Magnetism Question 81
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
Options:
A) $ 4T $
B) $ T/4 $
C) $ T/2 $
D) $ T $
Show Answer
Answer:
Correct Answer: C
Solution:
When magnet of length l is cut into four equal parts. then $ {m}’=\frac{m}{2} $ and $ {l}’=\frac{l}{2};\ \ \
Therefore \ {M}’=\frac{m}{2}\times \frac{l}{2}=\frac{ml}{4}=\frac{M}{4} $ New moment of inertia $ {I}’=\frac{\text{w}l^{2}}{12}=\frac{\frac{\text{w}}{4}.{{( \frac{1}{2} )}^{2}}}{12}=\frac{1}{16}.\frac{\text{w}l^{2}}{12} $ Here w is the mass of magnet.
$ \
Therefore ,{I}’=\frac{1}{16}I $ ; Time period of each part $ {T}’=2\pi \sqrt{\frac{{{I}’}}{{M}‘B _{H}}} $
$ =2\pi \sqrt{\frac{I/16}{(M/4)B _{H}}}=2\pi \sqrt{\frac{I}{4MB _{H}}}=\frac{T}{2} $
BETA
