1
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
A series L-R circuit is connected to a battery of emf V. If the circuit is switched on at t = 0, then the time at which the energy stored in the inductor reaches $$\left( {{1 \over n}} \right)$$ times of its maximum value, is :
A
$${L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n + 1}}} \right)$$
B
$${L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n - 1}}} \right)$$
C
$${L \over R}\ln \left( {{{\sqrt n + 1} \over {\sqrt n - 1}}} \right)$$
D
$${L \over R}\ln \left( {{{\sqrt n - 1} \over {\sqrt n }}} \right)$$
2
JEE Main 2020 (Online) 4th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Choose the correct option relating wave lengths of different parts of electromagnetic wave spectrum:
A
$$\lambda $$radio waves > $$\lambda $$micro waves > $$\lambda $$visible > $$\lambda $$x-rays
B
$$\lambda $$visible > $$\lambda $$x-rays > $$\lambda $$radio waves > $$\lambda $$micro waves
C
$$\lambda $$visible < $$\lambda $$micro waves < $$\lambda $$radio waves < $$\lambda $$x-rays
D
$$\lambda $$x-rays < $$\lambda $$micro waves < $$\lambda $$radio waves < $$\lambda $$visible
3
JEE Main 2020 (Online) 3rd September Evening Slot
MCQ (Single Correct Answer)
+4
-1
The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is
$$\overrightarrow E = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$$.
The magnetic field $$\overrightarrow B $$ , at the moment t = 0 is :
A
$$\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat j$$
B
$$\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat k$$
C
$$\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat k$$
D
$$\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat j$$
4
JEE Main 2020 (Online) 3rd September Morning Slot
MCQ (Single Correct Answer)
+4
-1
The magnetic field of a plane electromagnetic wave is
$$\overrightarrow B = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat i$$ T
where c = 3 $$ \times $$ 108 ms–1 is the speed of light. The corresponding electric field is :
A
$$\overrightarrow E = - {10^{ - 6}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m
B
$$\overrightarrow E = - 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m
C
$$\overrightarrow E = 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m
D
$$\overrightarrow E = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$
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