Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

The co-ordinates of a moving particle at any time 't' are given by x = $$\alpha $$t^{3} and y = βt^{3}. The speed to the particle at time 't' is given by

A

$$3t\sqrt {{\alpha ^2} + {\beta ^2}} $$

B

$$3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}} $$

C

$${t^2}\sqrt {{\alpha ^2} + {\beta ^2}} $$

D

$$\sqrt {{\alpha ^2} + {\beta ^2}} $$

Given that $$x = \alpha {t^3}\,\,\,\,$$ and $$\,\,\,\,y = \beta {t^3}$$

$$\therefore$$ $${v_x} = {{dx} \over {dt}} = 3\alpha {t^2}\,\,\,\,$$

and$$\,\,\,\,\,{v_y} = {{dy} \over {dt}} = 3\beta {t^2}$$

$$\therefore$$ $$v = \sqrt {v_x^2 + v_y^2} $$

$$ = \sqrt {9{\alpha ^2}{t^4} + 9{\beta ^2}{t^4}} $$

$$ = 3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}} $$

$$\therefore$$ $${v_x} = {{dx} \over {dt}} = 3\alpha {t^2}\,\,\,\,$$

and$$\,\,\,\,\,{v_y} = {{dy} \over {dt}} = 3\beta {t^2}$$

$$\therefore$$ $$v = \sqrt {v_x^2 + v_y^2} $$

$$ = \sqrt {9{\alpha ^2}{t^4} + 9{\beta ^2}{t^4}} $$

$$ = 3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}} $$

2

MCQ (Single Correct Answer)

A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an
angle of $$30^\circ $$ with the horizontal. How far from the throwing point will the ball be at the height
of 10 m from the ground?
$$\left[ {g = 10m/{s^2},\sin 30^\circ = {1 \over 2},\cos 30^\circ = {{\sqrt 3 } \over 2}} \right]$$

A

5.20 m

B

4.33 m

C

2.60 m

D

8.66 m

From the figure it is clear that maximum horizontal range

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

$$ = {{{{\left( {10} \right)}^2}\sin \left( {2 \times {{30}^ \circ }} \right)} \over {10}}$$

$$ = 5\sqrt 3 $$ = 8.66 m

3

MCQ (Single Correct Answer)

A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the
same car is moving at a speed of 100 km/hr, the minimum stopping distance is

A

12 m

B

18 m

C

24 m

D

6 m

u = 50 km/hr = $${{50 \times 1000} \over {3600}}$$ m/s = $${{125} \over 9}$$ m/s, v = 0, s = 6 m, $$a$$ = ?

$$\therefore$$ 0

$$ \Rightarrow $$ $$a = - {{{u^2}} \over {2s}}$$

$$ \Rightarrow $$ $$a = - {{{{\left( {{{125} \over 9}} \right)}^2}} \over {2 \times 6}}$$ = $$-$$16 m/s

u = 100 km/hr = $${{100 \times 1000} \over {3600}}$$ m/s = $${{250} \over 9}$$ m/s, v = 0, $$a$$ = $$-$$16, s = ?

$$\therefore$$ 0

$$ \Rightarrow $$ $$s = - {{{u^2}} \over {2a}}$$

$$ \Rightarrow $$ $$s = - {{{{\left( {{{250} \over 9}} \right)}^2}} \over {2 \times -16}}$$ = 24 m

4

MCQ (Single Correct Answer)

Speeds of two identical cars are $$u $$ and $$4$$$$u $$ at the specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is

A

$$1:1$$

B

$$1:4$$

C

$$1:8$$

D

$$1:16$$

Given that initial speed of two cars $$u$$ and $$4u$$ and final speed $$v$$ is 0 for both car. Both car is gradually slowing down so acceleration = $$(-a)$$

So formula becomes, 0 = $${u^2}$$ - 2$$a$$s.

$$ \Rightarrow {u^2} = 2as$$

For first car $${u^2} = 2a{s_1}$$ ..........(i)

For second car $${\left( {4u} \right)^2} = 2a{s_2}$$ ..........(ii)

Dividing $$(i)$$ and $$(ii),$$

$${{{u^2}} \over {16{u^2}}} = {{2a{s_1}} \over {2a{s_2}}}$$

$$ \Rightarrow {1 \over {16}} = {{{s_1}} \over {{s_2}}}$$

So formula becomes, 0 = $${u^2}$$ - 2$$a$$s.

$$ \Rightarrow {u^2} = 2as$$

For first car $${u^2} = 2a{s_1}$$ ..........(i)

For second car $${\left( {4u} \right)^2} = 2a{s_2}$$ ..........(ii)

Dividing $$(i)$$ and $$(ii),$$

$${{{u^2}} \over {16{u^2}}} = {{2a{s_1}} \over {2a{s_2}}}$$

$$ \Rightarrow {1 \over {16}} = {{{s_1}} \over {{s_2}}}$$

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