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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)

Very nice