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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2010

MCQ (Single Correct Answer)
In the circuit shown below, the key $$K$$ is closed at $$t=0.$$ The current through the battery is
A
$${{V{R_1}{R_2}} \over {\sqrt {R_1^2 + R_2^2} }}$$ at $$t=0$$ and $${V \over {{R_2}}}$$ at $$t = \infty $$
B
$${V \over {{R_2}}}$$ at $$\,t = 0$$ and $${{V\left( {{R_1} + {R_2}} \right)} \over {{R_1}{R_2}}}$$ at $$t = \infty $$
C
$${V \over {{R_2}}}$$ at $$\,t = 0$$ and $${{V{R_1}{R_2}} \over {\sqrt {R_1^2 + R_2^2} }}$$ at $$t = \infty $$
D
$${{V\left( {{R_1} + {R_2}} \right)} \over {{R_1}{R_2}}}$$ at $$t=0$$ and $${V \over {{R_2}}}$$ at $$t = \infty $$

Explanation

At $$t=0,$$ no current will flow through $$L$$ and $${R_1}$$

$$\therefore$$ Current through battery $$ = {V \over {{R_2}}}$$

At $$t = \infty ,$$

effective resistance, $${{\mathop{\rm R}\nolimits} _{eff}} = {{{R_1}{R_2}} \over {{R_1} + {R_2}}}$$

$$\therefore$$ Current through battery $$ = {V \over {{{\mathop{\rm R}\nolimits} _{eff}}}} = {{V\left( {{R_1} + {R_2}} \right)} \over {{R_1}{R_2}}}$$
2

AIEEE 2010

MCQ (Single Correct Answer)
A rectangular loop has a sliding connector $$PQ$$ of length $$l$$ and resistance $$R$$ $$\Omega $$ and it is moving with a speed $$v$$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $${I_1},{I_2}$$ and $$I$$ are
A
$${I_1} = - {I_2} = {{Blv} \over {6R}},\,\,I = {{2Blv} \over {6R}}$$
B
$${I_1} = {I_2} = {{Blv} \over {3R}},\,\,I = {{2Blv} \over {3R}}$$
C
$${I_1} = {I_2} = I = {{Blv} \over R}$$
D
$${I_1} = {I_2} = {{Blv} \over {6R}},I = {{Blv} \over {3R}}$$

Explanation



Due to the movement of resistor $$R,$$ an $$emf$$ equal to $$Blv$$ will be induced in it as shown in figure clearly,

$$I = {I_1} + {I_2}$$
<
Also, $${I_1} = {I_2}$$

Solving the circuit, we get

$${I_1} = {I_2} = {{Blv} \over {3R}}$$

and $$I = 2{I_1} = {{2Blv} \over {3R}}$$
3

AIEEE 2010

MCQ (Single Correct Answer)
In a series $$LCR$$ circuit $$R = 200\Omega $$ and the voltage and the frequency of the main supply is $$220V$$ and $$50$$ $$Hz$$ respectively. On taking out the capacitance from the circuit the current lags behind the voltage by $${30^ \circ }.$$ On taking out the inductor from the circuit the current leads the voltage by $${30^ \circ }.$$ The power dissipated in the $$LCR$$ circuit is
A
$$305$$ $$W$$
B
$$210$$ $$W$$
C
$$zero$$ $$W$$
D
$$242$$ $$W$$

Explanation

When capacitance is taken out, the circular is $$LR.$$

$$\therefore$$ $$\tan \phi = {{\omega L} \over R}$$

$$ \Rightarrow \omega L = R\,\tan \phi $$

$$ = 200 \times {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}$$

Again, when inductor is taken out, the circuit is $$CR.$$

$$\therefore$$ $$\tan \phi = {1 \over {\omega CR}}$$

$$ \Rightarrow {1 \over {\omega c}} = R\tan \phi $$

$$ = 200 \times {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}$$

Now, $$Z = \sqrt {{R^2} + {{\left( {{1 \over {\omega C}} - \omega L} \right)}^2}} $$

$$ = \sqrt {{{\left( {200} \right)}^2} + {{\left( {{{200} \over {\sqrt 3 }} - {{200} \over {\sqrt 3 }}} \right)}^2}} = 200\Omega $$

Power dissipated $$ = {V_{rms}}{I_{rms}}\cos \phi $$

$$ = {V_{rms}}.{{{V_{rms}}} \over Z}.{R \over Z}$$

$$\left( {\,\,\,} \right.$$ as $$\left. {\,\,\cos \phi = {R \over Z}\,\,\,} \right)$$

$$ = {{{V^2}rmsR} \over {{Z^2}}} = {{{{\left( {220} \right)}^2} \times 200} \over {{{\left( {200} \right)}^2}}}$$

$$ = {{220 \times 220} \over {200}} = 242\,W$$
4

AIEEE 2009

MCQ (Single Correct Answer)
An inductor of inductance $$L=400$$ $$mH$$ and resistors of resistance $${R_1} = 2\Omega $$ and $${R_2} = 2\Omega $$ are connected to a battery of $$emf$$ $$12$$ $$V$$ as shown in the figure. The internal resistance of the battery is negligible. The switch $$S$$ is closed at $$t=0.$$ The potential drop across $$L$$ as a function of time is :
A
$${{12} \over t}{e^{ - 3t}}V$$
B
$$6\left( {1 - {e^{ - t/0.2}}} \right)V$$
C
$$12{e^{ - 5t}}V$$
D
$$6{e^{ - 5t}}V$$

Explanation

Growth in current in $$L{R_2}$$ branch when switch is closed is given by

$$i = {E \over {{R_2}}}\left[ {1 - {e^{ - {R_2}t/L}}} \right]$$

$$ \Rightarrow {{di} \over {dt}} = {E \over {{R_2}}}.{{{R_2}} \over L}.{e^{ - {R_{2t/L}}}}$$

$$ = {E \over L}{e^{{{{R_2}t} \over L}}}$$

Hence, potential drop across

$$L = \left( {{E \over L}{e^{ - {R_2}t/L}}} \right)L = E{e^{ - {R_2}t/L}}$$

$$ = 12{e^{ - {{2t} \over {400 \times {{10}^{ - 3}}}}}} = 12{e^{ - 5t}}V$$

Questions Asked from Alternating Current and Electromagnetic Induction

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