Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A particle of mass $$10$$ $$g$$ is kept on the surface of a uniform sphere of mass $$100$$ $$kg$$ and radius $$10$$ $$cm.$$ Find the work to be done against the gravitational force between them to take the particle far away from the sphere (you may take $$G$$ $$ = 6.67 \times {10^{ - 11}}\,\,N{m^2}/k{g^2}$$)

A

$$3.33 \times {10^{ - 10}}\,J$$

B

$$13.34 \times {10^{ - 10}}\,J$$

C

$$6.67 \times {10^{ - 10}}\,J$$

D

$$6.67 \times {10^{ - 9}}\,J$$

We know, Work done = Difference in potential energy

$$\therefore$$ $$W = \Delta U = {U_f} - {U_i} = 0 - \left[ {{{ - GMm} \over R}} \right]$$

$$ \Rightarrow $$$$W = {{6.67 \times {{10}^{ - 11}} \times 100} \over {0.1}} \times {{10} \over {1000}}$$

$$ = 6.67 \times {10^{ - 10}}J$$

$$\therefore$$ $$W = \Delta U = {U_f} - {U_i} = 0 - \left[ {{{ - GMm} \over R}} \right]$$

$$ \Rightarrow $$$$W = {{6.67 \times {{10}^{ - 11}} \times 100} \over {0.1}} \times {{10} \over {1000}}$$

$$ = 6.67 \times {10^{ - 10}}J$$

2

MCQ (Single Correct Answer)

The time period of an earth satellite in circular orbit is independent of

A

both the mass and radius of the orbit

B

radius of its orbit

C

the mass of the satellite

D

neither the mass of the satellite nor the radius of its orbit

For satellite, gravitational force = centripetal force

$$\therefore$$ $${{m{v^2}} \over {R + x}} = {{GmM} \over {{{\left( {R + x} \right)}^2}}}$$

$$x=$$ height of satellite from earth surface

$$m=$$ mass of satellite

$$ \Rightarrow {v^2} = {{GM} \over {\left( {R + x} \right)}}$$ or $$v = \sqrt {{{GM} \over {R + x}}} $$

We know, $$T = {{2\pi } \over \omega }$$

$$T = {{2\pi \left( {R + x} \right)} \over v} $$ [ as $$\omega = {v \over r}$$ ]

$$= {{2\pi \left( {R + x} \right)} \over {\sqrt {{{GM} \over {R + x}}} }}$$

which is independent of mass of satellite

$$\therefore$$ $${{m{v^2}} \over {R + x}} = {{GmM} \over {{{\left( {R + x} \right)}^2}}}$$

$$x=$$ height of satellite from earth surface

$$m=$$ mass of satellite

$$ \Rightarrow {v^2} = {{GM} \over {\left( {R + x} \right)}}$$ or $$v = \sqrt {{{GM} \over {R + x}}} $$

We know, $$T = {{2\pi } \over \omega }$$

$$T = {{2\pi \left( {R + x} \right)} \over v} $$ [ as $$\omega = {v \over r}$$ ]

$$= {{2\pi \left( {R + x} \right)} \over {\sqrt {{{GM} \over {R + x}}} }}$$

which is independent of mass of satellite

3

MCQ (Single Correct Answer)

If $$g$$ is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass $$m$$ raised from the surface of the earth to a height equal to the radius $$R$$ of the earth is

A

$${1 \over 4}mgR$$

B

$$2mgR$$

C

$${1 \over 2}mgR$$

D

$$mgR$$

Gravitational potential energy on the earth surface of a body

U = $$-{{GmM} \over R}$$

And at the height h from the earth surface the potential energy

$${U_h} = - {{GmM} \over {R + h}}$$ = $$ - {{GmM} \over {2R}}$$ [ as h = R ]

So the gain in the potential energy

$$\Delta U = {U_h} - U$$

$$\therefore$$ $$\Delta U = {{ - GmM} \over {2R}} + {{GmM} \over R};$$

$$ \Rightarrow $$ $$\Delta U = {{GmM} \over {2R}}$$

Now $${{GM} \over {{R^2}}} = g;$$ $$\,\,\,$$ $$\therefore$$ $${\mkern 1mu} {{GM} \over R} = gR$$

$$\therefore$$ $$\Delta U = {1 \over 2}mgR$$

U = $$-{{GmM} \over R}$$

And at the height h from the earth surface the potential energy

$${U_h} = - {{GmM} \over {R + h}}$$ = $$ - {{GmM} \over {2R}}$$ [ as h = R ]

So the gain in the potential energy

$$\Delta U = {U_h} - U$$

$$\therefore$$ $$\Delta U = {{ - GmM} \over {2R}} + {{GmM} \over R};$$

$$ \Rightarrow $$ $$\Delta U = {{GmM} \over {2R}}$$

Now $${{GM} \over {{R^2}}} = g;$$ $$\,\,\,$$ $$\therefore$$ $${\mkern 1mu} {{GM} \over R} = gR$$

$$\therefore$$ $$\Delta U = {1 \over 2}mgR$$

4

MCQ (Single Correct Answer)

Suppose the gravitational force varies inversely as the n^{th} power of distance. Then the time period of a planet in circular orbit of radius $$R$$ around the sun will be proportional to

A

$${R^n}$$

B

$${R^{\left( {{{n - 1} \over 2}} \right)}}$$

C

$${R^{\left( {{{n + 1} \over 2}} \right)}}$$

D

$${R^{\left( {{{n - 2} \over 2}} \right)}}$$

For moving a planet around the sun in the circular orbit,

The necessary centripetal force = Gravitational force exerted on it

$$\therefore$$ $${{m{v^2}} \over r} = {{GMm} \over {{R^n}}}$$

$$ \Rightarrow $$ $$v = \sqrt {{{GM} \over {{R^{n - 1}}}}} $$

We know, $$T = {{2\pi R} \over v}$$

$$ = 2\pi R \times \sqrt {{{{R^{n - 1}}} \over {GM}}} $$

= $$2\pi \times \sqrt {{{{R^2} \times {R^{n - 1}}} \over {GM}}} $$

= $$2\pi \times {{{R^{{{n + 1} \over 2}}}} \over {\sqrt {GM} }}$$

$$\therefore$$ $$T \propto {R^{{{ \left( {n + 1} \right)} \over 2}}}$$

The necessary centripetal force = Gravitational force exerted on it

$$\therefore$$ $${{m{v^2}} \over r} = {{GMm} \over {{R^n}}}$$

$$ \Rightarrow $$ $$v = \sqrt {{{GM} \over {{R^{n - 1}}}}} $$

We know, $$T = {{2\pi R} \over v}$$

$$ = 2\pi R \times \sqrt {{{{R^{n - 1}}} \over {GM}}} $$

= $$2\pi \times \sqrt {{{{R^2} \times {R^{n - 1}}} \over {GM}}} $$

= $$2\pi \times {{{R^{{{n + 1} \over 2}}}} \over {\sqrt {GM} }}$$

$$\therefore$$ $$T \propto {R^{{{ \left( {n + 1} \right)} \over 2}}}$$

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