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1

JEE Main 2021 (Online) 27th August Morning Shift

Numerical
The alternating current is given by $$i = \left\{ {\sqrt {42} \sin \left( {{{2\pi } \over T}t} \right) + 10} \right\}A$$

The r.m.s. value of of this current is ................. A.
Your Input ________

Answer

Correct Answer is 11

Explanation

$$f_{rms}^2 = f_{1\,rms}^2 + f_{2\,rms}^2$$

$$ = {\left( {{{\sqrt {42} } \over {\sqrt 2 }}} \right)^2} + {10^2}$$

$$ = 121 \Rightarrow {f_{rms}}$$ = 11 A
2

JEE Main 2021 (Online) 27th July Evening Shift

Numerical
In the given figure the magnetic flux through the loop increases according to the relation $$\phi$$B(t) = 10t2 + 20t, where $$\phi$$B is in milliwebers and t is in seconds.

The magnitude of current through R = 2$$\Omega$$ resistor at t = 5 s is ___________ mA.

Your Input ________

Answer

Correct Answer is 60

Explanation

$$\left| \in \right| = {{d\phi } \over {dt}}$$ = 20t + 20 mV

$$\left| i \right| = {{\left| \in \right|} \over R}$$ = 10t + 10 mA

at t = 5

$$\left| i \right|$$ = 60 mA
3

JEE Main 2021 (Online) 27th July Morning Shift

Numerical
Consider an electrical circuit containing a two way switch 'S'. Initially S is open and then T1 is connected to T2. As the current in R = 6$$\Omega$$ attains a maximum value of steady state level, T1 is disconnected from T2 and immediately connected to T3. Potential drop across r = 3$$\Omega$$ resistor immediately after T1 is connected to T3 is __________ V. (Round off to the Nearest Integer)

Your Input ________

Answer

Correct Answer is 3

Explanation

What T1 and T2 are connected, then the steady state current in the inductor $$I = {6 \over 6} = 1A$$

When T1 and T3 are connected then current through inductor remains same. So potential difference across 3$$\Omega$$

V = Ir = 1 $$\times$$ 3 = 3 Volt
4

JEE Main 2021 (Online) 25th July Evening Shift

Numerical
Two circuits are shown in the figure (a) & (b). At a frequency of ____________ rad/s the average power dissipated in one cycle will be same in both the circuits.

Your Input ________

Answer

Correct Answer is 500

Explanation

For figure (a)

$${P_{avg}} = {{v_{rms}^2} \over R}$$

$${{v_{rms}^2} \over {{Z^2}}} \times R = {{v_{rms}^2} \over R} \times 1$$

$${R^2} = {Z^2}$$

$$25 = {\left( {\sqrt {{{({x_C} - {x_L})}^2} + {5^2}} } \right)^2}$$

$$ = 25{({x_C} - {x_L})^2} + 25$$

$${x_C} = {x_L} \Rightarrow {1 \over {\omega C}} = \omega L$$

$${\omega ^2} = {1 \over {LC}} = {{{{10}^6}} \over {0.1 \times 40}}$$

$$\omega = 500$$

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