Question

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in the vibration magnetometer is 2s. 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:
$\begin{aligned} & (A)\dfrac{2}{3}s \\\ & (B)\sqrt{\dfrac{2}{3}}s \\\ & (C)\dfrac{3}{2}s \\\ & (D)\sqrt{\dfrac{3}{2}}s \\\ \end{aligned}$

Answer 1

It has been given that the magnet has been cut in three equal parts, this means that their weight distribution and length distribution must be the same. Also, we will use the formula for finding the time period of oscillation of a magnet oscillating under the effect of Earth’s magnetic field.

Complete answer:
Let the time period initially be ${{T}_{1}}$ and the time period after the magnet has been cut into three equal parts be ${{T}_{2}}$ . Also, let the mass of the magnet be $m$ and its length be $l$ .
Now, time period of a magnet is given by the following formula:
$\Rightarrow T=2\pi \sqrt{\dfrac{I}{MH}}$
Here,
$I$ is the moment of inertia of the magnet about the point of oscillation.
$M$ is the product of its pole strength (say P) and length.
$H$ is the horizontal component of Earth’s magnetic field.
Now, we shall calculate all these terms before and after the magnet was cut.
So, let $({{I}_{1}})and({{I}_{2}})$ be the moment of inertia before and after it was cut. Then,
$\Rightarrow {{I}_{1}}=\dfrac{1}{12}M{{L}^{2}}$
$\begin{aligned} & \Rightarrow {{I}_{2}}=\left[ \dfrac{1}{12}\times \dfrac{M}{3}\times {{\left( \dfrac{L}{3} \right)}^{2}} \right]\times 3 \\\ & \Rightarrow {{I}_{2}}=\left( \dfrac{1}{12}M{{L}^{2}} \right)\times \dfrac{1}{9} \\\ & \Rightarrow {{I}_{2}}=\dfrac{{{I}_{1}}}{9} \\\ \end{aligned}$
Now for M, we know the pole strength remains the same even after cutting the magnet.
Therefore,
$\begin{aligned} & \Rightarrow {{M}_{1}}=P\times l \\\ & \Rightarrow {{M}_{1}}=Pl \\\ \end{aligned}$
And,
$\begin{aligned} & \Rightarrow {{M}_{2}}=\left( P\times \dfrac{l}{3} \right)\times 3 \\\ & \Rightarrow {{M}_{2}}=Pl \\\ & \Rightarrow {{M}_{2}}={{M}_{1}} \\\ \end{aligned}$
Also, the horizontal component of Earth’s magnetic field will remain the same. Therefore, we can write using the values of above terms:
$\begin{aligned} & \Rightarrow \dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{2\pi \sqrt{\dfrac{{{I}_{2}}}{{{M}_{2}}H}}}{2\pi \sqrt{\dfrac{{{I}_{1}}}{{{M}_{1}}H}}} \\\ & \Rightarrow \dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{\sqrt{\dfrac{\dfrac{{{I}_{1}}}{9}}{{{M}_{1}}H}}}{\sqrt{\dfrac{{{I}_{1}}}{{{M}_{1}}H}}} \\\ & \Rightarrow \dfrac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\dfrac{1}{9}} \\\ & \Rightarrow \dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{1}{3} \\\ \end{aligned}$
Therefore, using the value of ${{T}_{1}}$ equal to 2 seconds, we get:
$\begin{aligned} & \Rightarrow {{T}_{2}}=\dfrac{{{T}_{1}}}{3}s \\\ & \Rightarrow {{T}_{2}}=\dfrac{2}{3}s \\\ \end{aligned}$
Hence, the time period when the magnet has been cut in three equal parts comes out to be$\dfrac{2}{3}s$.

Note:
These are some very less known and very less used formulas. Also, the number of questions from this section of Physics has been low to this date so one should always be careful to not let these topics slide as these topics are yet to be explored and good questions can be framed out of these topics.