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The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is 2 s. The magnet is cut along its length into three equal parts and three parts are then placed on each other with their like poles together. The time period of this combination will be

  • Option 1)

    2 s

  • Option 2)

    \frac{2}{3}\; s\;

  • Option 3)

    \; \; \; (2\sqrt{3})s\; \;

  • Option 4)

    \; \left ( \frac{2}{\sqrt{3}} \right )s


Answers (1)


As we learnt in

Time Period of Freely Suspended Magnet -

T= 2\pi \sqrt{\frac{I}{MB}}

- wherein

I - Moment of Inertia



For a vibrating magnet,   T=2\pi \sqrt{\frac{I}{MB}}   where  I=ml^{2}/12,\: M=xl,\; x=pole\; strength\; of\; magnet

I'=\left ( \frac{m}{3} \right )\left ( \frac{l}{3} \right )^{2}\times \frac{3}{12}=\frac{ml^{2}}{9\times 12}=\frac{I}{9}       (For three pieces together)

M'=(x)\left ( \frac{l}{3} \right )\times 3=xl=M       (For three pieces together)

\therefore \; \; \; T'=2\pi \sqrt{\frac{I'}{M'B}}=2\pi \sqrt{\frac{I/9}{MB}}

=\frac{1}{3}\times 2\pi \sqrt{\frac{I}{MB}}=\frac{T}{3}

\therefore \; \; \; T'=\frac{T}{3}=\frac{2}{3}sec

Option 1)

2 s


Option 2)

\frac{2}{3}\; s\;


Option 3)

\; \; \; (2\sqrt{3})s\; \;


Option 4)

\; \left ( \frac{2}{\sqrt{3}} \right )s


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