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*Simulator*

**Previous Years AIEEE/JEE Mains Questions**

(1) $6\left(1-\mathrm{e}^{-\mathrm{t} / 0.2}\right) \mathrm{V}$

(2) $12 \mathrm{e}^{-5 \mathrm{t}} \mathrm{V}$

(3) $6 \mathrm{e}^{-5 \mathrm{t}} \mathrm{V}$

( 4)$\frac{12}{\mathrm{t}} e^{-3 \mathrm{t}} \mathrm{V}$

**[AIEEE – 2009]**

**Sol.**(2)

$I=I_{0} e^{-\frac{R T}{L}}$

(1) $\frac{V\left(R_{1}+R_{2}\right)}{R_{1} R_{2}}$ at $t=0$ and $\frac{V}{R_{2}}$ at $t=\infty$

(2) $\frac{\mathrm{VR}_{1} \mathrm{R}_{2}}{\sqrt{\mathrm{R}_{1}^{2}+\mathrm{R}_{2}^{2}}}$ at $\mathrm{t}=0$ and $\frac{\mathrm{V}}{\mathrm{R}_{2}}$ at $\mathrm{t}=\infty$

(3) $\frac{\mathrm{V}}{\mathrm{R}_{2}}$ at $\mathrm{t}=0$ and $\frac{\mathrm{V}\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)}{\mathrm{R}_{1} \mathrm{R}_{2}}$ at $\mathrm{t}=\infty$

(4) $\frac{\mathrm{V}}{\mathrm{R}_{2}}$ at $\mathrm{t}=0$ and $\frac{\mathrm{VR}_{1} \mathrm{R}_{2}}{\sqrt{\mathrm{R}_{1}^{2}+\mathrm{R}_{2}^{2}}}$ at $\mathrm{t}=\infty$

**[AIEEE – 2010]**

**Sol.**(3)

At t = 0 inductor behaves as broken wire then

at $\mathrm{t}=\infty$ Inductor behaves as conducting wire

(1) $\mathrm{I}_{1}=\mathrm{I}_{2}=\frac{\mathrm{B} \ell \mathrm{v}}{6 \mathrm{R}}, \mathrm{I}=\frac{\mathrm{B} \ell \mathrm{v}}{3 \mathrm{R}}$

(2) $\mathrm{I}_{1}=-\mathrm{I}_{2}=\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}}, \mathrm{I}=\frac{2 \mathrm{B} \ell \mathrm{v}}{\mathrm{R}}$

(3) $\mathrm{I}_{1}=\mathrm{I}_{2}=\frac{\mathrm{B} \ell \mathrm{v}}{3 \mathrm{R}}, \quad \mathrm{I}=\frac{2 \mathrm{B} \ell \mathrm{v}}{3 \mathrm{R}}$

(4) $\mathrm{I}_{1}=\mathrm{I}_{2}=\mathrm{I}=\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}}$

**[AIEEE – 2010]**

**Sol.**(3)

Circuit can be reduced as

(1) 0.50 mV

(2) 0.15 mV

(3) 1 mV

(4) 0.75 Mv

**[AIEEE – 2011]**

**Sol.**(2)

$e=\mathrm{B} \ell \mathrm{v}=\left(5 \times 10^{-5}\right)(2)(1.50)=0.15 \mathrm{mV}$

(1) 6.0 mV

(2) 3 mV

(3) 4.5 mV

(4) 1.5 mV

**[AIEEE – 2011]**

**Sol.**(2)

(1) Electromagnetic induction in the aluminium plate giving rise to electromagnetic damping

(2) Development of air current when the plate is placed

(3) Induction of electrical charge on the plate

(4) Shielding of magnetic lines of force as aluminium is a paramagnetic material

**[AIEEE – 2012]**

**Sol.**(1)

Due to conducting nature of Al eddy currents are produced

**$\omega$**on a horizontal table with one end of the string fixed. If there is a vertical magnetic field ‘B’ in the region, the e.m.f. induced across the ends of the rod is :

(1) $\frac{2 \mathrm{B} \omega \mathrm{l}^{2}}{2}$

(2) $\frac{3 \mathrm{B} \omega \mathrm{l}^{2}}{2}$

(3) $\frac{4 \mathrm{B} \omega \mathrm{l}^{2}}{2}$

(4) $\frac{5 \mathrm{B} \omega \mathrm{l}^{2}}{2}$

**[JEE Main-2013]**

**Sol.**(4)

E.M.F. induced $=\frac{1}{2} \mathrm{B} \omega\left[(3 \ell)^{2}-(2 \ell)^{2}\right]$

$=\frac{1}{2} \mathrm{B} \omega\left[5 \ell^{2}\right]$

$=\frac{5}{2} \mathrm{B} \omega \ell^{2}$

(1) $9.1 \times 10^{-11}$ weber

(2) $6 \times 10^{-11}$ weber

(3) $3.3 \times 10^{-11}$ weber

(4) $6.6 \times 10^{-9}$ weber

** [JEE Main-2013]**

**Sol.**(1)

(1) –1

(2) $\frac{1-\mathrm{e}}{\mathrm{e}}$

(3) $\frac{\mathrm{e}}{1-\mathrm{e}}$

(4) 1

** [JEE Main-2014]**

**Sol.**(1)

$\mathrm{V}_{\mathrm{R}}+\mathrm{V}_{\mathrm{L}}=0$

$\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{V}_{\mathrm{L}}}=-1$

(1) 6.7 mA

(2) 0.67 mA

(3) 100 mA

(4) 67 mA

**[JEE Main-2015]**

**Sol.**(2)

Decay of current

$=6.66 \times 10^{-4}$

$=0.666 \times 10^{-3}$

$=0.67 \mathrm{mA}$

**[JEE Main-2015]**

**Sol.**(3)

As damping is happening its amplitude would vary as

The oscillations decay exponentially and will be proportional to $\mathrm{e}^{-\gamma t}$ where $\gamma$ depends inversely on L.

So as inductance increases decay becomes slower

$\therefore$ for

(1) 250 Wb

(2) 275 Wb

(3) 200 Wb

(4) 225 Wb

**[JEE Advance-2017]**

**Sol.**(1)