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1

AIEEE 2004

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
$$1200$$ $$J$$

Explanation

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.
2

AIEEE 2004

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
$${\log _e}x$$

Explanation

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 $${x^2}$$.

$$\therefore$$ $$\Delta K \propto {x^2}$$
3

AIEEE 2003

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

Explanation

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

$$\therefore$$ $$\int\limits_0^x {dx = \sqrt {{{2c} \over m}} } \times \int\limits_0^t {{t^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} dt$$
$$x = \sqrt {{{2c} \over m}} \times {{2{t^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over 3} \Rightarrow x \propto {t^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
4

AIEEE 2003

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
$$0.1$$ $$J$$

Explanation

The elastic potential energy

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

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

Questions Asked from Work Power & Energy

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