Question

A thin magnetic needle oscillates in a horizontal plane with a period T. It is broken into n equal parts. The time period of each part will be

• A. nT
• B. n2T
• C. $\frac { \mathrm { T } } { \mathrm { n } }$
• D.

Answer 1

Answer: (C)

$\frac { \mathrm { T } } { \mathrm { n } }$

Answer 2

: Since for each part of magnetic needle

mass , $\left( \mathrm { m } ^ { \prime } \right) = \frac { \mathrm { m } } { \mathrm { n } }$ length,

Then moment of intertia

$\left( l ^ { \prime } \right) = \frac { \operatorname { mass } \times ( \text { length } ) ^ { 2 } } { 12 }$

$= \frac { 1 } { 12 } \frac { \mathrm { m } } { \mathrm { n } } \left( \frac { \mathrm { l } } { \mathrm { n } } \right) ^ { 2 } = \frac { \mathrm { ml } ^ { 2 } } { 12 } \cdot \frac { 1 } { \mathrm { n } ^ { 3 } } = \frac { \mathrm { I } ^ { \prime } } { \mathrm { n } ^ { 3 } }$

Now magnetization, (M) = pole strength ×length=M/n the time period of each particle,