Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

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 $$(1)$$ ]

2

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

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)

The relation between time t and distance x is t = ax^{2} + bx where a and b are constants.
The acceleration is

A

2bv^{3}

B

-2abv^{2}

C

2av^{2}

D

-2av^{3}

Given $$t = a{x^2} + bx;$$

Differentiate with respect to time $$(t)$$

$${d \over {dt}}\left( t \right) = a{d \over {dt}}\left( {{x^2}} \right) + b{{dx} \over {dt}}$$

$$ = a.2x{{dx} \over {dt}} + b.{{dx} \over {dt}}$$

$$1 = 2axv + bv = v\left( {2ax + b} \right)$$

[as $${{dx} \over {dt}} = v$$]

$$ \Rightarrow 2ax + b = {1 \over v}.$$

Again differentiating,

$$2a{{dx} \over {dt}} + 0 = - {1 \over {{v^2}}}{{dv} \over {dt}}$$

$$ \Rightarrow {{dv} \over {dt}} = f = - 2a{v^3}$$

(as $${{dv} \over {dt}} = f$$ = Acceleration )

Differentiate with respect to time $$(t)$$

$${d \over {dt}}\left( t \right) = a{d \over {dt}}\left( {{x^2}} \right) + b{{dx} \over {dt}}$$

$$ = a.2x{{dx} \over {dt}} + b.{{dx} \over {dt}}$$

$$1 = 2axv + bv = v\left( {2ax + b} \right)$$

[as $${{dx} \over {dt}} = v$$]

$$ \Rightarrow 2ax + b = {1 \over v}.$$

Again differentiating,

$$2a{{dx} \over {dt}} + 0 = - {1 \over {{v^2}}}{{dv} \over {dt}}$$

$$ \Rightarrow {{dv} \over {dt}} = f = - 2a{v^3}$$

(as $${{dv} \over {dt}} = f$$ = Acceleration )

On those following papers in MCQ (Single Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

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Units & Measurements *keyboard_arrow_right*

Motion *keyboard_arrow_right*

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