Question

Time period for a magnet is T. If it is divided in four equal parts along its axis and perpendicular to its axis as shown then time period for each part will be

• A. $4 T$
• B.
• C.
• D. $T$

Answer 1

Answer: (C)

Answer 2

When magnet of length l is cut into four equal parts. then $l ^ { \prime } = \frac { l } { 2 } ; \therefore M ^ { \prime } = \frac { m } { 2 } \times \frac { l } { 2 } = \frac { m l } { 4 } = \frac { M } { 4 }$

New moment of inertia $I ^ { \prime } = \frac { \mathrm { w } l ^ { 2 } } { 12 } = \frac { \frac { \mathrm { w } } { 4 } \cdot \left( \frac { 1 } { 2 } \right) ^ { 2 } } { 12 } = \frac { 1 } { 16 } \cdot \frac { \mathrm { w } l ^ { 2 } } { 12 }$

Here w is the mass of magnet.

$\therefore I ^ { \prime } = \frac { 1 } { 16 } I$; Time period of each part $T ^ { \prime } = 2 \pi \sqrt { \frac { I ^ { \prime } } { M ^ { \prime } B _ { H } } }$

$= 2 \pi \sqrt { \frac { I / 16 } { ( M / 4 ) B _ { H } } } = 2 \pi \sqrt { \frac { I } { 4 M B _ { H } } } = \frac { T } { 2 }$