Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A body of mass $$' m ',$$ acceleration uniformly from rest to $$'{v_1}'$$ in time $${T}$$. The instantaneous power delivered to the body as a function of time is given by

A

$${{m{v_1}{t^2}} \over {{T}}}$$

B

$${{mv_1^2t} \over {T^2}}$$

C

$${{m{v_1}t} \over {{T}}}$$

D

$${{mv_1^2t} \over {{T}}}$$

Assume acceleration of body be $$a$$

$$\therefore$$ $${v_1} = 0 + a{T} \Rightarrow a = {{{v_1}} \over {{T}}}$$

$$\therefore$$ $$v = at \Rightarrow v = {{{v_1}t} \over {{T}}}$$

$${P_{inst}} = \overrightarrow F .\overrightarrow v = \left( {m\overrightarrow a } \right).\overrightarrow v $$

$$= \left( {{{m{v_1}} \over {{T}}}} \right)\left( {{{{v_1}t} \over {{T}}}} \right)$$

$$ = m{\left( {{{{v_1}} \over {{T}}}} \right)^2}t$$

$$\therefore$$ $${v_1} = 0 + a{T} \Rightarrow a = {{{v_1}} \over {{T}}}$$

$$\therefore$$ $$v = at \Rightarrow v = {{{v_1}t} \over {{T}}}$$

$${P_{inst}} = \overrightarrow F .\overrightarrow v = \left( {m\overrightarrow a } \right).\overrightarrow v $$

$$= \left( {{{m{v_1}} \over {{T}}}} \right)\left( {{{{v_1}t} \over {{T}}}} \right)$$

$$ = m{\left( {{{{v_1}} \over {{T}}}} \right)^2}t$$

2

MCQ (Single Correct Answer)

A force $$\overrightarrow F = \left( {5\overrightarrow i + 3\overrightarrow j + 2\overrightarrow k } \right)N$$ is applied over a particle which displaces it from its origin to the point $$\overrightarrow r = \left( {2\overrightarrow i - \overrightarrow j } \right)m.$$ The work done on the particle in joules is

A

$$+10$$

B

$$+7$$

C

$$-7$$

D

$$+13$$

Work done when the particle displaces from the origin,

$$W = \overrightarrow F .\overrightarrow x $$

$$= \left( {5\widehat i + 3\widehat j + 2\widehat k} \right).\left( {2\widehat i - \widehat j} \right)$$

$$=10-3=7$$ J

$$W = \overrightarrow F .\overrightarrow x $$

$$= \left( {5\widehat i + 3\widehat j + 2\widehat k} \right).\left( {2\widehat i - \widehat j} \right)$$

$$=10-3=7$$ J

3

MCQ (Single Correct Answer)

A uniform chain of length $$2$$ $$m$$ is kept on a table such that a length of $$60$$ $$cm$$ hangs freely from the edge of the table. The total mass of the chain is $$4$$ $$kg.$$ What is the work done in pulling the entire chain on the table?

A

$$12$$ $$J$$

B

$$3.6$$ $$J$$

C

$$7.2$$ $$J$$

D

$$1200$$ $$J$$

Mass of hanging part $$(m') = {4 \over 2} \times \left( {0.6} \right)kg$$ = 1.2 kg

Let at the surface $$PE=0$$

Center of mass of hanging part $$=0.3$$ $$m$$ below the surface of the table

$${U_i} = - m'gx = - 1.2 \times 10 \times 0.30$$ = - 3.6 J

$$\Delta U = m'gx = 3.6 J = $$ Work done in putting the entire chain on the table.

Let at the surface $$PE=0$$

Center of mass of hanging part $$=0.3$$ $$m$$ below the surface of the table

$${U_i} = - m'gx = - 1.2 \times 10 \times 0.30$$ = - 3.6 J

$$\Delta U = m'gx = 3.6 J = $$ Work done in putting the entire chain on the table.

4

MCQ (Single Correct Answer)

A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $$x$$ is proportional to

A

$$x$$

B

$${e^x}$$

C

$${x^2}$$

D

$${\log _e}x$$

Given that, retardation $$ \propto $$ displacement

$$ \Rightarrow $$ $$a=-kx$$

But we know $$a = v{{dv} \over {dx}}\,\,\,\,\,\,\,\,\,$$

$$\therefore$$ $${{vdv} \over {dx}} = - kx $$

$$\Rightarrow \int\limits_{{v_1}}^{{v_2}} v \,dv = - k\int\limits_0^x {xdx} $$

$$\left( {v_2^2 - v_1^2} \right) = - k{{{x^2}} \over 2}$$

$$ \Rightarrow {1 \over 2}m\left( {v_2^2 - v_1^2} \right) = {1 \over 2}mk\left( {{{ - x^2} \over 2}} \right)$$

$$\therefore$$ Loss in kinetic energy is proportional to $${x^2}$$.

$$\therefore$$ $$\Delta K \propto {x^2}$$

$$ \Rightarrow $$ $$a=-kx$$

But we know $$a = v{{dv} \over {dx}}\,\,\,\,\,\,\,\,\,$$

$$\therefore$$ $${{vdv} \over {dx}} = - kx $$

$$\Rightarrow \int\limits_{{v_1}}^{{v_2}} v \,dv = - k\int\limits_0^x {xdx} $$

$$\left( {v_2^2 - v_1^2} \right) = - k{{{x^2}} \over 2}$$

$$ \Rightarrow {1 \over 2}m\left( {v_2^2 - v_1^2} \right) = {1 \over 2}mk\left( {{{ - x^2} \over 2}} \right)$$

$$\therefore$$ Loss in kinetic energy is proportional to $${x^2}$$.

$$\therefore$$ $$\Delta K \propto {x^2}$$

On those following papers in MCQ (Single Correct Answer)

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Units & Measurements *keyboard_arrow_right*

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