*Simulator***Previous Years AIEEE/JEE Mains Questions**

Q. The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is :-
(1) $\frac{\mathrm{R}} {2}$ (2) $\sqrt{2} \mathrm{R}$ (3) 2R (4) $\frac{\mathrm{R}}{\sqrt{2}}$

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**Sol.**(3) $\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}} \Rightarrow\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}=\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=9$ $\Rightarrow 1+\frac{\mathrm{h}}{\mathrm{R}}=3 \Rightarrow \mathrm{h}=2 \mathrm{R}$

Q. Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is :-
$(1)-\frac{6 \mathrm{Gm}}{\mathrm{r}}$
$(2)-\frac{9 \mathrm{Gm}}{\mathrm{r}}$
(3) zero
$(4)-\frac{4 \mathrm{Gm}}{\mathrm{r}}$

**[AIEEE – 2011]**
Q. Two particles of equal mass ‘m’ go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is:-
(1) $\sqrt{\frac{\mathrm{Gm}}{\mathrm{R}}}$ (2) $\sqrt{\frac{\mathrm{Gm}}{4 \mathrm{R}}}$ (3) $\sqrt{\frac{\mathrm{Gm}}{3 \mathrm{R}}}$ (4) $\sqrt{\frac{\mathrm{Gm}}{2 \mathrm{R}}}$

**[AIEEE-2011]**
Q. The mass of a spaceship is 1000 kg. It is to be launched from the earth’s surface out into free space. The value of ‘g’ and ‘R’ (radius of earth) are 10 m/s2 and 6400 km respectively. The required energy for this work will be :-
(1) $6.4 \times 10^{10}$ Joules
(2) $6.4 \times 10^{11}$ Joules
(3) $6.4 \times 10^{8}$ Joules
(4) $6.4 \times 10^{9}$ Joules

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**Sol.**(1) $\mathrm{PE}_{\mathrm{i}}+\mathrm{KE}_{\mathrm{i}}=\mathrm{PE}_{\mathrm{f}}+\mathrm{KE}_{\mathrm{f}}$ $-\mathrm{mgR}+\mathrm{KE}_{\mathrm{i}}=0+0$ $\mathrm{KE}_{\mathrm{i}}=+\mathrm{mgR}=1000 \times 10 \times 6.4 \times 10^{6}$ work done $=6.4 \times 10^{10} \mathrm{J}$

Q. What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R ?
(1) $\frac{5 \mathrm{GmM}}{6 \mathrm{R}}$ (2) $\frac{2 \mathrm{GmM}}{3 \mathrm{R}}$ (3) $\frac{\mathrm{GmM}}{2 \mathrm{R}}$ (4) $\frac{\mathrm{GmM}}{3 \mathrm{R}}$

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**Sol.**(1) From energy conservation $\frac{\mathrm{GMm}}{\mathrm{R}}+\mathrm{KE}=\frac{-\mathrm{GMm}}{3 \mathrm{R}}+\frac{1}{2} \mathrm{mV}^{2} \ldots(\mathrm{i})$ …(i) From force balance at A, $\frac{\mathrm{GMm}}{(3 \mathrm{R})^{2}}=\frac{\mathrm{mv}^{2}}{3 \mathrm{R}} \Rightarrow \mathrm{V}^{2}=\frac{\mathrm{GM}}{3 \mathrm{R}}$ ………..(ii) from (i) & (ii) $\mathrm{KE}_{\text {suface }}=\frac{5}{6} \frac{\mathrm{GMm}}{\mathrm{R}}$

Q. Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is :
(1) $\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$
(2) $\frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$
(3) $\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$
(4) $\sqrt{2 \sqrt{2} \frac{\mathrm{GM}}{\mathrm{R}}}$

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**Sol.**(2) Net force on one particle $\mathrm{F}_{\mathrm{net}}=\mathrm{F}_{1}+2 \mathrm{F}_{2} \cos 45^{\circ}=$ Centripetal force $\Rightarrow \frac{\mathrm{GM}^{2}}{(2 \mathrm{R})^{2}}+\left[\frac{2 \mathrm{GM}^{2}}{(\sqrt{2} \mathrm{R})^{2}} \cos 45^{\circ}\right]=\frac{\mathrm{MV}^{2}}{\mathrm{R}}$ $\mathrm{V}=\frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$ $\mathrm{V}=-\frac{\mathrm{GM}}{2 \mathrm{R}^{3}}\left[3 \mathrm{R}^{2}-\frac{\mathrm{R}^{2}}{4}\right]+\frac{3 \mathrm{G}}{2} \frac{\mathrm{M}}{8 \frac{\mathrm{R}}{2}}$ $=\frac{-11 \mathrm{GM}}{8 \mathrm{R}}+\frac{3 \mathrm{GM}}{8 \mathrm{R}}=-\frac{\mathrm{GM}}{\mathrm{R}}$

Q. From a solid sphere of mass M and radius R, a spherical portion of radius $\frac{\mathrm{R}}{2}$ is removed, as shown in the figure. Taking gravitational potential V = 0 at r = $\infty$, the potential at the centre of the cavity thus formed is : (G =

*gravitational constant*) (1) $\frac{-2 \mathrm{GM}}{3 \mathrm{R}}$ (2) $\frac{-2 \mathrm{GM}}{\mathrm{R}}$ (3) $\frac{-\mathrm{GM}}{2 \mathrm{R}}$ (4) $\frac{-\mathrm{GM}}{\mathrm{R}}$**[JEE-Mains 2015]****Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...**

**Sol.**(4) By principle of superosition $\mathrm{V}=-\frac{\mathrm{GM}}{2 \mathrm{R}^{3}}\left[3 \mathrm{R}^{2}-\frac{\mathrm{R}^{2}}{4}\right]+\frac{3 \mathrm{G}}{2} \frac{\mathrm{M}}{8 \frac{\mathrm{R}}{2}}$ $=\frac{-11 \mathrm{GM}}{8 \mathrm{R}}+\frac{3 \mathrm{GM}}{8 \mathrm{R}}=-\frac{\mathrm{GM}}{\mathrm{R}}$

Q. A satellite is reolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R ; h << R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to : (Neglect the effect of atmosphere).
(1) $\sqrt{\mathrm{gR}}(\sqrt{2}-1)$
(2) $\sqrt{2 \mathrm{gR}}$
(3) $\sqrt{\mathrm{gR}}$
(4) $\sqrt{\mathrm{gR} / 2}$

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**Sol.**(1) $\mathrm{V}_{0}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ or $\sqrt{\mathrm{gR}}$ $\mathrm{V}_{\mathrm{e}} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ or $\sqrt{2 \mathrm{gR}}$ $\therefore$ Increase in velocity $=\sqrt{\mathrm{gR}}[\sqrt{2}-1]$

Q. The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth’s radius) :-

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**Sol.**(2) $\mathrm{g}=\frac{\mathrm{GMx}}{\mathrm{R}^{3}}$ inside the Earth (straight line) $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{r}^{2}}$ outside the Earth where M is Mass of Earth option (2)

solution text isnt clear at all

It’s good , but add some more questions

need more questions

Questions till 2020 would be great sir.

we want latest years questions but it helps a lot

THNX SIR…..WANT MORE

YES SIR

Thanks for this content

amazing ques bank

Thanks for this content

great yaar……..

Thanks and want more questions

Want more anyways thanks so much

All topics are not covered but still nice collection

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Thanks for this kind of helping hand.Waiting for more♥

Thanks it was really helpful.

Jee adv?

Need some more questions

NICE.NEED SOME MORE QUESTIONS

thanks .. it was enlightening

Thanks , it’s very useful….

Thanks for your collection, more questions would be great

Kamino kaha hai previous year questiond