*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)

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YES SIR

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Need some more questions

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Kamino kaha hai previous year questiond