Magnetism Question 112
Question: The period of oscillations of a magnetic needle in a magnetic field is 1.0 sec. If the length of the needle is halved by cutting it, the time period will be
[MP PMT/PET 1998]
Options:
A) 1.0 sec
B) 0.5 sec
C) 0.25 sec
D) 2.0 sec
Show Answer
Answer:
Correct Answer: B
Solution:
$ T=2\pi \sqrt{\frac{I}{MB}}=2\pi \sqrt{\frac{\text{w}l^{2}/12}{\text{Pole}\ \text{strength}\times 2l\times B}} $
$ \
Therefore T\propto \sqrt{Wl} $
$ \
Therefore ,\frac{T_2}{T_1}=\sqrt{\frac{\text{w}_2}{\text{w}_1}\times \frac{l_2}{l_1}}$=
$\sqrt{\frac{\text{w}_1/2}{\text{w} _{1}}\times \frac{l_1/2}{l_1}}=\frac{1}{2} $
$ \Rightarrow T _{2}=\frac{T _{1}}{2}=0.5\ s $
BETA
