Gravitation – JEE Main Previous Year Questions with Solutions


JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

 

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) $\frac{\mathrm{R}}{\sqrt{2}}$

[AIEEE – 2009]

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]

Sol. (2)


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]

Sol. (2)


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

[AIEEE-2012]

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) $\frac{\mathrm{GmM}}{3 \mathrm{R}}$

[JEE-Mains 2013]

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) $\sqrt{2 \sqrt{2} \frac{\mathrm{GM}}{\mathrm{R}}}$

[JEE-Mains 2014]

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]

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) $\sqrt{\mathrm{gR} / 2}$

[JEE-Mains 2016]

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) :-

[JEE-Mains 2017]

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)


Administrator

Leave a comment

Please enter comment.
Please enter your name.


Comments
  • August 9, 2020 at 9:58 am

    Thanks it was really helpful.

  • Idk
    July 22, 2020 at 11:56 pm

    Jee adv?

  • June 22, 2020 at 6:32 am

    Need some more questions

  • June 10, 2020 at 2:41 pm

    NICE.NEED SOME MORE QUESTIONS

  • June 5, 2020 at 3:41 pm

    thanks .. it was enlightening

  • May 15, 2020 at 8:34 am

    Thanks , it’s very useful….

  • February 20, 2020 at 11:46 am

    Kamino kaha hai previous year questiond