Q. 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
A) $1.0\ sec$
B) $0.5\ sec$
C) $0.25\ sec$
D) $2.0\ sec$
Explanation
(b) $T = 2\pi\sqrt{\frac{I}{MB}} = 2\pi\sqrt{\frac{wl^{2}/12}{Pole\ strength \times 2l \times B}}$
$\therefore T \propto \sqrt{Wl}$
$\therefore\frac{T_{2}}{T_{1}} = \sqrt{\frac{w_{2}}{w_{1}} \times \frac{l_{2}}{l_{1}}} = \sqrt{\frac{w_{1}/2}{w_{2}} \times \frac{l_{1}/2}{l_{1}}} = \frac{1}{2}$
$\Rightarrow T_{2} = \frac{T_{1}}{2} = 0.5\ sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.