Q.40 Physics ⭐ 360° Option Elimination Active

NEET 2021 Physics Question 40 Solution

Topic: Alternating Current · Official NTA Booklet Code M1

Question

A series LCR circuit containing 5.0 H inductor, 80 μF capacitor and 40 Ω resistor is connected to 230 V variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be:

✓

Verified Against NTA Final Answer Key

Option (3) : 46 rad/s and 54 rad/s

📚 Alternating Current Medium

📋 Given Parameters & Problem Setup

Inductance (L) 5.0 H

Capacitance (C) 80 μF = 80 × 10^-6 F

Resistance (R) 40 Ω

RMS Voltage 230 V

⚡

The 30-Second Exam Hall Shortcut: ⚡ Half-Power Frequencies: ω0 ± R/(2L)

⏱️ Save ~25 seconds

Resonant frequency $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5 \times 80 \times 10^{-6}}} = \frac{1}{0.02} = 50\text{ rad/s}$ . Half bandwidth $\Delta\omega = \frac{R}{2L} = \frac{40}{2 \times 5} = 4\text{ rad/s}$ . Half-power frequencies are $\omega = 50 \pm 4 = 46\text{ rad/s}$ and $54\text{ rad/s}$ ! Match Option (3)!

1 Compute resonant angular frequency:

\omega_0 = \frac{1}{\sqrt{L C}} = \frac{1}{\sqrt{400 \times 10^{-6}}} = \frac{1}{20 \times 10^{-3}} = 50\text{ rad/s}

2 Compute half-bandwidth:

\Delta\omega = \frac{R}{2L} = \frac{40\ \Omega}{2(5.0\text{ H})} = 4\text{ rad/s}

3 Lower half-power frequency:

\omega_1 = \omega_0 - \Delta\omega = 50 - 4 = 46\text{ rad/s}

4 Upper half-power frequency:

\omega_2 = \omega_0 + \Delta\omega = 50 + 4 = 54\text{ rad/s}

5 Match Option (3).

🛡️

Why You Can 100% Trust This Shortcut: Canonical series LCR resonance half-power bandwidth definitions from NCERT Alternating Current.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Definition of Half-Power Frequencies in LCR Circuits

In an LCR series circuit, power dissipated is $P = I_{\text{rms}}^2 R$ . At resonance ( $\omega = \omega_0$ ), power is maximum: $P_{\text{max}} = \frac{V^2}{R}$ . The half-power frequencies $\omega_1$ and $\omega_2$ are the frequencies where power drops to half of maximum ( $P = P_{\text{max}} / 2$ ):

\omega_1 = \omega_0 - \Delta\omega, \quad \omega_2 = \omega_0 + \Delta\omega, \quad \text{where } \Delta\omega = \frac{R}{2L}

Step 2

Calculating the Resonant Angular Frequency (ω0)

Substitute $L = 5.0\text{ H}$ and $C = 80\ \mu\text{F} = 80 \times 10^{-6}\text{ F}$ :

\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5.0 \times 80 \times 10^{-6}}} = \frac{1}{\sqrt{400 \times 10^{-6}}} = \frac{1}{2 \times 10^{-2}} = 50\text{ rad/s}

Step 3

Calculating the Half-Bandwidth (Δω)

Substitute $R = 40\ \Omega$ and $L = 5.0\text{ H}$ :

\Delta\omega = \frac{R}{2L} = \frac{40\ \Omega}{2 \times 5.0\text{ H}} = \frac{40}{10} = 4\text{ rad/s}

Step 4

Computing the Half-Power Angular Frequencies

Evaluate the two frequencies:

\omega_1 = 50 - 4 = 46\text{ rad/s}, \quad \omega_2 = 50 + 4 = 54\text{ rad/s}

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 1

✗ Incorrect

Verdict & Scientific Reason:

Uses an arbitrary bandwidth of $\pm 25\text{ rad/s}$ .

⚠️

Student Fallacy (Why candidates pick this trap): Guessing wide bandwidth.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Contains the resonant frequency itself (50 rad/s), where power is maximum, not half-maximum.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing resonant frequency with half-power points.

3 3

✓ 100% Correct

Verdict & Scientific Reason:

Exact half-power frequencies derived from $\omega_0 \pm \frac{R}{2L} = 50 \pm 4\text{ rad/s}$ .

💡

Why this is airtight: None; textbook bandwidth calculation.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Result of using full bandwidth $R/L = 8$ instead of half-bandwidth $R/2L = 4$ ( $50 \pm 8$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Using

\Delta\omega = R/L

instead of

R/2L

⚠️

NTA Trap Alert: Full Bandwidth vs. Half Bandwidth Trap

The full bandwidth is $2\Delta\omega = \frac{R}{L} = 8\text{ rad/s}$ . Students who add and subtract the full bandwidth ( $50 \pm 8$ ) obtain $42\text{ rad/s}$ and $58\text{ rad/s}$ (Option 4). The shift on either side of $\omega_0$ is the HALF-bandwidth $\Delta\omega = \frac{R}{2L} = 4\text{ rad/s}$ .

🛡️ How to avoid the -1 mark:

\omega_1 = \omega_0 - \frac{R}{2L}

and

\omega_2 = \omega_0 + \frac{R}{2L}

. Divide

R/L

by 2!

📖

Core NEET Syllabus Standard

NCERT Physics · Alternating Current Standard Benchmark

Core Scientific Principle Tested:

"The frequencies at which power is half the resonant value are $\omega = \omega_0 \pm \Delta\omega$ , where $\Delta\omega = \frac{R}{2L}$ ."

Syllabus Focus: Sharpness of Resonance and Bandwidth

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Physics Question 40 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Half-Power Frequencies: ω0 ± R/(2L)

Step-by-Step Rigorous Workflow & Mathematical Proof

Definition of Half-Power Frequencies in LCR Circuits

Calculating the Resonant Angular Frequency (ω0)

Calculating the Half-Bandwidth (Δω)

Computing the Half-Power Angular Frequencies

360° Option Elimination Engine Key Feature

NTA Trap Alert: Full Bandwidth vs. Half Bandwidth Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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