Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A particle of mass $$100g$$ is thrown vertically upwards with a speed of $$5$$ $$m/s$$. The work done by the force of gravity during the time the particle goes up is

A

$$-0.5J$$

B

$$-1.25J$$

C

$$1.25J$$

D

$$0.5J$$

Kinetic energy at point of throwing is converted into potential energy of the particle during rise.

$$K.E = {1 \over 2}m{v^2} = {1 \over 2} \times 0.1 \times 25 = 1.25\,J$$

$$W = - mgh = - \left( {{1 \over 2}m{v^2}} \right) = - 1.25\,J$$

$$\left[ \, \right.$$ As we know, $$mgh = {1 \over 2}m{v^2}$$ by energy conservation $$\left. \, \right]$$

$$K.E = {1 \over 2}m{v^2} = {1 \over 2} \times 0.1 \times 25 = 1.25\,J$$

$$W = - mgh = - \left( {{1 \over 2}m{v^2}} \right) = - 1.25\,J$$

$$\left[ \, \right.$$ As we know, $$mgh = {1 \over 2}m{v^2}$$ by energy conservation $$\left. \, \right]$$

2

MCQ (Single Correct Answer)

A ball of mass $$0.2$$ $$kg$$ is thrown vertically upwards by applying a force by hand. If the hand moves $$0.2$$ $$m$$ while applying the force and the ball goes upto $$2$$ $$m$$ height further, find the magnitude of the force. (consider $$g = 10\,m/{s^2}$$).

A

$$4N$$

B

$$16$$ $$N$$

C

$$20$$ $$N$$

D

$$22$$ $$N$$

According to energy conservation law,

Work done by the hand and due to gravity = total change in the kinetic energy

Initially the the ball is at rest and finally at top its velocity become zero so total change in kinetic energy $$\Delta K$$ = 0

$${W_{hand}} + {W_{gravity}} = \Delta K$$

[Here distance covered would be 0.2 meter for force by hand as force is applied while ball is in contact with hand.

And gravity will still work while ball is in contact with hand so total distance due to gravity would be 2 + 0.2 = 2.2 meter.]

$$ \Rightarrow F\left( {0.2} \right) - \left( {0.2} \right)\left( {10} \right)\left( {2.2} \right)$$ $$ = 0 \Rightarrow F = 22\,N$$

$$\therefore$$ Option (D) is correct.

Work done by the hand and due to gravity = total change in the kinetic energy

Initially the the ball is at rest and finally at top its velocity become zero so total change in kinetic energy $$\Delta K$$ = 0

$${W_{hand}} + {W_{gravity}} = \Delta K$$

[Here distance covered would be 0.2 meter for force by hand as force is applied while ball is in contact with hand.

And gravity will still work while ball is in contact with hand so total distance due to gravity would be 2 + 0.2 = 2.2 meter.]

$$ \Rightarrow F\left( {0.2} \right) - \left( {0.2} \right)\left( {10} \right)\left( {2.2} \right)$$ $$ = 0 \Rightarrow F = 22\,N$$

$$\therefore$$ Option (D) is correct.

3

MCQ (Single Correct Answer)

A mass of $$M$$ $$kg$$ is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of $${45^ \circ }$$ with the initial vertical direction is

A

$$Mg\left( {\sqrt 2 + 1} \right)$$

B

$$Mg\sqrt 2 $$

C

$${{Mg} \over {\sqrt 2 }}$$

D

$$Mg\left( {\sqrt 2 - 1} \right)$$

From work energy theorem we can say,

Work done by tension $$+$$ work done by force (applied) $$+$$ Work done by gravitational force $$=$$ change in kinetic energy

Here Work done by tension is zero

$$ \Rightarrow 0 + F \times AB - Mg \times AC = 0$$

$$ \Rightarrow F = Mg\left( {{{AC} \over {AB}}} \right) = Mg\left[ {{{1 - {1 \over {\sqrt 2 }}} \over {{1 \over 2}}}} \right]$$

[ as $$AB = \ell \sin {45^ \circ } = {\ell \over {\sqrt 2 }}$$

and $$AC = OC - OA = \ell - \ell \,\cos \,{45^ \circ } = \ell \left( {1 - {1 \over {\sqrt 2 }}} \right)$$

where $$\ell = $$ length of the string. ]

$$ \Rightarrow F = Mg\left( {\sqrt 2 - 1} \right)$$

Work done by tension $$+$$ work done by force (applied) $$+$$ Work done by gravitational force $$=$$ change in kinetic energy

Here Work done by tension is zero

$$ \Rightarrow 0 + F \times AB - Mg \times AC = 0$$

$$ \Rightarrow F = Mg\left( {{{AC} \over {AB}}} \right) = Mg\left[ {{{1 - {1 \over {\sqrt 2 }}} \over {{1 \over 2}}}} \right]$$

[ as $$AB = \ell \sin {45^ \circ } = {\ell \over {\sqrt 2 }}$$

and $$AC = OC - OA = \ell - \ell \,\cos \,{45^ \circ } = \ell \left( {1 - {1 \over {\sqrt 2 }}} \right)$$

where $$\ell = $$ length of the string. ]

$$ \Rightarrow F = Mg\left( {\sqrt 2 - 1} \right)$$

4

MCQ (Single Correct Answer)

The block of mass $$M$$ moving on the frictionless horizontal surface collides with the spring of spring constant $$k$$ and compresses it by length $$L.$$ The maximum momentum of the block after collision is

A

$${{k{L^2}} \over {2M}}$$

B

$$\sqrt {Mk} \,\,L$$

C

$${{M{L^2}} \over k}$$

D

Zero

Elastic energy stored in the spring = $${1 \over 2}k{L^2}$$

And kinetic energy of the block = $${1 \over 2}M{v^2}$$

$$\therefore$$ $${1 \over 2}M{v^2} = {1 \over 2}k{L^2}$$

$$ \Rightarrow v = \sqrt {{k \over M}} .L$$

$$\therefore$$ Momentum $$ = M \times v = M \times \sqrt {{k \over M}} .L = \sqrt {kM} .L$$

And kinetic energy of the block = $${1 \over 2}M{v^2}$$

$$\therefore$$ $${1 \over 2}M{v^2} = {1 \over 2}k{L^2}$$

$$ \Rightarrow v = \sqrt {{k \over M}} .L$$

$$\therefore$$ Momentum $$ = M \times v = M \times \sqrt {{k \over M}} .L = \sqrt {kM} .L$$

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