Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

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,

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

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)

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

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