Time period of Oscillation, T=2πMBI ⇒41=2πM×0.0499.8×10−6⇒161=4π2×M×49×10−39.8×10−6⇒M=49×10−34π2×9.8×10−6×16=494π2×9.8×16×10−3=12.8π2×10−3×10−2×102=1280π2×10−5Am2
NEET UG 2024 — Physics Electromagnetism
2024mcqmedium
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×10−5Am2, then the value of ' x ' is : 
Official previous-year question
Held on 30 Apr 2024 · Verified 9 Jul 2026.
Options
- A
128π2
- B
50π2
- C
1280π2
- D
5π2
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