Question

A magnetic needle has negligible thickness and breadth as compared to its length. It oscillates in a horizontal plane with a period T. The period of oscillation of each part obtained on breaking magnet into n parts which are equal and perpendicular to the length is
(A) $T$
(B) $\dfrac{T}{n}$
(C) $Tn$
(D) $\dfrac{1}{{Tn}}$

Answer 1

Hint Divide the mass of magnetic needle into $n$ equal parts.
$M = \dfrac{{{M_0}}}{n}$
Similarly, divide its length into $n$equal parts.
Also, use the formula:
$T = 2\pi \sqrt {\dfrac{I}{{mB}}} $

Complete step-by-step answer:
A magnetic needle has negligible breadth and thickness. This needle oscillates in the horizontal plane.
Let the mass of needle be $M$and length be $L$
According to the question, it is given that magnetic needle is broken into $n$ equal parts
Mass$ = M = \dfrac{{M'}}{n} \cdots (i)$
Length$ = L = \dfrac{{L'}}{n} \cdots (ii)$
We know that,
$I = \dfrac{{M{L^2}}}{{\sqrt 2 }}$
Putting the value of $M$and $L$from equation $(i)$and $(ii)$respectively, we get
\[
  I' = \dfrac{1}{{\sqrt 2 }} \times \dfrac{{M'}}{n} \times \dfrac{{\left( {L{'^2}} \right)}}{n} \\
  I' = \dfrac{{M'L{'^2}}}{{{n^2}\sqrt 2 }} \\
  I' = \dfrac{I}{{{n^2}}} \\
 \]
Here, $I$is the moment of inertia
We know that,
$T = 2\pi \sqrt {\dfrac{I}{{mB}}} $
$\therefore T' = 2\pi \sqrt {\dfrac{{I'}}{{m'B}}} $
where, $m = $Magnetic moment
$
  T' = 2\pi \sqrt {\dfrac{I}{{{n^3}mB}}} \Rightarrow \dfrac{1}{n}T \\
  T' = \dfrac{T}{n} \\
 $

Therefore, option (B) is the correct answer.

Note The magnetic has negligible breadth and thickness that’s why the breadth and thickness were not used in finding the perpendicular to the length.
When something divides other into equal parts, we write them as:
$\dfrac{D}{P}$ (where, D is the quantity which is cut into parts and P is the number in which D is cut)