Q147 / 160 · TS EAMCET 2008 · Physics

With a standard rectangular bar magnet the time period of a vibration magnetometer is $4 \mathrm{~s}$. The bar magnet is cut parallel to its length into four equal pieces. The time period of vibration magnetometer when one piece is used (in second) (bar magnet breadth is small) is

  • A
    $16$
  • B
    $8$
  • Correct answer
    C
    $4$
  • D
    $2$

Solution

Time period of magnet, $T=2 \pi \sqrt{\frac{I}{M B}}$ When magnet is cut parallel to its length into four equal pieces. Then new magnetic moment, $M^{\prime}=\frac{M}{4}$ New moment of inertia, $I^{\prime}=\frac{I}{4}$ $\therefore \quad$ New time period, $T^{\prime}=2 \pi \sqrt{\frac{I^{\prime}}{M^{\prime} B}}$ $\Rightarrow \quad T=T^{\prime}=4 \mathrm{~s}$

Hence, the correct answer is option: 'C'.

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