Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by :

$${{dv} \over {dt}} = - 2.5\sqrt v $$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be :

$${{dv} \over {dt}} = - 2.5\sqrt v $$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be :

A

2 s

B

4 s

C

8 s

D

1 s

Given $${{dv} \over {dt}} = - 2.5\sqrt v $$

$$\Rightarrow {{dv} \over {\sqrt v }} = - 2.5dt$$

On integrating, $$\int_{6.25}^0 {{v^{ - {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \,dv = - 2.5\int_0^t {dt} $$

$$ \Rightarrow \left[ {{{{v^{ + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over {\left( {{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right)}}} \right]_{6.25}^0 = - 2.5\left[ t \right]_0^t$$

$$ \Rightarrow - 2{\left( {6.25} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}} = - 2.5t$$

$$ \Rightarrow t = 2\,sec$$

$$\Rightarrow {{dv} \over {\sqrt v }} = - 2.5dt$$

On integrating, $$\int_{6.25}^0 {{v^{ - {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \,dv = - 2.5\int_0^t {dt} $$

$$ \Rightarrow \left[ {{{{v^{ + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over {\left( {{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right)}}} \right]_{6.25}^0 = - 2.5\left[ t \right]_0^t$$

$$ \Rightarrow - 2{\left( {6.25} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}} = - 2.5t$$

$$ \Rightarrow t = 2\,sec$$

2

MCQ (Single Correct Answer)

A point $$P$$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of $$P$$ is such that it sweeps out a length $$s = {t^3} + 5,$$ where $$s$$ is in metres and $$t$$ is in seconds. The radius of the path is $$20$$ $$m.$$ The acceleration of $$'P'$$ when $$t=2$$ $$s$$ is nearly.

A

$$13m/{s_2}$$

B

$$12m/{s^2}$$

C

$$7.2m{s^2}$$

D

$$14m/{s^2}$$

Given $$s = {t^3} + 5 $$

$$\Rightarrow$$ Speed,$$\,\,\,v = {{ds} \over {dt}} = 3{t^2}$$

Tangential acceleration $${a_t} = {{dv} \over {dt}} = 6t$$

Radial acceleration $${a_c} = {{{v^2}} \over R} = {{9{t^4}} \over R}$$

At $$\,\,\,\,t = 2s,\,\,{a_t} = 6 \times 2 = 12\,\,m/{s^2}$$

$${a_c} = {{9 \times 16} \over {20}} = 7.2\,\,m/{s^2}$$

$$\therefore$$ Net acceleration

$$ = \sqrt {a_t^2 + a_c^2} $$

$$= \sqrt {{{\left( {12} \right)}^2} + {{\left( {7.2} \right)}^2}} $$

$$ = \sqrt {144 + 51.84} $$

$$ = \sqrt {195.84} $$

$$ = 14\,m/{s^2}$$

$$\Rightarrow$$ Speed,$$\,\,\,v = {{ds} \over {dt}} = 3{t^2}$$

Tangential acceleration $${a_t} = {{dv} \over {dt}} = 6t$$

Radial acceleration $${a_c} = {{{v^2}} \over R} = {{9{t^4}} \over R}$$

At $$\,\,\,\,t = 2s,\,\,{a_t} = 6 \times 2 = 12\,\,m/{s^2}$$

$${a_c} = {{9 \times 16} \over {20}} = 7.2\,\,m/{s^2}$$

$$\therefore$$ Net acceleration

$$ = \sqrt {a_t^2 + a_c^2} $$

$$= \sqrt {{{\left( {12} \right)}^2} + {{\left( {7.2} \right)}^2}} $$

$$ = \sqrt {144 + 51.84} $$

$$ = \sqrt {195.84} $$

$$ = 14\,m/{s^2}$$

3

MCQ (Single Correct Answer)

A small particle of mass $$m$$ is projected at an angle $$\theta $$ with the $$x$$-axis with an initial velocity $${v_0}$$ in the $$x$$-$$y$$ plane as shown in the figure. At a time $$t < {{{v_0}\sin \theta } \over g},$$ the angular momentum of the particle is ................,

where $$\widehat i,\widehat j$$ and $$\widehat k$$ are unit vectors along $$x,y$$ and $$z$$-axis respectively.

where $$\widehat i,\widehat j$$ and $$\widehat k$$ are unit vectors along $$x,y$$ and $$z$$-axis respectively.

A

$$ - mg\,{v_0}{t^2}\cos \theta \widehat j$$

B

$$mg\,{v_0}t\cos \theta \widehat k$$

C

$$ - {1 \over 2}mg\,{v_0}{t^2}\cos \,\theta \widehat k$$

D

$${1 \over 2}mg\,{v_0}{t^2}\cos \theta \widehat i$$

Position vector of the particle from the original at any time t is

$$\overrightarrow r = {v_0}t\cos \theta \widehat i + \left( {{v_0}t\sin \theta - {1 \over 2}g{t^2}} \right)\widehat j$$

As Velocity vector, $$\overrightarrow v = {{d\overrightarrow r } \over {dt}}$$

$$\therefore$$ $$\overrightarrow v = {d \over {dt}}\left( {{v_0}t\cos \theta \widehat i + \left( {{v_0}t\sin \theta - {1 \over 2}g{t^2}} \right)\widehat j} \right)$$

$$ = {v_0}\cos \theta \widehat i + \left( {{v_0}\sin \theta - gt} \right)\widehat j$$

Angular momentum of the particle about the origin is

$$\overrightarrow L = m\left( {\overrightarrow r \times \overrightarrow v } \right)$$

$$\eqalign{ & \overrightarrow L = m\left[ {{v_0}\,\cos \theta t\widehat i + \left( {{v_0}\,\sin \theta t - {1 \over 2}g{t^2}} \right)\widehat j} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \times \left[ {{v_0}\,\cos \,\theta \widehat i + \left( {{v_0}\,\sin \theta - gt} \right)\widehat j} \right] \cr} $$

$$ = m{v_0}\cos \theta t\left[ { - {1 \over 2}gt} \right]\widehat k$$

$$ = - {1 \over 2}mg{v_0}{t^2}\cos \theta \widehat k$$

$$\overrightarrow r = {v_0}t\cos \theta \widehat i + \left( {{v_0}t\sin \theta - {1 \over 2}g{t^2}} \right)\widehat j$$

As Velocity vector, $$\overrightarrow v = {{d\overrightarrow r } \over {dt}}$$

$$\therefore$$ $$\overrightarrow v = {d \over {dt}}\left( {{v_0}t\cos \theta \widehat i + \left( {{v_0}t\sin \theta - {1 \over 2}g{t^2}} \right)\widehat j} \right)$$

$$ = {v_0}\cos \theta \widehat i + \left( {{v_0}\sin \theta - gt} \right)\widehat j$$

Angular momentum of the particle about the origin is

$$\overrightarrow L = m\left( {\overrightarrow r \times \overrightarrow v } \right)$$

$$\eqalign{ & \overrightarrow L = m\left[ {{v_0}\,\cos \theta t\widehat i + \left( {{v_0}\,\sin \theta t - {1 \over 2}g{t^2}} \right)\widehat j} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \times \left[ {{v_0}\,\cos \,\theta \widehat i + \left( {{v_0}\,\sin \theta - gt} \right)\widehat j} \right] \cr} $$

$$ = m{v_0}\cos \theta t\left[ { - {1 \over 2}gt} \right]\widehat k$$

$$ = - {1 \over 2}mg{v_0}{t^2}\cos \theta \widehat k$$

4

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

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

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