Question

A thin rectangular bar magnet suspended freely has a period of oscillation of $4\text{s}$. What will be the period of oscillation if the magnet is broken into two halves $($each having half the original length$)$ and one of the pieces is made to oscillate in the same field?

• A. $2\text{s}$
• B. $4\text{s}$
• C. $6\text{s}$
• D. $8\text{s}$

Answer 1

The correct option is A $2\text{s}$

We know that time period is given by,

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

When the magnet is broken into two equal halves, perpendicular to its length,

$M^{{}^{\prime }}=\frac{M}{2}$

If $A$ is the cross-sectional area and $\rho$ is the density of the material, the moment of inertia of the original magnet is,

$I=\frac{1}{12}M{L}^2$

$\Rightarrow I=\frac{1}{12}\left[\right(2\rho Al)\times (2l{)}^2]$

After breaking into two halves,

$I^{{}^{\prime }}=\frac{1}{12}\left[\right(\rho Al)\times (l^2\left)\right]=\frac{I}{8}$

So,

$T^{{}^{\prime }}=2\pi \sqrt{\frac{I^{{}^{\prime }}}{M^{{}^{\prime }}B}}=2\pi \sqrt{\frac{I/8}{\left(M/2\right)B}}=\pi \sqrt{\frac{I}{MB}}$

$\Rightarrow T^{{}^{\prime }}=\frac{T}{2}=\frac{4}{2}$ $[$ From $\left(1\right)]$

$\Rightarrow T^{{}^{\prime }}=2\text{s}$