Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

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

2

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.

3

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 $${x^2}$$.

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

4

MCQ (Single Correct Answer)

A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time $$'t'$$ is proportional to

A

$${t^{3/4}}$$

B

$${t^{3/2}}$$

C

$${t^{1/4}}$$

D

$${t^{1/2}}$$

We know that $$F \times v = $$ Power

According to the question, power is constant.

$$\therefore$$ $$F \times v = c\,\,\,\,$$ where $$c=$$ constant

$$\therefore$$ $$m{{dv} \over {dt}} \times v = c$$ $$\,\,\,\,\left( \, \right.$$ $$\therefore$$ $$\left. {F = ma = {{mdv} \over {dt}}\,\,} \right)$$

$$\therefore$$ $$m\int\limits_0^v {vdv = c\int\limits_0^t {dt} } \,\,\,\,\,\,\,\,\,$$ $$\therefore$$ $${1 \over 2}m{v^2} = ct$$

$$\therefore$$ $$v = \sqrt {{{2c} \over m}} \times {t^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$

$${{dx} \over {dt}} = \sqrt {{{2c} \over m}} \times {t^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}\,\,\,\,$$ where $$v = {{dx} \over {dt}}$$

$$\therefore$$ $$\int\limits_0^x {dx = \sqrt {{{2c} \over m}} } \times \int\limits_0^t {{t^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} dt$$

$$x = \sqrt {{{2c} \over m}} \times {{2{t^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over 3} \Rightarrow x \propto {t^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$

According to the question, power is constant.

$$\therefore$$ $$F \times v = c\,\,\,\,$$ where $$c=$$ constant

$$\therefore$$ $$m{{dv} \over {dt}} \times v = c$$ $$\,\,\,\,\left( \, \right.$$ $$\therefore$$ $$\left. {F = ma = {{mdv} \over {dt}}\,\,} \right)$$

$$\therefore$$ $$m\int\limits_0^v {vdv = c\int\limits_0^t {dt} } \,\,\,\,\,\,\,\,\,$$ $$\therefore$$ $${1 \over 2}m{v^2} = ct$$

$$\therefore$$ $$v = \sqrt {{{2c} \over m}} \times {t^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$

$${{dx} \over {dt}} = \sqrt {{{2c} \over m}} \times {t^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}\,\,\,\,$$ where $$v = {{dx} \over {dt}}$$

$$\therefore$$ $$\int\limits_0^x {dx = \sqrt {{{2c} \over m}} } \times \int\limits_0^t {{t^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} dt$$

$$x = \sqrt {{{2c} \over m}} \times {{2{t^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over 3} \Rightarrow x \propto {t^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$

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