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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2007

MCQ (Single Correct Answer)
In an $$a.c.$$ circuit the voltage applied is $$E = {E_0}\,\sin \,\omega t.$$ The resulting current in the circuit is $$I = {I_0}\sin \left( {\omega t - {\pi \over 2}} \right).$$ The power consumption in the circuit is given by
A
$$P = \sqrt 2 {E_0}{I_0}$$
B
$$P = {{{E_0}{I_0}} \over {\sqrt 2 }}$$
C
$$P=zero$$
D
$$P = {{{E_0}{I_0}} \over 2}$$

Explanation

KEY CONCEPT : We know that power consumed in a.c. circuit is given by,

$$P = {E_{rms}}{I_{rms}}\cos \phi $$

Here, $$E = {E_0}\sin \omega t$$

$$I = {I_0}\sin \left( {\omega t - {\pi \over 2}} \right)$$

which implies that the phase difference, $$\phi = {\pi \over 2}$$

$$\therefore$$ $$P = {E_{rms}}.{I_{rms}}.\cos {\pi \over 2} = 0$$

$$\left( {\,\,} \right.$$ as $$\left. {\,\,\cos {\pi \over 2} = 0\,\,} \right)$$
2

AIEEE 2006

MCQ (Single Correct Answer)
The $$rms$$ value of the electric field of the light coming from the Sun is $$720$$ $$N/C.$$ The average total energy density of the electromagnetic wave is
A
$$4.58 \times {10^{ - 6}}\,J/{m^3}$$
B
$$6.37 \times {10^{ - 9}}\,J/{m^3}$$
C
$$81.35 \times {10^{ - 12}}\,J/{m^3}$$
D
$$3.3 \times {10^{ - 3}}\,J/{m^3}$$

Explanation

$${E_{rms}} = 720$$

The average total energy density

$$ = {1 \over 2}{ \in _0}\,E_0^2 = {1 \over 2}{ \in _0}{\left[ {\sqrt 2 {E_{rms}}} \right]^2} = { \in _0}\,E_{rms}^2$$

$$ = 8.85 \times {10^{ - 12}} \times {\left( {720} \right)^2}$$

$$ = 4.58 \times {10^{ - 6}}\,J/{m^3}$$
3

AIEEE 2006

MCQ (Single Correct Answer)
An inductor $$(L=100$$ $$mH)$$, a resistor $$\left( {R = 100\,\Omega } \right)$$ and a battery $$\left( {E = 100V} \right)$$ are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points $$A$$ and $$B$$. The current in the circuit $$1$$ $$ms$$ after the short circuit is
A
$$1/eA$$
B
$$eA$$
C
$$0.1$$ $$A$$
D
$$1$$ $$A$$

Explanation

Initially, when steady state is achieved,

$$i = {E \over R}$$

Let $$E$$ is short circuited at a $$t=0.$$ Then

At $$t=0,$$ $${i_0} = {E \over R}$$

Let during decay of current at any time the current

flowing is $$ - L{{di} \over {dt}} - iR = 0$$

$$ \Rightarrow {{di} \over i} = - {R \over L}dt$$

$$ \Rightarrow \int\limits_{{i_0}}^i {{{di} \over i}} = \int\limits_0^t { - {R \over L}dt} $$

$$ \Rightarrow {\log _e}{i \over {{i_0}}} = - {R \over L}t$$

$$ \Rightarrow i = {i_0}\,{e^{ - {R \over L}t}}$$

$$ \Rightarrow i = {E \over R}{e^{{R \over L}t}} = {{100} \over {100}}{e^{{{ - 100 \times {{10}^{ - 3}}} \over {100 \times {{10}^{ - 3}}}}}} = {1 \over e}$$
4

AIEEE 2006

MCQ (Single Correct Answer)
The flux linked with a coil at any instant $$'t'$$ is given by
$$\phi = 10{t^2} - 50t + 250$$
The induced $$emf$$ at $$t=3s$$ is
A
$$-190$$ $$V$$
B
$$-10$$ $$V$$
C
$$10$$ $$V$$
D
$$190$$ $$V$$

Explanation

$$\phi = 10{t^2} - 50t + 250$$

$$e = - {{d\phi } \over {dt}} = - \left( {20t - 50} \right)$$

$${e_{t = 3}} = - 10\,V$$

Questions Asked from Alternating Current and Electromagnetic Induction

On those following papers in MCQ (Single Correct Answer)
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