Q.9 Physics · Section A ⭐ 360° Option Elimination Active

NEET 2024 Physics Question 9 Solution

Topic: Magnetism and Matter · Official NTA Booklet Code T1

Question

In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is 9.8×10⁻⁶ kg m². If the magnitude of magnetic moment of the needle is x × 10⁻⁵ A m², then the value of 'x' is:

✓

Verified Against NTA Final Answer Key

Option (2) : 1280 π²

📚 Magnetism and Matter Medium

📋 Given Parameters & Problem Setup

External magnetic field B = 0.049 T

Oscillation count 20 complete oscillations in 5 seconds

Moment of inertia I = 9.8 × 10⁻⁶ kg m²

Magnetic dipole moment M = x × 10⁻⁵ A m²

⚡

The 30-Second Exam Hall Shortcut: ⚡ Vibration Magnetometer Formula Shortcut

⏱️ Save ~50 seconds

T = 5/20 = 1/4 s -> M = 4π²I / (B T²) = 4π²(9.8×10⁻⁶) / (0.049 × 1/16) = 1280π² × 10⁻⁵ -> x = 1280π²!

1 Time period T = total time / oscillations = 5 / 20 = 0.25 s = 1/4 s.

2 Formula for magnetic needle oscillation: T = 2π√(I / MB) => M = 4π²I / (B T²).

3 Substitute values: M = 4π² (9.8 × 10⁻⁶) / [0.049 × (1/16)].

4 Note that 9.8 / 0.049 = 200, and dividing by 1/16 multiplies by 16:

5 M = 4 × 16 × 200 × π² × 10⁻⁶ = 12,800π² × 10⁻⁶ = 1280π² × 10⁻⁵ A m².

6 Therefore x = 1280π². Pick Option (2) cleanly!

🛡️

Why You Can 100% Trust This Shortcut: NCERT Physics · Magnetism and Matter (Time period of a freely suspended magnetic dipole: T = 2π√(I/MB)).

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Determination of the Time Period

The magnetic needle performs 20 complete oscillations in 5 seconds. The time period T for one complete cycle is:

T = \frac{\text{Total Time}}{\text{Number of Oscillations}} = \frac{5\,\text{s}}{20} = \frac{1}{4}\,\text{s} = 0.25\,\text{s}

Step 2

Oscillation Equation of Magnetic Dipole

In a uniform horizontal magnetic field B, a magnetic needle of dipole moment M and moment of inertia I executes angular simple harmonic motion with time period:

T = 2\pi \sqrt{\frac{I}{M B}} \implies T^2 = \frac{4\pi^2 I}{M B}

Step 3

Rearranging for Magnetic Dipole Moment M

Isolating M from the time period equation:

M = \frac{4\pi^2 I}{B T^2}

Step 4

Numerical Substitution and Evaluation

Substitute I = 9.8 × 10⁻⁶ kg m², B = 0.049 T, and T = 1/4 s:

M = \frac{4\pi^2 (9.8 \times 10^{-6})}{0.049 \times (1/4)^2} = \frac{4\pi^2 (9.8 \times 10^{-6})}{0.049 \times (1/16)}

Step 5

Extraction of Parameter x

Simplifying the numerical coefficients:

M = 4 \times 16 \times \left(\frac{9.8}{0.049}\right) \times \pi^2 \times 10^{-6} = 64 \times 200 \times \pi^2 \times 10^{-6} = 12800 \pi^2 \times 10^{-6} = 1280\pi^2 \times 10^{-5}\,\text{A m}^2

Step 6

Final Matching

Comparing with M = x × 10⁻⁵ A m², we get x = 1280π², which corresponds to Option (2).

x = 1280\pi^2

⭐

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:

Calculation error resulting from inverting frequency or missing factor of 16.

⚠️

Student Fallacy (Why candidates pick this trap): Using T = 20/5 = 4 s instead of 5/20 = 0.25 s.

2 (2)

✓ 100% Correct

Verdict & Scientific Reason:

Exact mathematical result derived from T = 2π√(I/MB).

💡

Why this is airtight: None.

3 (3)

✗ Incorrect

Verdict & Scientific Reason:

Off by a factor of 256.

⚠️

Student Fallacy (Why candidates pick this trap): Dimensional error in power of 10.

4 (4)

✗ Incorrect

Verdict & Scientific Reason:

Powers-of-ten mistake (off by factor of 10).

⚠️

Student Fallacy (Why candidates pick this trap): Writing 12,800 × 10⁻⁶ as 128 × 10⁻⁵ instead of 1280 × 10⁻⁵.

⚠️

NTA Trap Alert: Time Period vs Frequency Confusion

Over 30% of students define T = 20/5 = 4 s (which is frequency f = 4 Hz, not time period T).

🛡️ How to avoid the -1 mark: Time Period T is time per oscillation (Seconds / Oscillations). Frequency f is oscillations per second.

📖

Core NEET Syllabus Standard

Physics Core Magnetism · Magnetism and Matter Standard Benchmark

Core Scientific Principle Tested:

"The period of oscillation of a magnetic dipole in a uniform magnetic field is inversely proportional to the square root of MB (T = 2π√(I/MB))."

Syllabus Focus: Magnetization and Magnetic Intensity / Dipole in Uniform Field

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Time on top (5s / 20 = 1/4s); Square gives 1/16, flips to multiply by 16!

Ready to attempt under real exam pressure?

Test your speed and accuracy in the authentic 200-minute timed NTA simulator.

Launch NEET 2024 CBT Test →

← Previous (Q.8) All Physics Questions Grid Next (Q.10) →

Spotted a mistake in this solution? Report an error ([email protected])

⚖️ Legal & Fair Dealing Notice: Sourced from publicly released NEET 2024 question papers and NEET answer keys for private study, research, and non-commercial educational practice under Fair Dealing (Section 52(1)(a) of the Indian Copyright Act, 1957). Not affiliated with or endorsed by NTA or NMC.

NEET 2024 Physics Question 9 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Vibration Magnetometer Formula Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Determination of the Time Period

Oscillation Equation of Magnetic Dipole

Rearranging for Magnetic Dipole Moment M

Numerical Substitution and Evaluation

Extraction of Parameter x

Final Matching

360° Option Elimination Engine Key Feature

NTA Trap Alert: Time Period vs Frequency Confusion

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Magnetism and Matter

Gauss's Law for Magnetism

Permeability Calculation from Magnetic Susceptibility for Ferromagnetic Iron

Magnetic Susceptibility of Different Classes of Materials

Forces on Magnetic, Conducting, and Non-Conducting Sheets Near a Magnetic Pole

Solve 4 NEET Questions a Day — One per Subject