Question

A very long bar magnet of time period $T$ is cut into four equal parts by cutting it perpendicular to the length. Then the time period of one small part is :

• A. $T$
• B. $\frac{T}{16}$
• C. $\frac{T}{4}$
• D. $\frac{T}{8}$

Answer 1

Answer: (C)

$\frac{T}{4}$

Time period of vibration $T\propto \sqrt{\frac{I}{M}}$

where $I$ is the moment of inertia and $M$ is the magnetic moment.

$\therefore$ $\frac{T_1}{T_2}=\sqrt{\frac{I_1{M}_2}{I_2{M}_1}}$When the magnet is broken into 4 pieces, new magnetic moment of each piecebecomes ${\frac{1}{4}}^{th}$ of the initial magnetic moment.

$\therefore M_1=4M_2$

Moment of inertial also changes.

Initial moment of inertia $I_1=\frac{m{l}^2}{12}$

Mass of each piece $m_2=\frac{m}{4}$

Length of each piece $l_2=\frac{l}{4}$

Thus moment of inertia of each piece $I_2=\frac{\frac{m}{4}{\left(\frac{l}{4}\right)}^2}{12}$

$\therefore \frac{T_1}{T_2}=\sqrt{\frac{I_1{M}_2}{I_2\times 4M_2}}=\sqrt{\frac{1\times 64}{4}}=4$

$⟹T_2=T_1\times \frac{1}{4}=\frac{T_1}{4}=\frac{T}{4}$