 JEE Mains Previous Years Questions with Solutions

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JEE Main 2014 (Offline)

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

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

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