The period of an oscillation of a freely suspended magnetic needle does not depend upon
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Length of suspension wire
When a freely suspended magnetic needle is turned from its rest direction, the earth’s field exerts a restoring torque \( mB\sin\theta \), and the needle oscillates about that direction. For small oscillations its period is
\( \displaystyle T = 2\pi\sqrt{\frac{I}{mB}} \)
where I is the moment of inertia of the needle (set by its mass and dimensions), m is its dipole moment and B is the earth’s field. The suspension wire only supports the needle; its length does not enter the formula, so the period does not depend on the length of the suspension wire.
Note: A similar question was also asked in CUET (UG) 2023 on 05-06-2023 (Slot 3).
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
The Dipole in a Uniform Magnetic Field – “… This is done by placing a small compass needle of known magnetic moment m and moment of inertia \( \mathcal{I} \) and allowing it to oscillate in the magnetic field.”
The Dipole in a Uniform Magnetic Field – “… This represents a simple harmonic motion. The square of the angular frequency is \( \omega^2 = mB/\mathcal{I} \) and the time period is,
\( \displaystyle T = 2\pi\sqrt{\frac{\mathcal{I}}{mB}} \qquad (5.4) \)”