Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

For a particle in uniform circular motion the acceleration $$\overrightarrow a $$at a point P(R, θ) on the circle of radius R is (here θ is measured from the x–axis)

A

$$ - {{{v^2}} \over R}\cos \theta \widehat i + {{{v^2}} \over R}\sin \theta \widehat j$$

B

$$ - {{{v^2}} \over R}\sin \theta \widehat i + {{{v^2}} \over R}\cos \theta \widehat j$$

C

$$ - {{{v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j$$

D

$${{{v^2}} \over R}\widehat i + {{{v^2}} \over R}\widehat j$$

For a particle in uniform circular motion,

$${a_c} = {{{v^2}} \over R}$$ towards the center of the circle

From figure, $$\overrightarrow a = {a_c}\cos \theta \left( { - \widehat i} \right) + {a_c}\sin \theta \left( { - \widehat j} \right)$$

$$ = {{ - {v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j$$

$${a_c} = {{{v^2}} \over R}$$ towards the center of the circle

From figure, $$\overrightarrow a = {a_c}\cos \theta \left( { - \widehat i} \right) + {a_c}\sin \theta \left( { - \widehat j} \right)$$

$$ = {{ - {v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j$$

2

MCQ (Single Correct Answer)

Consider a rubber ball freely falling from a height $$h=4.9$$ $$m$$ onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic.

Then the velocity as a function of time and the height as a function of time will be :

Then the velocity as a function of time and the height as a function of time will be :

A

B

C

D

For downward motion $$v=-gt$$

The velocity of the rubber ball increases in downward direction and we get a straight line between $$v$$ and $$t$$ with a negative slope.

Also applying $$y - {y_0} = ut + {1 \over 2}a{t^2}$$

We get $$y - h = - {1 \over 2}g{t^2} \Rightarrow y = h - {1 \over 2}g{t^2}$$

The graph between $$y$$ and $$t$$ is a parabola with $$y=h$$ at $$t=0.$$ As time increases $$y$$ decreases.

**For upward motion.**

The ball suffer elastic collision with the horizontal elastic plate therefore the direction of velocity is reversed and the magnitude remains the same. Here $$v=u-gt$$ where $$u$$ is the velocity just after collision. As $$t$$ increases, $$v$$ decreases. We get a straight line between $$v$$ and $$t$$ with negative slope.

Also $$y = ut - {1 \over 2}g{t^2}$$

All these characteristics are represented by graph $$(c).$$

The velocity of the rubber ball increases in downward direction and we get a straight line between $$v$$ and $$t$$ with a negative slope.

Also applying $$y - {y_0} = ut + {1 \over 2}a{t^2}$$

We get $$y - h = - {1 \over 2}g{t^2} \Rightarrow y = h - {1 \over 2}g{t^2}$$

The graph between $$y$$ and $$t$$ is a parabola with $$y=h$$ at $$t=0.$$ As time increases $$y$$ decreases.

The ball suffer elastic collision with the horizontal elastic plate therefore the direction of velocity is reversed and the magnitude remains the same. Here $$v=u-gt$$ where $$u$$ is the velocity just after collision. As $$t$$ increases, $$v$$ decreases. We get a straight line between $$v$$ and $$t$$ with negative slope.

Also $$y = ut - {1 \over 2}g{t^2}$$

All these characteristics are represented by graph $$(c).$$

3

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

4

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 $$(b).$$

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