Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A particle is moving with velocity $$\overrightarrow v = k\left( {y\widehat i + x\widehat j} \right)$$, where K is a constant. The general equation for its path is

A

y = x^{2} + constant

B

y^{2} = x + constant

C

xy = constant

D

y^{2} = x^{2} + constant

$$\overrightarrow v = k\left( {y\widehat i + x\widehat j} \right)$$ ........(1)

Also $$\overrightarrow v = {v_x}\widehat i + {v_y}\widehat j$$

$$\overrightarrow v = {{dx} \over {dt}}\widehat i + {{dy} \over {dt}}\widehat j$$ ........(2)

Equating (1) and (2), we get

$${{dx} \over {dt}} = ky\,\,\,\,\,\,$$ .......(3)

and $$\,\,\,\,\,{{dy} \over {dt}} = kx$$ ......(4)

Dividing (3) and (4), we get

$${{dy} \over {dx}} = {x \over y} $$

$$\Rightarrow ydy = xdx$$

Integrating both sides of above equation, we get

$$\int {ydy} = \int {xdx} $$

$$ \Rightarrow {y^2} = {x^2} + $$ constant

Also $$\overrightarrow v = {v_x}\widehat i + {v_y}\widehat j$$

$$\overrightarrow v = {{dx} \over {dt}}\widehat i + {{dy} \over {dt}}\widehat j$$ ........(2)

Equating (1) and (2), we get

$${{dx} \over {dt}} = ky\,\,\,\,\,\,$$ .......(3)

and $$\,\,\,\,\,{{dy} \over {dt}} = kx$$ ......(4)

Dividing (3) and (4), we get

$${{dy} \over {dx}} = {x \over y} $$

$$\Rightarrow ydy = xdx$$

Integrating both sides of above equation, we get

$$\int {ydy} = \int {xdx} $$

$$ \Rightarrow {y^2} = {x^2} + $$ constant

2

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

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

3

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 $$(c).$$

4

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

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Units & Measurements *keyboard_arrow_right*

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