Question

A magnetic needle of negligible breadth and thickness compared to its length, oscillates in a horizontal plane with a period $T$ . The period of oscillation of each part obtained on breaking the magnet into $n$ equal parts and perpendicular to the length is

• A. $T$
• B. $\frac{T}{n}$
• C. $nT$
• D. $\frac{1}{n}$

Answer 1

Answer: (B)

$T=2\pi \sqrt{\frac{I}{MB}}$
on breaking n equal parts perpendicular to length
$I\prime =\frac{1}{12}\left(\frac{m}{n}\right){\left(\frac{l}{n}\right)}^2$ of each part
$I\prime =\frac{I}{n^3},M\prime =\frac{M}{n}$
$\frac{T^{\prime }}{T}=\sqrt{\frac{I^{\prime }}{M^{\prime }}\times \frac{M}{I}}=\sqrt{\frac{1}{n^3}\times n}=\frac{1}{n}\Rightarrow T^{\prime }=\frac{T}{n}$