \,\phi = \sin \phi or \mu = 2\,\tan \,\phi 4 AIEEE 2005 MCQ (Single Correct Answer) A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion? A 2.0 cm B 3.0 cm C 1.0 cm D 1.5 cm Explanation Let K be the initial kinetic energy and F be the resistive force. Then according to work-energy theorem,$$W = \Delta K$\$

i.e., $3F = {1 \over 2}m{v^2} - {1 \over 2}m{\left( {{v \over 2}} \right)^2}...\left( 1 \right)$

Let the bullet will penetrate x cm more before coming to rest.

$\therefore$ $Fx = {1 \over 2}m{\left( {{v \over 2}} \right)^2} - {1 \over 2}m{\left( 0 \right)^2}...\left( 2 \right)$

Dividing eq. $(1)$ and $(2)$ we get,

${x \over 3} = {1 \over 3}$ or x = 1 cm