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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2005

MCQ (Single Correct Answer)
The change in the value of $$g$$ at a height $$h$$ above the surface of the earth is the same as at a depth $$d$$ below the surface of earth. When both $$d$$ and $$h$$ are much smaller than the radius of earth, then which one of the following is correct?
A
$$d = {{3h} \over 2}$$
B
$$d = {h \over 2}$$
C
$$d = h$$
D
$$d = 2\,h$$

Explanation

At height h acceleration due to gravity, $${g_h} = g\left[ {1 - {{2h} \over R}} \right];$$

At depth d acceleration due to gravity, $${g_d} = g\left[ {1 - {d \over R}} \right]$$

According to the question,

$${g_h} = {g_d}$$

$$\therefore$$ $$g\left[ {1 - {{2h} \over R}} \right]$$ = $$g\left[ {1 - {d \over R}} \right]$$

$$ \Rightarrow {{2hg} \over R} = {{dg} \over R}$$

$$ \Rightarrow $$ $$d=2h$$
2

AIEEE 2005

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

Explanation

We know, Work done = Difference in potential energy

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

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

$$ = 6.67 \times {10^{ - 10}}J$$
3

AIEEE 2004

MCQ (Single Correct Answer)
Suppose the gravitational force varies inversely as the nth 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
$${R^{\left( {{{n - 2} \over 2}} \right)}}$$

Explanation

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

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

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

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

$$\therefore$$ $$T \propto {R^{{{ \left( {n + 1} \right)} \over 2}}}$$
4

AIEEE 2004

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

Explanation

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

Questions Asked from Gravitation

On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
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JEE Main 2019 (Online) 11th January Evening Slot (1)
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JEE Main 2016 (Offline) (1)
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AIEEE 2011 (1)
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AIEEE 2009 (1)
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AIEEE 2008 (2)
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