Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A particle is moving eastwards with a velocity of 5 m/s. In 10 seconds the velocity
changes to 5 m/s northwards. The average acceleration in this time is

A

$${1 \over 2}m{s^{ - 2}}$$ towards north

B

$${1 \over {\sqrt 2 }}m{s^{ - 2}}$$ towards north-east

C

$${1 \over {\sqrt 2 }}m{s^{ - 2}}$$ towards north-west

D

zero

Average acceleration

$$ = {{change\,\,in\,\,velocioty} \over {time\,\,{\mathop{\rm int}} erval}}$$

$$ = {{\Delta \overrightarrow v } \over t}$$

$${\overrightarrow v _1} = +5\widehat i$$, towards east direction.

$$\overrightarrow {{v_2}} = +5\widehat j$$, towards north direction

$$\Delta \overrightarrow v = \overrightarrow {{v_2}} - \overrightarrow {{v_1}} $$ = $$5\widehat i - 5\widehat j$$

$$\therefore$$ $$\,\,\,\,\overrightarrow a = {{5\widehat j - 5\widehat i} \over {10}} = {{\widehat j - \widehat i} \over 2}$$

$$\therefore$$ $$\,\,\,\, a = {{\sqrt {{1^2} + {{\left( { - 1} \right)}^2}} } \over 2} = {{\sqrt 2 } \over 2} = {1 \over {\sqrt 2 }}m{s^{ - 2}}$$

$$\tan \theta = {{{v_2}} \over {{v_1}}} = {5 \over 5} = 1$$

$$\therefore$$ $$\,\,\,\,\theta = {45^ \circ }$$

Therefore the direction is North-west.

2

MCQ (Single Correct Answer)

A car starting from rest accelerates at the rate f through a distance S, then continues
at constant speed for time t and then decelerates at the rate $${f \over 2}$$ to come to rest. If the
total distance traversed is 15 S, then

A

$$S = {1 \over 6}f{t^2}$$

B

$$S = ft$$

C

$$S = {1 \over 4}f{t^2}$$

D

$$S = {1 \over 72}f{t^2}$$

Initially car starts from rest so u = 0.

Now distance from $$A$$ to $$B$$,

$$\,\,\,\,\,\,\,\,\,\,$$ $$ S = {1 \over 2}ft_1^2 $$

$$\Rightarrow ft_1^2 = 2S$$

Distance from $$B$$ to $$C$$ $$ = \left( {f{t_1}} \right)t$$

In B to C velocity is constant and v = $${f{t_1}}$$

Distance from $$C$$ to $$D$$

$$\,\,\,\,\,\,\,\,\,\,$$ $$ = {{{u^2}} \over {2a}} = {{{{\left( {f{t_1}} \right)}^2}} \over {2\left( {f/2} \right)}} = ft_1^2 = 2S$$

$$ \Rightarrow S + f\,{t_1}t + 2S = 15S $$

$$\Rightarrow f\,{t_1}t = 12S$$ ........(1)

But$$\,\,\,\,\,\,\,\,\,$$ $${1 \over 2}f\,t_1^2 = S$$ .........(2)

On dividing the above two equations, we get $${t_1} = {t \over 6}$$

$$ \Rightarrow S = {1 \over 2}f{\left( {{t \over 6}} \right)^2} = {{f\,{t^2}} \over {72}}$$

Now distance from $$A$$ to $$B$$,

$$\,\,\,\,\,\,\,\,\,\,$$ $$ S = {1 \over 2}ft_1^2 $$

$$\Rightarrow ft_1^2 = 2S$$

Distance from $$B$$ to $$C$$ $$ = \left( {f{t_1}} \right)t$$

In B to C velocity is constant and v = $${f{t_1}}$$

Distance from $$C$$ to $$D$$

$$\,\,\,\,\,\,\,\,\,\,$$ $$ = {{{u^2}} \over {2a}} = {{{{\left( {f{t_1}} \right)}^2}} \over {2\left( {f/2} \right)}} = ft_1^2 = 2S$$

$$ \Rightarrow S + f\,{t_1}t + 2S = 15S $$

$$\Rightarrow f\,{t_1}t = 12S$$ ........(1)

But$$\,\,\,\,\,\,\,\,\,$$ $${1 \over 2}f\,t_1^2 = S$$ .........(2)

On dividing the above two equations, we get $${t_1} = {t \over 6}$$

$$ \Rightarrow S = {1 \over 2}f{\left( {{t \over 6}} \right)^2} = {{f\,{t^2}} \over {72}}$$

3

MCQ (Single Correct Answer)

A ball is thrown from a point with a speed ν_{0} at an angle of projection θ. From the same point
and at the same instant person starts running with a constant speed $${{{v_0}} \over 2}$$ to catch the ball.
Will the person be able to catch the ball? If yes, what should be the angle of projection θ?

A

No

B

Yes, $$30^\circ $$

C

Yes, $$60^\circ $$

D

Yes, $$45^\circ $$

Yes, the person can catch the ball when horizontal velocity is equal to the horizontal component of ball's velocity, the motion of ball will be only in vertical direction with respect to person for that,

$${{{v_0}} \over 2} = {v_0}\cos \theta \,\,\,\,$$

or $$\cos \theta = {1 \over 2}$$

$$ \Rightarrow \cos \theta = \cos 60^\circ $$

$$ \Rightarrow \theta = 60^\circ $$

$${{{v_0}} \over 2} = {v_0}\cos \theta \,\,\,\,$$

or $$\cos \theta = {1 \over 2}$$

$$ \Rightarrow \cos \theta = \cos 60^\circ $$

$$ \Rightarrow \theta = 60^\circ $$

4

MCQ (Single Correct Answer)

An automobile travelling with speed of 60 km/h, can brake to stop within a distance of 20 m.
If the car is going twice as fast, i.e 120 km/h, the stopping distance will be

A

60 m

B

40 m

C

20 m

D

80 m

Assume $$a$$ be the retardation for both the vehicle then

In case of automobile,

$$u_1^2 - 2a{s_1} = 0$$

$$ \Rightarrow u_1^2 = 2a{s_1}$$

And in case for car,

$$u_2^2 = 2a{s_2}$$

$$\therefore$$ $${\left( {{{{u_2}} \over {{u_1}}}} \right)^2} = {{{s_2}} \over {{s_1}}}$$

$$ \Rightarrow {\left( {{{120} \over {60}}} \right)^2} = {{{s_2}} \over {20}}$$

$$ \Rightarrow$$ s_{2} = 80 m

In case of automobile,

$$u_1^2 - 2a{s_1} = 0$$

$$ \Rightarrow u_1^2 = 2a{s_1}$$

And in case for car,

$$u_2^2 = 2a{s_2}$$

$$\therefore$$ $${\left( {{{{u_2}} \over {{u_1}}}} \right)^2} = {{{s_2}} \over {{s_1}}}$$

$$ \Rightarrow {\left( {{{120} \over {60}}} \right)^2} = {{{s_2}} \over {20}}$$

$$ \Rightarrow$$ s

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