The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is 2s . The magnent is cut into three equal parts of length (L)/(3) each,and three parts

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Correct answer: B. (2)/(3)s

The time period of oscillations of magnet

T = 2\pi\sqrt{\left(\dfrac{I}{M{B}_{H}}\right)} \ldots \left(i\right)

Where

I =

moment of inertia of magnet

= \dfrac{m{L}^{2}}{12}

(m, being the mass of magnet)

Now magnet is cut into three equal parts.

When the three equal parts of magnet are placed on one another with their like poles together, then

{I}^{′} = \dfrac{1}{12}\left(\dfrac{m}{3}\right) \times {\left(\dfrac{L}{3}\right)}^{2} \times 3

= \dfrac{1}{12}\dfrac{m{L}^{2}}{9} = \dfrac{I}{9}

and

{M}^{′} =

pole strength

\times \dfrac{L}{3} \times 3 = M

Hence,

⇒ {T}^{′} = \dfrac{1}{3} \times T

{T}^{′} = \dfrac{2}{3}s

The length of a magnet is large compared to its width and breadth....

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