Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

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

$$2\,\tan \,\,\phi $$

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 $$\mu = 2\,\tan \,\phi $$

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 $$\mu = 2\,\tan \,\phi $$

2

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

$$1.5$$ $$cm$$

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

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

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

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

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

3

MCQ (Single Correct Answer)

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particles takes place in a plane. It follows that

A

its kinetic energy is constant

B

is acceleration is constant

C

its velocity is constant

D

it moves in a straight line

Work done by such force is always zero when a force of constant magnitude always at right angle to the velocity of a particle when the motion of the particle takes place in a plane.

$$\therefore$$ From work-energy theorem, $$ \Delta K = 0$$

$$\therefore$$ $$K$$ remains constant.

$$\therefore$$ From work-energy theorem, $$ \Delta K = 0$$

$$\therefore$$ $$K$$ remains constant.

4

MCQ (Single Correct Answer)

A body of mass $$' m ',$$ acceleration uniformly from rest to $$'{v_1}'$$ in time $${T}$$. The instantaneous power delivered to the body as a function of time is given by

A

$${{m{v_1}{t^2}} \over {{T}}}$$

B

$${{mv_1^2t} \over {T^2}}$$

C

$${{m{v_1}t} \over {{T}}}$$

D

$${{mv_1^2t} \over {{T}}}$$

Assume acceleration of body be $$a$$

$$\therefore$$ $${v_1} = 0 + a{T} \Rightarrow a = {{{v_1}} \over {{T}}}$$

$$\therefore$$ $$v = at \Rightarrow v = {{{v_1}t} \over {{T}}}$$

$${P_{inst}} = \overrightarrow F .\overrightarrow v = \left( {m\overrightarrow a } \right).\overrightarrow v $$

$$= \left( {{{m{v_1}} \over {{T}}}} \right)\left( {{{{v_1}t} \over {{T}}}} \right)$$

$$ = m{\left( {{{{v_1}} \over {{T}}}} \right)^2}t$$

$$\therefore$$ $${v_1} = 0 + a{T} \Rightarrow a = {{{v_1}} \over {{T}}}$$

$$\therefore$$ $$v = at \Rightarrow v = {{{v_1}t} \over {{T}}}$$

$${P_{inst}} = \overrightarrow F .\overrightarrow v = \left( {m\overrightarrow a } \right).\overrightarrow v $$

$$= \left( {{{m{v_1}} \over {{T}}}} \right)\left( {{{{v_1}t} \over {{T}}}} \right)$$

$$ = m{\left( {{{{v_1}} \over {{T}}}} \right)^2}t$$

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