- Home
- Standard 12
- Physics
The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2 \,s. $ The magnet is cut along its length into three equal parts and three parts are then placed on each other with their like poles together. The time period of this combination will be
$2 \,s$
$2/3 \,s$
$2\sqrt 3 \,s$
$2/\sqrt 3 \,s$
Solution
(b) Initially, the time period of the magnet
$T = 2 = 2\pi \sqrt {\frac{I}{{MB}}} $ ….. $(i)$
For each part, it’s moment of inertia $ = \frac{I}{{27}}$ and magnetic moment $ = \frac{M}{3}$
$\therefore $ Moment of inertia of system ${I_s} = \frac{I}{{27}} \times 3 = \frac{I}{9}$
Magnetic moment of system ${M_s} = \frac{M}{3} \times 3 = M$
Time period of system
${T_s} = 2\pi \sqrt {\frac{{{I_s}}}{{{M_s}B}}} = \frac{1}{3} \times 2\pi \sqrt {\frac{I}{{MB}}} = \frac{T}{3} = \frac{1}{3}\,sec $
