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*Simulator*

**Previous Years AIEEE/JEE Main Questions**

(1) 2

(2) 1

(3) 1/2

(4) 1/4

**[AIEEE-2010]**

**Sol.**(3)

(1) 1

(2) 4

(3) 3

(4) 2

**[AIEEE – 2010]**

**Sol.**(4)

Therefore K =2

(1) $2.7 \times 10^{6} \Omega$

(2) $3.3 \times 10^{7} \Omega$

(3) $1.3 \times 10^{4} \Omega$

(4) $1.7 \times 10^{5} \Omega$

** [AIEEE-2011]**

**Sol.**(2)

Neon bulb is filled with gas, so its resistance is infinite, hence no current flows through it.

Now, $\mathrm{V}_{\mathrm{c}}=\mathrm{E}\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{RC}}\right)$

$\Rightarrow 120=200\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{RC}}\right)$

$\Rightarrow \mathrm{e}^{-\mathrm{t} / \mathrm{RC}}=\frac{2}{5} \Rightarrow \mathrm{t}=\mathrm{RCln} 2.5$

$\Rightarrow \mathrm{R}=\frac{\mathrm{t}}{\mathrm{Cln} 2.5}=\frac{\mathrm{t}}{2.303 \mathrm{Clog} 2.5}$

$=2.7 \times 10^{6} \Omega$

(1) 20 second

(2) 10 second

(3) 5 second

(4) 2.5 second

**[AIEEE-2011]**

**Sol.**(4)

(1) 100 sec and 150 sec

(2) 150 sec and 200 sec

(3) 0 and 50 sec

(4) 50 sec and 100 sec

** [AIEEE-2012]**

**Sol.**(1)

$\mathrm{t}=0.37 \% \mathrm{of} \mathrm{V}_{0}$

$=0.37 \times 25=9.25$ volt

where is in between 100 and 150 sec.

(1) $5 \mathrm{C}_{1}=3 \mathrm{C}_{2}$

(2) $3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$

(3) $3 \mathrm{C}_{1}+5 \mathrm{C}_{2}=0$

(4) $9 \mathrm{C}_{1}=4 \mathrm{C}_{2}$

**[JEE-Mains-2013]**

**Sol.**(2)

Common voltage $=\frac{\mathrm{C}_{1} \mathrm{v}_{1}-\mathrm{C}_{2} \mathrm{V}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$

(positive plate of one capacitor is connected with negative plate of second capacitor)

$\Rightarrow 120 \mathrm{C}_{1}=200 \mathrm{C}_{2} \Rightarrow 3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$

to :

(1) $3 \times 10^{4} \mathrm{C} / \mathrm{m}^{2}$

(2) $6 \times 10^{4} \mathrm{C} / \mathrm{m}^{2}$

(3) $6 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}$

(4) $3 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}$

**[JEE-Mains-2014]**

**Sol.**(3)

$\frac{\sigma}{\mathrm{K} \varepsilon_{0}}=3 \times 10^{4}$

$\frac{\sigma}{2.25 \times 8.86 \times 10^{-12}}=3 \times 10^{4}$

$\sigma=6 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}$

**[JEE-Mains-2015]**

**Sol.**(4)

$\mathrm{C}_{\mathrm{eq}}=\frac{3 \mathrm{C}}{3+\mathrm{C}} \quad \mathrm{Q}=\mathrm{Ceq} \mathrm{E}$

$\mathrm{Q}_{2}=\frac{2 \mathrm{CE}}{(\mathrm{C}+3)}=\frac{3 \mathrm{CE}}{(\mathrm{C}+3)}$

$=2 \mathrm{E}\left[1-\frac{3}{\mathrm{C}+3}\right]$

$\mathrm{Q}_{2 / \mathrm{when}} \mathrm{C}=1_{\mathrm{HF}}=2 \mathrm{E}\left[\frac{1}{4}\right]=\frac{\mathrm{E}}{2}$

$\mathrm{Q}_{2 / \mathrm{when}} \mathrm{C}=3_{1 \mathrm{F}}=2 \mathrm{E}\left[\frac{1}{2}\right]=\mathrm{E}$

$\frac{\mathrm{dQ}}{\mathrm{d} \mathrm{C}}=-\frac{2 \mathrm{E} \cdot 3}{(\mathrm{C}+3)}(-1)=\frac{6 \mathrm{E}}{(\mathrm{C}+3)^{2}}>0$

$\frac{\mathrm{d}^{2} \theta}{\mathrm{d} \mathrm{C}^{2}}=\frac{6 \mathrm{E}(-2)}{(\mathrm{C}+3)^{3}}=-\frac{12 \mathrm{E}}{(\mathrm{C}+3)^{2}}<0$

(1) 480 N/C

(2) 240 N/C

(3) 360 N/C

(4) 420 N/C

**[JEE-Mains-2016]**

**Sol.**(4)

$\mathrm{Q}=24+18=42 \mu \mathrm{c}$

$\mathrm{E}=\frac{\mathrm{KQ}}{\mathrm{r}^{2}}$

$\Rightarrow \mathrm{E}=\frac{9 \times 10^{9} \times 42 \times 10^{-6}}{(30)^{2}}=420 \mathrm{N} / \mathrm{C}$

(1) $\mathrm{CE} \frac{\mathrm{r}_{2}}{\left(\mathrm{r}+\mathrm{r}_{2}\right)}$

(2) $\mathrm{CE} \frac{\mathrm{r}_{1}}{\left(\mathrm{r}_{1}+\mathrm{r}\right)}$

(3) CE

(4) $\mathrm{CE} \frac{\mathrm{r}_{1}}{\left(\mathrm{r}_{2}+\mathrm{r}\right)}$

**[JEE-Mains-2017]**

**Sol.**(1)

It steady state, current through AB = 0

(1) 24

(2) 32

(3) 2

(4) 16

** [JEE-Mains-2017]**

**Sol.**(2)

To hold 1 KV potential differenceminimum four capacitors are required in series

$\Rightarrow \quad \mathrm{C}_{1}=\frac{1}{4}$ for one series. So for $\mathrm{Ceq}$ to be $2 \mu \mathrm{F}, 8$ parallel combinations are required.

$\Rightarrow$ Minimum no. of capacitors $=8 \times 4=32$

(1) 0.3 n C

(2) 2.4 n C

(3) 0.9 n C

(4) 1.2 n C

**[JEE-Mains-2018]**

**Sol.**(4)

$\mathrm{Q}=(\mathrm{k} \mathrm{C}) \mathrm{V}$

$=\left(\frac{5}{3} \times 90 \mathrm{pF}\right)(20 \mathrm{V})$

= 3000 pC

= 3nC

induced charges on dielectric

$\mathrm{Q}_{\mathrm{ind}}=\mathrm{Q}\left(1-\frac{1}{\mathrm{K}}\right)=3 \mathrm{nC}\left(1-\frac{3}{5}\right)=1.2 \mathrm{nC}$