Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A particle has an initial velocity $$3\widehat i + 4\widehat j$$ and an acceleration of $$0.4\widehat i + 0.3\widehat j$$. Its speed after 10 s is:

A

$$7\sqrt 2 $$ units

B

7 units

C

8.5 units

D

10 units

Given $$\overrightarrow u = 3\widehat i + 4\widehat j,\,\,\overrightarrow a = 0.4\widehat i + 0.3\widehat j,\,\,t = 10s$$

$$\overrightarrow v = \overrightarrow u + \overrightarrow a t $$

$$= 3\widehat i + 4\widehat j + \left( {0.4\widehat i + 0.3\widehat j} \right) \times 10$$

$$ = 7\widehat i + 7\widehat j$$

We know speed is equal to magnitude of velocity.

$$\therefore$$ $$\left| {\overrightarrow v } \right| = \sqrt {{7^2} + {7^2}} = 7\sqrt 2 \,\,\,$$ units

$$\overrightarrow v = \overrightarrow u + \overrightarrow a t $$

$$= 3\widehat i + 4\widehat j + \left( {0.4\widehat i + 0.3\widehat j} \right) \times 10$$

$$ = 7\widehat i + 7\widehat j$$

We know speed is equal to magnitude of velocity.

$$\therefore$$ $$\left| {\overrightarrow v } \right| = \sqrt {{7^2} + {7^2}} = 7\sqrt 2 \,\,\,$$ units

2

MCQ (Single Correct Answer)

A body is at rest at $$x=0.$$ At $$t=0,$$ it starts moving in the positive $$x$$-direction with a constant acceleration. At the same instant another body passes through $$x=0$$ moving in the positive $$x$$ direction with a constant speed. The position of the first body is given by $${x_1}\left( t \right)$$ after time $$'t';$$ and that of the second body by $${x_2}\left( t \right)$$ after the same time interval. Which of the following graphs correctly describes $$\left( {{x_1} - {x_2}} \right)$$ as a function of time $$'t'$$ ?

A

B

C

D

$${x_1} = 0 + {1 \over 2}a{t^2} \Rightarrow {x_1} = {1 \over 2}a{t^2}$$

$${x_2} = vt$$

$$\therefore$$ $${x_1} - {x_2} = {1 \over 2}a{t^2} - vt$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {{d\left( {{x_1} - {x_2}} \right)} \over {dt}} = at - v$$

at $$t=0,$$ $$\,\,\,\,\,\,{x_1} - {x_2} = 0$$, so graph should start from origin.

For $$at < v;$$ the slope is negative that means $${x_1} - {x_2}$$ < 0 so initially velocity of 1st body is less than second body and velocity of 1st body is increasing gradually.

For $$at = v;$$ the slope is zero. So $${x_1} - {x_2}$$ = 0 it means here velocity of both the bodies are same.

For $$at > v;$$ the slope is positive. So $${x_1} - {x_2}$$ > 0 it means here velocity of first body is greater than second body.

We know the relation between distance and time is.

$$S = ut + {1 \over 2}a{t^2}$$, which is a equation parabola. So the graph should be a parabola.

These characteristics are represented by graph $$(b).$$

3

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

4

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 $$(1)$$ ]

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