1
JEE Main 2025 (Online) 29th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged.

Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor $\left(R / \sqrt{R^2+\omega^2 L^2}\right)$, where $\omega$ is frequency of the supply across resistor $R$ and inductor $L$. If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage.

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

A

(A) is true but (R) is false

B

Both $(\mathbf{A})$ and $(\mathbf{R})$ are true and $(\mathbf{R})$ is the correct explanation of $(\mathbf{A})$ 3.

C

Both $(\mathbf{A})$ and $(\mathbf{R})$ are true but $(\mathbf{R})$ is not the correct explanation of $(\mathbf{A})$

D

(A) is false but (R) is true

2
JEE Main 2025 (Online) 28th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
JEE Main 2025 (Online) 28th January Evening Shift Physics - Alternating Current Question 21 English

In the circuit shown here, assuming threshold voltage of diode is negligibly small, then voltage $ V_{AB} $ is correctly represented by:

A
JEE Main 2025 (Online) 28th January Evening Shift Physics - Alternating Current Question 21 English Option 1
B
JEE Main 2025 (Online) 28th January Evening Shift Physics - Alternating Current Question 21 English Option 2
C

$ V_{AB} $ would be zero at all times.

D
JEE Main 2025 (Online) 28th January Evening Shift Physics - Alternating Current Question 21 English Option 4
3
JEE Main 2025 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

An alternating current is given by $\mathrm{I}=\mathrm{I}_{\mathrm{A}} \sin \omega \mathrm{t}+\mathrm{I}_{\mathrm{B}} \cos \omega \mathrm{t}$. The r.m.s current will be

A
$\frac{\sqrt{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}}{2}$
B
$\sqrt{\frac{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}{2}}$
C
$\frac{\left|\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}\right|}{\sqrt{2}}$
D
$\sqrt{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}$
4
JEE Main 2025 (Online) 22nd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is $I_0$. If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be

A
$\frac{\mathrm{I}_0}{2}$
B
$\frac{\mathrm{I}_0}{\sqrt{2}}$
C
$2 \mathrm{I}_0$
D
$\mathrm{I_0}$

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