Magnetism Question 100
Question: The time period of a freely suspended magnet is 4 seconds. If it is broken in length into two equal parts and one part is suspended in the same way, then its time period will be
[NCERT 1984; CPMT 1991; MP PMT 1994; MH CET 2004]
Options:
A) 4 sec
B) 2 sec
C) 0.5 sec
D) 0.25 sec
Show Answer
Answer:
Correct Answer: B
Solution:
$ T=2\pi \sqrt{\frac{I}{MB _{H}}}=4\sec $ When magnet is cut into two equal halves, then New magnetic moment $ {M}’=\frac{M}{2} $ New moment of inertia $ {I}’=\frac{( \text{w}/2 ){{( l/2 )}^{2}}}{12}=\frac{1}{8}.\frac{\text{w}l^{2}}{12} $ Where w is the initial mass of the magnet But $ I=\frac{\text{w}l^{2}}{12};\ \ \
Therefore \ {I}’=\frac{I}{8} $
$ \
Therefore $ New time period $ {T}’=2\pi \sqrt{\frac{{{I}’}}{{M}‘B _{H}}} $
$ =2\pi \sqrt{\frac{I/8}{( M/2 )B _{H}}}=\frac{1}{2}2\pi \sqrt{\frac{I}{M _{H}}} $
$ =\frac{1}{2}\times T=\frac{1}{2}\times 4=2\sec $
BETA
