Question

A thin magnetic needle oscillates in a horizontal plane with a period \(\mathrm{T}\). It is broken into n equal parts. The time period of each part will be (a) \(\mathrm{T}\) (b) \(\mathrm{n}^{2} \mathrm{~T}\) (c) \((\mathrm{T} / \mathrm{n})\) (d) \(\left(\mathrm{T} / \mathrm{n}^{2}\right)\)

Short answer

The time period of oscillation for each part of the new needle is given by \(T_{n} = \frac{T}{n^{\frac{3}{2}}}\). However, this exact expression is not present among the given options. The closest option to the obtained expression is (d) \(\frac{T}{n^2}\). There could be an error in the question, but based on the available options, the best answer is (d).

Step-by-step solution

Recall the time period formula for a magnetic needle

The time period of oscillation for a magnetic needle is given by the formula \(T = 2\pi \sqrt{\frac{I}{k}}\), where \(T\) is the time period, \(I\) is the moment of inertia of the magnetic needle and k is the torsional constant. Step 2: Express the moment of inertia for the original magnetic needle

Express the moment of inertia for the original magnetic needle

Because the needle is thin, its moment of inertia can be expressed as \(I = m_{orig}\frac{l^{2}}{12}\), where \(m_{orig}\) denotes the mass of the original needle and \(l\) is the length of the original needle. Step 3: Find the moment of inertia for a part of the new needle

Find the moment of inertia for a part of the new needle

We are given that the original needle is broken into n equal parts, so each new part has mass \(m = \frac{m_{orig}}{n}\) and length \(l_{n} = \frac{l}{n}\). The moment of inertia for a part of the new needle can be expressed as \[I_{n} = m\frac{l_{n}^{2}}{12} = \left(\frac{m_{orig}}{n}\right) \frac{\left(\frac{l}{n}\right)^2}{12} = \frac{m_{orig} l^2}{12n^3}\] Step 4: Find the time period for each part of the new needle

Find the time period for each part of the new needle

Plug the moment of inertia of each part, \(I_{n}\), into the time period formula: \[\begin{aligned} T_{n} &= 2\pi \sqrt{\frac{I_{n}}{k}} \\ &= 2\pi \sqrt{\frac{\frac{m_{orig} l^2}{12n^3}}{k}} \\ &= \frac{2\pi \sqrt{\frac{m_{orig} l^2}{12}}}{n^{\frac{3}{2}}}\sqrt{\frac{1}{k}} \end{aligned}\] Since the original time period \(T = 2\pi \sqrt{\frac{I}{k}} = 2\pi \sqrt{\frac{m_{orig}l^2}{12k}}\), we can rewrite the time period for each part as: \[T_{n} = \frac{T}{n^{\frac{3}{2}}}\] Step 5: Match the answer with the options

Match the answer with the options

The time period of each part does not match any of the given options exactly so there seems to be a mistake in the options. However, the closest option is (d) \(T/n^2\), so it is possible there is an error in the question. In this case, we would still choose (d) as the best available answer.