Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

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 $$v = {{dx} \over {dt}}$$

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

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

2

MCQ (Single Correct Answer)

A spring of spring constant $$5 \times {10^3}\,N/m$$ is stretched initially by $$5$$ $$cm$$ from the unstretched position. Then the work required to stretch it further by another $$5$$ $$cm$$ is

A

$$12.50$$ $$N$$-$$m$$

B

$$18.75$$ $$N$$-$$m$$

C

$$25.00$$ $$N$$-$$m$$

D

$$625$$ $$N$$-$$m$$

Given $$k = 5 \times {10^3}N/m$$

Work done when a spring stretched from x_{1} cm to x_{2} cm,

$$W = {1 \over 2}k\left( {x_2^2 - x_1^2} \right) $$

$$= {1 \over 2} \times 5 \times {10^3}\left[ {{{\left( {0.1} \right)}^2} - {{\left( {0.05} \right)}^2}} \right]$$

$$ = {{5000} \over 2} \times 0.15 \times 0.05 = 18.75\,\,Nm$$

Work done when a spring stretched from x

$$W = {1 \over 2}k\left( {x_2^2 - x_1^2} \right) $$

$$= {1 \over 2} \times 5 \times {10^3}\left[ {{{\left( {0.1} \right)}^2} - {{\left( {0.05} \right)}^2}} \right]$$

$$ = {{5000} \over 2} \times 0.15 \times 0.05 = 18.75\,\,Nm$$

3

MCQ (Single Correct Answer)

A wire suspended vertically from one of its ends is stretched by attaching a weight of $$200N$$ to the lower end. The weight stretches the wire by $$1$$ $$mm.$$ Then the elastic energy stored in the wire is

A

$$0.2$$ $$J$$

B

$$10$$ $$J$$

C

$$20$$ $$J$$

D

$$0.1$$ $$J$$

The elastic potential energy

$$ = {1 \over 2} \times $$ Force $$ \times $$ extension

$$= {1 \over 2} \times 200 \times 0.001 = 0.1\,J$$

$$ = {1 \over 2} \times $$ Force $$ \times $$ extension

$$= {1 \over 2} \times 200 \times 0.001 = 0.1\,J$$

4

MCQ (Single Correct Answer)

Consider the following two statements:

$$A.$$$$\,\,\,\,\,\,$$ Linear momentum of a system of particles is zero

$$B.$$$$\,\,\,\,\,\,$$ Kinetic energy of a system of particles is zero.

then

$$A.$$$$\,\,\,\,\,\,$$ Linear momentum of a system of particles is zero

$$B.$$$$\,\,\,\,\,\,$$ Kinetic energy of a system of particles is zero.

then

A

$$A$$ does not imply $$B$$ and $$B$$ does not imply $$A$$

B

$$A$$ implies $$B$$ but $$B$$ does not imply $$A$$

C

$$A$$ does not imply $$B$$ but $$B$$ implies $$A$$

D

$$A$$ implies $$B$$ and $$B$$ implies $$A$$

Kinetic energy of a system of particle is zero this is possible only when the speed of each particles is zero. And if speed of each particle is zero, the linear momentum of the system of particle has to be zero.

Also the linear momentum of the system may be zero even when the particles are moving in different direction. This is because linear momentum is a vector quantity. In this case the kinetic energy of the system of particles will not be zero.

$$\therefore$$ $$A$$ does not implies $$B$$ but $$B$$ implies $$A.$$

On those following papers in MCQ (Single Correct Answer)

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Units & Measurements *keyboard_arrow_right*

Motion *keyboard_arrow_right*

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