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A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is ${2^{5/4}}$ $seconds$. One of the magnets is removed and if the other magnet oscillates in the same field, then the time period in seconds is
${2^{1/4}}$
${2^{1/2}}$
$2$
${2^{3/4}}$
Solution
(c)Initially magnetic moment of system
${M_1} = \sqrt {{M^2} + {M^2}} = 2M$ and moment of inertia
${I_1} = I + I = 2I.$
Finally when one of the magnet is removed then
${M_2} = M$ and ${I_2} = I$
So $T = 2\pi \sqrt {\frac{I}{{M\;{B_H}}}} $
$\frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{{I_1}}}{{{I_2}}} \times \frac{{{M_2}}}{{{M_1}}}} = \sqrt {\frac{{2I}}{I} \times \frac{M}{{\sqrt 2 M}}} $$ \Rightarrow {T_2} = \frac{{{2^{5/4}}}}{{{2^{1/4}}}} = 2\;\sec .$
