Question

The time period of a freely suspended magnet is $2sec$. If it is broken in length into two equal parts and one part is suspended in the same way, then its time period will be

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

Answer 1

The correct option is D

$1sec$

The explanation for the correct option:

Step 1: Given data

The time period of a freely suspended magnet, $\mathrm{T}=2\sec$

Number of cuts, $\mathrm{n}=2$

Step 2: To find

We have to determine the time period of a freely suspended magnet after the partition.

Step 3: Calculation

We know the time period of oscillation of a freely suspended magnet is,

$\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{I}}{\mathrm{MB}}}$

Here,

$\mathrm{I}$ is the moment of inertia.

$\mathrm{M}$ is the magnetic moment.

$\mathrm{B}$ is the horizontal component of the earth’s magnetic field.

From the above equation, whenever a freely suspended magnet is cut into n equivalent parts and one of them is suspended in an identical manner, its time duration is:

$\mathrm{T}\text{'}=\frac{\mathrm{T}}{\mathrm{n}}$

Here,

T' is the new time period.

T is the original time period.

n is the number of cuts.

By substituting the values in the formula, we get

$$\begin{gathered} =\frac{2}{2} \\ =1\sec \end{gathered}$$

Therefore, the time period will be $1sec$.