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**Simulator**

**Previous Years AIEEE/JEE Mains Questions**

(1) 1

$(2)-\frac{1}{\sqrt{2}}$

(3) $-2 \sqrt{2}$

(4) –1

**[AIEEE – 2009] **

**Sol.**(3)

**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]**

**Sol.**(4)

Electrostatics field is a conservative field work is independent of path.

(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]**

**Sol.**(2)

$\mathrm{q}=-100 \times 1.6 \times 10^{-19} \mathrm{C}$

$\Delta \mathrm{V}=-14 \mathrm{volt}$z

$\mathrm{W}=\mathrm{q} \Delta \mathrm{V}=2.24 \times 10^{-16} \mathrm{J}$

(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]**

**Sol.**(1)

(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]**

**Sol.**(4)

$\overrightarrow{\mathrm{E}}=\frac{2 \mathrm{k} \lambda}{\mathrm{r}}(-\hat{\mathrm{j}})=\frac{2}{\left(4 \pi \epsilon_{0} \mathrm{r}\right)} \frac{\mathrm{q}}{(\pi \mathrm{r})}(-\hat{\mathrm{j}})$

(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]**

**Sol.**(3)

Total charge

$Q=\int_{0}^{r} \rho d V=\int_{0}^{r} \rho_{0}\left(\frac{5}{4}-\frac{r}{R}\right) 4 \pi r^{2} d r$

$=4 \pi \rho_{0} \int_{0}^{r}\left(\frac{5 r^{2}}{4}-\frac{r^{3}}{R}\right) d r=4 \pi \rho_{0}\left[\frac{5 r^{3}}{12}-\frac{r^{4}}{4 R}\right]$

$\mathrm{E}=\frac{\mathrm{KQ}}{\mathrm{r}^{2}}=\frac{1}{4 \pi \epsilon_{0} \mathrm{r}^{2}} 4 \pi \rho_{0}\left[\frac{5}{12} \mathrm{r}^{3}-\frac{\mathrm{r}^{4}}{4 \mathrm{R}}\right]$

$=\frac{\rho_{0} r}{4 \in_{0}}\left[\frac{5}{3}-\frac{r}{R}\right]$

(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]**

**Sol.**(3)

$\tan \theta=\frac{\mathrm{F}}{\mathrm{M}_{9}} \quad(\text { since } \theta \text { small })$

(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]**

**Sol.**(2)

(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]**

**Sol.**(1)

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]**

**Sol.**(4)

**[AIEEE – 2012]**

**Sol.**(4)

For uniformly charged sphere

$\mathrm{E}=\frac{\mathrm{Kqr}}{\mathrm{R}^{3}}(\mathrm{r}<\mathrm{R})$

$\mathrm{E}=\frac{\mathrm{Kq}}{\mathrm{R}^{2}} \quad(\mathrm{r}=\mathrm{R})$

$\mathrm{E}=\frac{\mathrm{Kq}}{\mathrm{r}^{2}} \quad(\mathrm{r}>\mathrm{R})$

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]**

**Sol.**(2)

(1) y (2) –y (3) $\frac{1}{y}$ $ (4)-\frac{1}{\mathrm{y}}$

**[AIEEE – 2013]**

**Sol.**(1)

(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]**

**Sol.**(4)

(1)–80 J

(2) 80 J

(3) 120 J

(4) –120 J

**Sol.**(1)

$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{0}=-10[8-0]=-80 \mathrm{V}$

(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]**

**Sol.**(2)

Gaussian surface at distance r from center

(1) $60^{\circ}$ (2) $90^{\circ}$ (3) $30^{\circ}$ (4) $45^{\circ}$

**[JEE-Main-2017]**

**Sol.**(1)

So from $\quad \vec{\tau}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{E}}$

**[JEE-Main-2018]**

**Sol.**(1)

Thankyou, it was very helpful

Thank you

Really esaral is the best site ……

U should give nice explanation and useful questions

Gjici

thank you but add some more questions of recent years

very nice…..

Thank you so much but add some questions on Gauss law

Thank u very much and please add some 2019and 2020 questions also

great work

Thank you sir

This is an easier way for JEE aspirants

Thank you very much……but please add some more questions

We need more

Awesome work.E Saral will soon be one of the best online platform for competitive exams

Thanks sir

Having kids

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

Thanks sir

Thanks for such a great helping

sir,please upload 2019 and 2020 questions also.

Thanks for providing the questions of previous years.

Thank you very much! Wish you succeed in your work and hope that work will be to provide the stuff like this.

Very nice