Question

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

• A. $\frac{2}{3} \, s$
• B. $\sqrt{\frac{2}{3} \, }s$
• C. $\frac{3}{2} \, s$
• D. $\sqrt{\frac{3}{2}} \, s$

Answer 1

Answer: (A)

The time period of oscillations of the magnet
$T=2\pi \sqrt{\frac{I}{M H}}$
Where, $I=$ moment of inertia of magnet
$= \, \frac{m l^{2}}{12}$ (m, being the mass of magnet)
$M=pole \, strength\times L$
And $H=$ horizontal component of Earth's magnetic field
When then three equal parts of magnet are placed on one another with their like poles together, then
$I^{′}=\frac{1}{12}\left(\frac{m}{3}\right) \, \left(\frac{L}{3}\right)^{2}\times 3=\frac{1}{12}\frac{m L^{2}}{9}=\frac{I}{9}$
And $M^{′}=Pole \, strength\times \frac{L}{3}\times 3=M \, $
Hence, $T=2\pi \sqrt{\left(\frac{\frac{I}{9}}{M H}\right)}\Rightarrow T^{′}=\frac{1}{3}\times T$
$T^{′}=\frac{2}{3} \, s$