Q. The magnet of a vibration magnetometer is heated so as to reduce its magnetic moment by 19%. By doing this the periodic time of the magnetometer will
A) Increase by 19%
B) Decrease by19%
C) Increase by 11%
D) Decrease by 21%
Explanation
(c) Time period, $T = 2\pi\sqrt{\frac{I}{MB_{H}}}$
$\Rightarrow \ \ T \propto \frac{1}{\sqrt{M}}$
$\Rightarrow \ \ \frac{T_{1}}{T_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$
If $\ \ M_{1} = 100$ then $M_{2} = (100 - 19) = 81$
So, $\frac{T_{1}}{T_{2}} = \sqrt{\frac{81}{100}} = \frac{9}{10}$
$\Rightarrow \ \ T_{2} = \frac{10}{9}T_{1} = 1.11\ T_{1}$
$\Rightarrow \ \ $Time period increase by 11%
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.