Question

A thin rectangular magnet suspended freely has a period of oscillation equal to $T$. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. if its period of oscillation is $T$', the ratio $T^{\prime }/T$ is

• A. $\frac{1}{2\sqrt{2}}$
• B. $1/2$
• C. $2$
• D. $1/4$

Answer 1

Answer: (B)

$1/2$

Time period for oscillating magnet is

$T=2\pi \sqrt{\frac{I}{MB}}.........\left(1\right)$

where I is the moment of inertia,

M is the magnetic moment

$M=x\times l$

$x=polestrength$

When the magnet is broken in two equal parts, the new magnetic dipole moment is

$M^{{}^{\prime }}=\frac{x\times l}{2}=\frac{M}{2}$

$I^{\prime }=\frac{M{L}^{12}}{2}$

$=\frac{\frac{M}{2}\times {\left(\frac{L}{2}\right)}^2}{12}$

$=\frac{M{L}^2}{12\times 8}$

$=\frac{I}{8}$

New time period is

$T^{\prime }=2\pi \sqrt{\frac{I^{\prime }}{M^{\prime }B}}$

$\frac{T^{\prime }}{T}=\sqrt{\frac{I^{\prime }}{M^{\prime }}\times \frac{M}{I}}$

Putting all the values we get

$\frac{T^{\prime }}{T}=\sqrt{\frac{1}{4}}=\frac{1}{2}$