Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

The velocity of a particle is v = v_{0} + gt + ft^{2}. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is

A

v_{0} + g/2 + f

B

v_{0} + 2g + 3f

C

v_{0} + g/2 + f/3

D

v_{0} + g + f

Given that, v = v_{0} + gt + ft^{2}

We know that, $$v = {{dx} \over {dt}} $$

$$\Rightarrow dx = v\,dt$$

Integrating, $$\int\limits_0^x {dx} = \int\limits_0^t {v\,dt} $$

or $$\,\,\,\,\,x = \int\limits_0^t {\left( {{v_0} + gt + f{t^2}} \right)} dt$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {{v_0}t + {{g{t^2}} \over 2} + {{f{t^3}} \over 3}} \right]_0^t$$

or $$\,\,\,\,\,x = {v_0}t + {{g{t^2}} \over 2} + {{f{t^3}} \over 3}$$

At $$t = 1,\,\,\,\,\,x = {v_0} + {g \over 2} + {f \over 3}.$$

We know that, $$v = {{dx} \over {dt}} $$

$$\Rightarrow dx = v\,dt$$

Integrating, $$\int\limits_0^x {dx} = \int\limits_0^t {v\,dt} $$

or $$\,\,\,\,\,x = \int\limits_0^t {\left( {{v_0} + gt + f{t^2}} \right)} dt$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {{v_0}t + {{g{t^2}} \over 2} + {{f{t^3}} \over 3}} \right]_0^t$$

or $$\,\,\,\,\,x = {v_0}t + {{g{t^2}} \over 2} + {{f{t^3}} \over 3}$$

At $$t = 1,\,\,\,\,\,x = {v_0} + {g \over 2} + {f \over 3}.$$

2

MCQ (Single Correct Answer)

A particle is projected at $$60^\circ $$ to the horizontal with a kinetic energy K. The kinetic energy at the
highest point is

A

K/2

B

K

C

Zero

D

K/4

Let $$u$$ be the velocity with which the particle is thrown and $$m$$ be the mass of the particle. Then

$$KE = {1 \over 2}m{u^2}.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

At the highest point the velocity is $$u$$ $$\cos \,{60^ \circ }$$ (only the horizontal component remains, the vertical component being zero at the top-most point).

Therefore kinetic energy at the highest point,

$${\left( {KE} \right)_H} = {1 \over 2}m{u^2}{\cos ^2}60^\circ $$

$$\,\,\,\,\,\,\,\,\,\,\,\, = {K \over 4}$$ [ From eq $$(1)$$ ]

$$KE = {1 \over 2}m{u^2}.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

At the highest point the velocity is $$u$$ $$\cos \,{60^ \circ }$$ (only the horizontal component remains, the vertical component being zero at the top-most point).

Therefore kinetic energy at the highest point,

$${\left( {KE} \right)_H} = {1 \over 2}m{u^2}{\cos ^2}60^\circ $$

$$\,\,\,\,\,\,\,\,\,\,\,\, = {K \over 4}$$ [ From eq $$(1)$$ ]

3

MCQ (Single Correct Answer)

A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity
'v' that varies as $$v = \alpha \sqrt x $$. The displacement of the particle varies with time as

A

t^{2}

B

t

C

t^{1/2}

D

t^{3}

Given $$v = \alpha \sqrt x$$

$$\therefore$$ $${{dx} \over {dt}} = \alpha \sqrt x $$

$$\Rightarrow {{dx} \over {\sqrt x }} = \alpha \,dt$$

On integrating we get,

$$\int\limits_0^x {{{dx} \over {\sqrt x }} = \alpha \int\limits_0^t {dt} } $$

$$\Rightarrow \left[ {{{2\sqrt x } \over 1}} \right]_0^x = \alpha \left[ t \right]_0^t$$

$$\Rightarrow 2\sqrt x = \alpha t $$

$$\Rightarrow x = {{{\alpha ^2}} \over 4}{t^2}$$

So $$x \propto {t^2}$$

$$\therefore$$ $${{dx} \over {dt}} = \alpha \sqrt x $$

$$\Rightarrow {{dx} \over {\sqrt x }} = \alpha \,dt$$

On integrating we get,

$$\int\limits_0^x {{{dx} \over {\sqrt x }} = \alpha \int\limits_0^t {dt} } $$

$$\Rightarrow \left[ {{{2\sqrt x } \over 1}} \right]_0^x = \alpha \left[ t \right]_0^t$$

$$\Rightarrow 2\sqrt x = \alpha t $$

$$\Rightarrow x = {{{\alpha ^2}} \over 4}{t^2}$$

So $$x \propto {t^2}$$

4

MCQ (Single Correct Answer)

A particle is moving eastwards with a velocity of 5 m/s. In 10 seconds the velocity
changes to 5 m/s northwards. The average acceleration in this time is

A

$${1 \over 2}m{s^{ - 2}}$$ towards north

B

$${1 \over {\sqrt 2 }}m{s^{ - 2}}$$ towards north-east

C

$${1 \over {\sqrt 2 }}m{s^{ - 2}}$$ towards north-west

D

zero

Average acceleration

$$ = {{change\,\,in\,\,velocioty} \over {time\,\,{\mathop{\rm int}} erval}}$$

$$ = {{\Delta \overrightarrow v } \over t}$$

$${\overrightarrow v _1} = +5\widehat i$$, towards east direction.

$$\overrightarrow {{v_2}} = +5\widehat j$$, towards north direction

$$\Delta \overrightarrow v = \overrightarrow {{v_2}} - \overrightarrow {{v_1}} $$ = $$5\widehat i - 5\widehat j$$

$$\therefore$$ $$\,\,\,\,\overrightarrow a = {{5\widehat j - 5\widehat i} \over {10}} = {{\widehat j - \widehat i} \over 2}$$

$$\therefore$$ $$\,\,\,\, a = {{\sqrt {{1^2} + {{\left( { - 1} \right)}^2}} } \over 2} = {{\sqrt 2 } \over 2} = {1 \over {\sqrt 2 }}m{s^{ - 2}}$$

$$\tan \theta = {{{v_2}} \over {{v_1}}} = {5 \over 5} = 1$$

$$\therefore$$ $$\,\,\,\,\theta = {45^ \circ }$$

Therefore the direction is North-west.

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