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Previous Years AIEEE/JEE Mains Questions
Q. A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then $\frac{\mathrm{Q}}{\mathrm{q}}$ equals :-
(1) 1
$(2)-\frac{1}{\sqrt{2}}$
(3) $-2 \sqrt{2}$
(4) –1
[AIEEE - 2009]
Ans. (3)
Q. Statement–1 : For a charged particle moving from point P to point Q the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q.
Statement–2 : The net work done by a conservative force on an object moving along closed loop is zero.
(1) Statement–1 is true, Statement–2 is true; Statement–2 is not the correct explanation of Statement–1
(2) Statement–1 is false, Statement–2 is true
(3) Statement–1 is true, Statement–2 is false
(4) Statement–1 is true, Statement–2 is true; Statement–2 is the correct explanation of Statement–
[AIEEE - 2009]
Ans. (4)
Electrostatics field is a conservative field work is independent of path.
Q. Two points P and Q are maintained at the potential of 10V and –4V, respectively. The work done in moving 100 electrons from P to Q is :-
(1) $-2.24 \times 10^{-16} \mathrm{J}$
(2) $2.24 \times 10^{-16} \mathrm{J}$
(3) $-9.60 \times 10^{-17} \mathrm{J}$
(4) $9.60 \times 10^{-17} \mathrm{J}$
[AIEEE - 2009]
Ans. (2)
$\mathrm{q}=-100 \times 1.6 \times 10^{-19} \mathrm{C}$
$\Delta \mathrm{V}=-14 \mathrm{volt}$z
$\mathrm{W}=\mathrm{q} \Delta \mathrm{V}=2.24 \times 10^{-16} \mathrm{J}$
Q. $\operatorname{Let} \mathrm{P}(\mathrm{r})=\frac{\mathrm{Q}}{\pi \mathrm{R}^{4}} \mathrm{r}$ be the charge density distribution for a solid sphere of radius R and total charge Q. For a point 'p' inside the sphere at distance $r_{1}$ from the centre of the sphere, the magnitude of electric field is :-
(1) $\frac{\mathrm{Qr}_{1}^{2}}{4 \pi \epsilon_{0} \mathrm{R}^{4}}$
(2) $\frac{\mathrm{Qr}_{1}^{2}}{3 \pi \epsilon_{0} \mathrm{R}^{4}}$
(3) 0
(4) $\frac{\mathrm{Q}}{4 \pi \epsilon_{0} \mathrm{r}_{1}^{2}}$
[AIEEE - 2009]
Ans. (1)
Q. A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field $\overrightarrow{\mathrm{E}}$ at the centre O is :-
(1) $\frac{\mathrm{q}}{2 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$
( 2)$\frac{\mathrm{q}}{4 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$
$(3)-\frac{\mathrm{q}}{4 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$
(4) $-\frac{\mathrm{q}}{2 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$
[AIEEE - 2010]
(1) $\frac{\mathrm{q}}{2 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$
( 2)$\frac{\mathrm{q}}{4 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$
$(3)-\frac{\mathrm{q}}{4 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$
(4) $-\frac{\mathrm{q}}{2 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$
[AIEEE - 2010]
Ans. (4)
$\overrightarrow{\mathrm{E}}=\frac{2 \mathrm{k} \lambda}{\mathrm{r}}(-\hat{\mathrm{j}})=\frac{2}{\left(4 \pi \epsilon_{0} \mathrm{r}\right)} \frac{\mathrm{q}}{(\pi \mathrm{r})}(-\hat{\mathrm{j}})$
Q. Let there be a spherically symmetric charge distribution with charge density varying as
(r) = $\rho_{0}\left(\frac{5}{4}-\frac{\mathrm{r}}{\mathrm{R}}\right)$ upto r = R, and (r) = 0 for r > R, where r is the distance from the origin. The electric field at a distance r (r < R) from the origion is given by :
( 1)$\frac{\rho_{0} \mathrm{r}}{3 \varepsilon_{0}}\left(\frac{5}{4}-\frac{\mathrm{r}}{\mathrm{R}}\right)$
(2) $\frac{4 \pi \rho_{0} \mathrm{r}}{3 \varepsilon_{0}}\left(\frac{5}{3}-\frac{\mathrm{r}}{\mathrm{R}}\right)$
(3) $\frac{\rho_{0} \mathrm{r}}{4 \varepsilon_{0}}\left(\frac{5}{3}-\frac{\mathrm{r}}{\mathrm{R}}\right)$
(4) $\frac{4 \rho_{0} \mathrm{r}}{3 \varepsilon_{0}}\left(\frac{5}{4}-\frac{\mathrm{r}}{\mathrm{R}}\right)$
[AIEEE - 2010]
Ans. (3)
Total charge
$Q=\int_{0}^{r} \rho d V=\int_{0}^{r} \rho_{0}\left(\frac{5}{4}-\frac{r}{R}\right) 4 \pi r^{2} d r$
$=4 \pi \rho_{0} \int_{0}^{r}\left(\frac{5 r^{2}}{4}-\frac{r^{3}}{R}\right) d r=4 \pi \rho_{0}\left[\frac{5 r^{3}}{12}-\frac{r^{4}}{4 R}\right]$
$\mathrm{E}=\frac{\mathrm{KQ}}{\mathrm{r}^{2}}=\frac{1}{4 \pi \epsilon_{0} \mathrm{r}^{2}} 4 \pi \rho_{0}\left[\frac{5}{12} \mathrm{r}^{3}-\frac{\mathrm{r}^{4}}{4 \mathrm{R}}\right]$
$=\frac{\rho_{0} r}{4 \in_{0}}\left[\frac{5}{3}-\frac{r}{R}\right]$
Q. Two identical charged spheres suspended from a common point by two massless string of length $\ell$are initially a distance d(d << $\ell$) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity v. Then as a function of distance x between them :-
(1) $\mathrm{v} \propto \mathrm{x}^{1 / 2}$
(2) $\mathrm{v} \propto \mathrm{x}$
(3) $\mathrm{v} \propto \mathrm{x}^{-1 / 2}$
(4) $\mathrm{v} \propto \mathrm{x}^{-1}$
[AIEEE - 2011]
Ans. (3)
$\tan \theta=\frac{\mathrm{F}}{\mathrm{M}_{9}} \quad(\text { since } \theta \text { small })$
Q. The electrostatic potential inside a charged spherical ball is given by $\phi=\mathrm{ar}^{2}+\mathrm{b}$where r is the distance from the centre; a, b are constant. Then the charge density inside the ball is :-
(1) $-24 \pi \mathrm{a} \in_{0}$
$(2)-6 \mathrm{a} \in_{0}$
(3) $-24 \pi \mathrm{a} \in_{0} \mathrm{r}$
$(4)-6 \mathrm{a} \in_{0} \mathrm{r}$
[AIEEE - 2011]
Ans. (2)
Q. Two positive charges of magnitude 'q' are placed at the ends of a side (side 1) of a square of side '2a'. Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge Q moves from the middle of side 1 to the centre of square, its kinetic energy at the centre of square is :-
(1) $\frac{1}{4 \pi \epsilon_{0}} \frac{2 \mathrm{qQ}}{\mathrm{a}}\left(1-\frac{1}{\sqrt{5}}\right)$
(2) zero
(3) $\frac{1}{4 \pi \epsilon_{0}} \frac{2 \mathrm{qQ}}{\mathrm{a}}\left(1+\frac{1}{\sqrt{5}}\right)$
(4) $\frac{1}{4 \pi \epsilon_{0}} \frac{2 \mathrm{qQ}}{\mathrm{a}}\left(1-\frac{2}{\sqrt{5}}\right)$
[AIEEE - 2011]
Ans. (1)
Q. This question has Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.
An insulating solid sphere of radius R has a uniformaly positive charge density . As a result of this uniform charge distribution there is a finite value of electric potential at the centre of the sphere, at the surface of the sphre and also at a point out side the sphere. The electric potential at infinity is zero.
Statement-1: When a charge 'q' is taken from the centre to the surface of the sphere, its potential energy changes by $\frac{\mathrm{q} \rho}{3 \epsilon_{0}}$
Statement-2 : The electric field at a distance r (r < R) from the centre of the sphere is $\frac{\rho \mathrm{r}}{3 \epsilon_{0}}$
(1) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of
Statement-1.
(2) Statement-1 is true, Statement-2 is true and Statement-2 is not the correct explanation of
statement-1.
(3) Statement-1 is true, Statement-2 is false
(4) Statement-1 is false, Statement-2 is true
[AIEEE - 2012]
Ans. (4)
Q. In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as a function of distance from the centre. The graph which would correspond to the above will be :-
[AIEEE - 2012]
[AIEEE - 2012]
Ans. (4)
For uniformly charged sphere
$\mathrm{E}=\frac{\mathrm{Kqr}}{\mathrm{R}^{3}}(\mathrm{r}<\mathrm{R})$
$\mathrm{E}=\frac{\mathrm{Kq}}{\mathrm{R}^{2}} \quad(\mathrm{r}=\mathrm{R})$
$\mathrm{E}=\frac{\mathrm{Kq}}{\mathrm{r}^{2}} \quad(\mathrm{r}>\mathrm{R})$
Q. Let $\left[\in_{0}\right]$ denote the dimensional formula of the permittivity of vacuum. If M = mass, L = Length,
T = Time and A = electric current, then :
(1) $\left[\in_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^{2} \mathrm{A}\right]$
(2) $\left[\in_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^{4} \mathrm{A}^{2}\right]$
(3) $\left[\in_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{L}^{2} \mathrm{T}^{-1} \mathrm{A}^{-2}\right]$
(4) $\left[\in_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{L}^{2} \mathrm{T}^{-1} \mathrm{A}\right]$
[AIEEE - 2013]
Ans. (2)
Q. Two charges, each equal to q, are kept at x = –a and x = a on the x-axis. A particle of mass m and charge $\mathrm{q}_{0}=\frac{\mathrm{q}}{2}$ is placed at the origin. If charge $q_{0}$ is given a small displacement (y << a) along the y-axis, the net force acting on the particle is proportional to
(1) y (2) –y (3) $\frac{1}{y}$ $ (4)-\frac{1}{\mathrm{y}}$
[AIEEE - 2013]
Ans. (1)
Q. A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at a distance L from the end A is :-
(1) $\frac{\mathrm{Q}}{8 \pi \epsilon_{0} \mathrm{L}}$
(2) $\frac{3 \mathrm{Q}}{4 \pi \in_{0} \mathrm{L}}$
(3) $\frac{\mathrm{Q}}{4 \pi \epsilon_{0} \mathrm{L} \ln 2}$
(4) $\frac{\mathrm{Q} \ln 2}{4 \pi \epsilon_{0} \mathrm{L}}$
[JEE-Main-2013]
(1) $\frac{\mathrm{Q}}{8 \pi \epsilon_{0} \mathrm{L}}$
(2) $\frac{3 \mathrm{Q}}{4 \pi \in_{0} \mathrm{L}}$
(3) $\frac{\mathrm{Q}}{4 \pi \epsilon_{0} \mathrm{L} \ln 2}$
(4) $\frac{\mathrm{Q} \ln 2}{4 \pi \epsilon_{0} \mathrm{L}}$
[JEE-Main-2013]
Ans. (4)
Q. Assume that an electric field $\overrightarrow{\mathrm{E}}=30 \mathrm{x}^{2} \hat{\mathrm{i}}$ exists in space. Then the potential difference $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{O}}$, where $\mathrm{V}_{\mathrm{O}}$ is the potential at the origin and $\mathrm{V}_{\mathrm{A}}$ the potential at x = 2 m is :-
(1)–80 J
(2) 80 J
(3) 120 J
(4) –120 J
Ans. (1)
$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{0}=-10[8-0]=-80 \mathrm{V}$
$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{0}=-10[8-0]=-80 \mathrm{V}$
Q. The region between two concentric spheres of radii 'a' and 'b', respectively (see figure), has volume charge density $\rho=\frac{\mathrm{A}}{\mathrm{r}}$, where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is :-
(1) $\frac{2 Q}{\pi a^{2}}$
(2) $\frac{\mathrm{Q}}{2 \pi \mathrm{a}^{2}}$
(3) $\frac{\mathrm{Q}}{2 \pi\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right)}$
(4) $\frac{2 \mathrm{Q}}{\pi\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)}$
[JEE-Main-2016]
(1) $\frac{2 Q}{\pi a^{2}}$
(2) $\frac{\mathrm{Q}}{2 \pi \mathrm{a}^{2}}$
(3) $\frac{\mathrm{Q}}{2 \pi\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right)}$
(4) $\frac{2 \mathrm{Q}}{\pi\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)}$
[JEE-Main-2016]
Ans. (2)
Gaussian surface at distance r from center
Q. An electric dipole has a fixed dipole moment $\overrightarrow{\mathrm{p}}$, which makes angle with respect to x-axis. When subjected to an electric field $\overrightarrow{\mathrm{E}}_{1}=\mathrm{E} \hat{\mathrm{i}}$, it experiences a torque $\overrightarrow{\mathrm{T}}_{1}=\tau \hat{\mathrm{k}}$. When subjected to another electric field $\overrightarrow{\mathrm{E}}_{2}=\sqrt{3} \mathrm{E}_{1} \hat{\mathrm{j}}$ it experiences torque . The angle $\theta$ is :
(1) $60^{\circ}$ (2) $90^{\circ}$ (3) $30^{\circ}$ (4) $45^{\circ}$
[JEE-Main-2017]
Ans. (1)
So from $\quad \vec{\tau}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{E}}$
Q. Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities $+\sigma,-\sigma$ and $+\sigma$ respectively. The potential of shell B is :-
[JEE-Main-2018]
[JEE-Main-2018]
Ans. (1)
Comments
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MUSKAAN MEHROTRA
Nov. 22, 2020, 3:52 p.m.
APPRECIATION; THANKU SO MUCH, It was really helpful for me. :>
REQUEST: Plz.... upload chapter wise segregated questions from 2019 and 2020 jee papers.
BEFORE JEE MAINS 2021 JANUARY.
Anam Bushra Granger
Aug. 19, 2020, 12:04 p.m.
The best site for solving questions because it gives time to solve without displaying the answer directly and provide very explaining solution
Rahul
Aug. 17, 2020, 6:05 p.m.
thank u esaral ,,,,,it was very helpful......hope to get such helps in upcoming days also,,,
De-morgan markonikov
May 25, 2020, 8:10 p.m.
Awesome work.E Saral will soon be one of the best online platform for competitive exams
harsha
May 3, 2020, 8:48 p.m.
Thank you so much sir it's so much useful
But I think that AIEEE 2010 solution was wrong and the process would be by taking Gaussian surface