Question

Magnets A and B are geometrically similar but the magnetic moment of A is twice that of B. If $T_1$ and $T_2$ are the time periods of oscillation when their like poles and unlike poles are kept together respectively, then $\frac{T_1}{T_2}$ will be -

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\sqrt{3}$

Answer 1

The correct option is C $\frac{1}{\sqrt{3}}$

We know that time period of oscillation of a magnet of magnetic moment ${}^{\prime }{m}^{\prime }$ and moment of inertia ${}^{\prime }{I}^{\prime }$ in a field ${}^{\prime }{B}^{\prime }$ is given by
$T=2\pi \sqrt{\frac{I}{mB}}$

Here, $B=B_H$ (horizontal component of earth's magnetic field.
$I=I_A+I_B$ = sum of MOIs of the two magnets
When like poles are kept together,
$m=m_A+m_B$
& when unlike poles are kept together,
$m=|m_A-m_B|$

$\therefore T_1=2\pi \sqrt{\frac{I_A+I_B}{(m_A+m_B)B_H}}$
$\&T_2=2\pi \sqrt{\frac{I_A+I_B}{(m_A-m_B)B_H}}$
So,
$\frac{T_1}{T_2}=\sqrt{\frac{m_A-m_B}{m_A+m_B}}=\sqrt{\frac{2m-m}{2m+m}}$
$\Rightarrow \frac{T_1}{T_2}=\frac{1}{\sqrt{3}}$