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JEE Mains Previous Years Questions with Solutions

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MCQ (Single Correct Answer)

For the given circuit the current i through the battery when the key in closed and the steady state has been reached is __________.

6 A

25 A

10 A

0 A

Explanation

We know in study state potential difference across inductor = 0

So, equivalent circuit becomes

$${1 \over {{R_{eq}}}} = {1 \over 3} + {1 \over 3} + {1 \over 3} = 1$$

$$\Rightarrow$$ R_eq = 1$$\Omega$$

$$\Rightarrow$$ Circuit becomes

$$ \Rightarrow i = {{30} \over 3} = 10$$ A

JEE Main 2021 (Online) 31st August Evening Shift

MCQ (Single Correct Answer)

Statement - I :

To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect a capacitor across the output parallel to the load R_L.

Statement - II :

To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect an inductor in series with R_L.

In the light of the above statements, choose the most appropriate answer from the options given below :

Statement I is true but Statement II is false

Statement I is false but Statement II is true

Both Statement I and Statement II are false

Both Statement I and Statement II are true

Explanation

To convert pulsating dc into steady dc both of mentioned method are correct.

JEE Main 2021 (Online) 31st August Morning Shift

MCQ (Single Correct Answer)

In an ac circuit, an inductor, a capacitor and a resistor are connected in series with X_L = R = X_C. Impedance of this circuit is :

2R²

Zero

R$$\sqrt 2 $$

Explanation

$$Z = \sqrt {{{({X_L} - {X_C})}^2} + {R^2}} = R$$ $$\because$$ X_L = X_C

Option (c)

JEE Main 2021 (Online) 31st August Morning Shift

MCQ (Single Correct Answer)

A small square loop of side 'a' and one turn is placed inside a larger square loop of side b and one turn (b >> a). The two loops are coplanar with their centres coinciding. If a current I is passed in the square loop of side 'b', then the coefficient of mutual inductance between the two loops is :

$${{{\mu _0}} \over {4\pi }}8\sqrt 2 {{{a^2}} \over b}$$

$${{{\mu _0}} \over {4\pi }}{{8\sqrt 2 } \over a}$$

$${{{\mu _0}} \over {4\pi }}8\sqrt 2 {{{b^2}} \over a}$$

$${{{\mu _0}} \over {4\pi }}{{8\sqrt 2 } \over b}$$

Explanation

$$B = \left[ {{{{\mu _0}} \over {4\pi }}{I \over {b/2}} \times 2\sin 45} \right] \times 4$$

$$\phi = 2\sqrt 2 {{{\mu _0}} \over \pi }{I \over b} \times {a^2}$$

$$\therefore$$ $$M = {\phi \over I} = {{2\sqrt 2 {\mu _0}{a^2}} \over {\pi b}} = {{{\mu _0}} \over {4\pi }}8\sqrt 2 {{{a^2}} \over b}$$

Option (a)