Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

Numerical

A particle is moving with constant acceleration 'a'. Following graph shows v^{2} versus x(displacement) plot. The acceleration of the particle is ___________ m/s^{2}.

Your Input ________

Correct Answer is **1**

y = mx + C

v^{2} = $${{20} \over {10}}$$x + 20

v^{2} = 2x + 20

2v$${{dv} \over {dx}}$$ = 2

$$\therefore$$ a = v$${{dv} \over {dx}}$$ = 1

v

v

2v$${{dv} \over {dx}}$$ = 2

$$\therefore$$ a = v$${{dv} \over {dx}}$$ = 1

2

Numerical

If the velocity of a body related to displacement x is given by $$\upsilon = \sqrt {5000 + 24x} $$ m/s, then the acceleration of the body is .................... m/s^{2}.

Your Input ________

Correct Answer is **12**

$$V = \sqrt {5000 + 24x} $$

$${{dV} \over {dx}} = {1 \over {2\sqrt {5000 + 24x} }} \times 24 = {{12} \over {\sqrt {5000 + 24x} }}$$

Now, $$a = V{{dV} \over {dx}}$$

$$ = \sqrt {5000 + 24x} \times {{12} \over {\sqrt {5000 + 24x} }}$$

a = 12 m/s^{2}

$${{dV} \over {dx}} = {1 \over {2\sqrt {5000 + 24x} }} \times 24 = {{12} \over {\sqrt {5000 + 24x} }}$$

Now, $$a = V{{dV} \over {dx}}$$

$$ = \sqrt {5000 + 24x} \times {{12} \over {\sqrt {5000 + 24x} }}$$

a = 12 m/s

3

Numerical

Two spherical balls having equal masses with radius of 5 cm each are thrown upwards along the same vertical direction at an interval of 3s with the same initial velocity of 35 m/s, then these balls collide at a height of ............... m. (Take g = 10 m/s^{2})

Your Input ________

Correct Answer is **50**

Image

When both balls will collied

y_{1} = y_{2}

$$35t - {1 \over 2} \times 10 \times {t^2} = 35(t - 3) - {1 \over 2} \times 10 \times {(t - 3)^2}$$

$$35t - {1 \over 2} \times 10 \times {t^2} = 35t - 105 - {1 \over 2} \times 10 \times {t^2} - {1 \over 2} \times 10 \times {3^2} + {1 \over 2} \times 10 \times 6t$$

0 = 150 $$-$$ 30 t

t = 5 sec

$$\therefore$$ Height at which both balls will collied

$$h = 35t - {1 \over 2} \times 10 \times {t^2}$$

$$ = 35 \times 5 - {1 \over 2} \times 10 \times {5^2}$$

h = 50 m

When both balls will collied

y

$$35t - {1 \over 2} \times 10 \times {t^2} = 35(t - 3) - {1 \over 2} \times 10 \times {(t - 3)^2}$$

$$35t - {1 \over 2} \times 10 \times {t^2} = 35t - 105 - {1 \over 2} \times 10 \times {t^2} - {1 \over 2} \times 10 \times {3^2} + {1 \over 2} \times 10 \times 6t$$

0 = 150 $$-$$ 30 t

t = 5 sec

$$\therefore$$ Height at which both balls will collied

$$h = 35t - {1 \over 2} \times 10 \times {t^2}$$

$$ = 35 \times 5 - {1 \over 2} \times 10 \times {5^2}$$

h = 50 m

4

Numerical

A swimmer wants to cross a river from point A to point B. Line AB makes an angle of 30$$^\circ$$ with the flow of river. Magnitude of velocity of the swimmer is same as that of the river. The angle $$\theta$$ with the line AB should be _________$$^\circ$$, so that the swimmer reaches point B.

Your Input ________

Correct Answer is **30**

Both velocity vectors are of same magnitude therefore resultant would pass exactly midway through them

$$\theta$$ = 30$$^\circ$$

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