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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
In the circuit shown here, the point $$'C'$$ is kept connected to point $$'A'$$ till the current flowing through the circuit becomes constant. Afterward, suddenly, point $$'C'$$ is disconnected from point $$'A'$$ and connected to point $$'B'$$ at time $$t=0.$$ Ratio of the voltage across resistance and the inductor at $$t=L/R$$ will be equal to :
A
$${e \over {1 - e}}$$
B
$$1$$
C
$$-1$$
D
$${{1 - e} \over e}$$

Explanation

Applying kirchhoffs law of voltage in closed loop

$$ - {V_R} - {V_C} = 0 \Rightarrow {{{V_R}} \over {{V_C}}} = - 1$$

2

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
The magnetic field in a travelling electromagnetic wave has a peak value of $$20$$ $$n$$$$T$$. The peak value of electric field strength is :
A
$$3V/m$$
B
$$6V/m$$
C
$$9V/m$$
D
$$12V/m$$

Explanation

From question,

$${B_0} = 20nT = 20 \times {10^{ - 9}}T$$

( as velocity of light in vacuum $$C = 3 \times {10^8}\,\,m{s^{ - 1}}$$ )

$${\overrightarrow E _0} = {\overrightarrow B _0} \times \overrightarrow C $$

$$\left| {{{\overrightarrow E }_0}} \right| = \left| {\overrightarrow B } \right|.\left| {\overrightarrow C } \right| = 20 \times {10^{ - 9}} \times 3 \times {10^8}$$

$$ = 6\,\,V/m.$$
3

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
In an $$LCR$$ circuit as shown below both switches are open initially. Now switch $${S_1}$$ is closed, $${S_2}$$ kept open. ($$q$$ is charge on the capacitor and $$\tau $$ $$=RC$$ is Capacitance time constant). Which of the following statement is correct ?
A
Work done by the battery is half of the energy dissipated in the resistor
B
$$t = \,\tau ,\,q = CV/2$$
C
At $$t = \,2\tau ,\,q = CV\left( {1 - {e^{ - 2}}} \right)$$
D
At $$t = \,2\tau ,\,q = CV\left( {1 - {e^{ - 1}}} \right)$$

Explanation

Charge on he capacitor at any time $$t$$ is given by

$$q = CV\left( {1 - {e^{t/\tau }}} \right)$$

at $$t = 2\tau $$

$$q = CV\left( {1 - {e^{ - 2}}} \right)$$
4

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
A metallic rod of length $$'\ell '$$ is tied to a string of length $$2$$$$\ell $$ and made to rotate with angular speed $$w$$ 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
A
$${{2B\omega \ell } \over 2}$$
B
$${{3B\omega \ell } \over 2}$$
C
$${{4B\omega {\ell ^2}} \over 2}$$
D
$${{5B\omega {\ell ^2}} \over 2}$$

Explanation

Here, induced $$e.m.f.$$



$$e = \int\limits_{2\ell }^{3\ell } {\left( {\omega x} \right)Bdx = B\omega } {{\left[ {{{\left( {3\ell } \right)}^2} - {{\left( {2\ell } \right)}^2}} \right]} \over 2}$$

$$ = {{5B{\ell ^2}\omega } \over 2}$$

Questions Asked from Alternating Current and Electromagnetic Induction

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