Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy
can throw the same stone up to will be

A

$$20\sqrt 2 $$ m

B

10 m

C

$$10\sqrt 2 $$ m

D

20 m

We know, $$R = {{{u^2}{{\sin }2}\theta } \over g}$$ and $$H = {{{u^2}{{\sin }^2}\theta } \over {2g}};$$

$${H_{\max }}\,\,$$ is possible when $$\theta = 90$$$$^\circ $$

$${H_{\max }} = {{{u^2}} \over {2g}} = 10 \Rightarrow {u^2} = 10g \times 2$$

As $$R = {{{u^2}\sin 2\theta } \over g}$$

Range is maximum when projectile is thrown at an angle $$45^\circ $$.

$$ \Rightarrow {R_{\max }} = {{{u^2}} \over g}$$

$${R_{\max }} = {{10 \times g \times 2} \over g} = 20$$ meter

$${H_{\max }}\,\,$$ is possible when $$\theta = 90$$$$^\circ $$

$${H_{\max }} = {{{u^2}} \over {2g}} = 10 \Rightarrow {u^2} = 10g \times 2$$

As $$R = {{{u^2}\sin 2\theta } \over g}$$

Range is maximum when projectile is thrown at an angle $$45^\circ $$.

$$ \Rightarrow {R_{\max }} = {{{u^2}} \over g}$$

$${R_{\max }} = {{10 \times g \times 2} \over g} = 20$$ meter

2

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

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

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

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

3

MCQ (Single Correct Answer)

A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the
fountain is v, the total area around the fountain that gets wet is :

A

$$\pi {{{v^4}} \over {{g^2}}}$$

B

$${\pi \over 2}{{{v^4}} \over {{g^2}}}$$

C

$$\pi {{{v^2}} \over {{g^2}}}$$

D

$$\pi {{{v^2}} \over g}$$

Maximum range of water coming out of fountain,

$${R_{\max }} = {{{v^2}\sin 2\theta } \over g} = {{{v^2}\sin {{90}^ \circ }} \over g} = {{{v^2}} \over g}$$

Total area around fountain,

$$A = \pi R_{\max }^2\,\, = \,\,\pi {{{v^4}} \over {{g^2}}}$$

$${R_{\max }} = {{{v^2}\sin 2\theta } \over g} = {{{v^2}\sin {{90}^ \circ }} \over g} = {{{v^2}} \over g}$$

Total area around fountain,

$$A = \pi R_{\max }^2\,\, = \,\,\pi {{{v^4}} \over {{g^2}}}$$

4

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

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

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

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

$$ = \sqrt {195.84} $$

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

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