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Chapter 13: Problem 1988

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

Expert verified
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

01

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
02

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
03

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
04

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
05

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.

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Most popular questions from this chapter

The time period of a freely suspended magnet is a 4 seconds. 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) \(4 \mathrm{sec}\) (b) \(2 \mathrm{sec}\) (c) \(0.5 \mathrm{sec}\) (d) \(0.25 \mathrm{sec}\)

A conducting circular loop of radius a carries a constant current I. It is placed in a uniform magnetic field \(\mathrm{B}^{-}\), such that \(\mathrm{B}^{-}\) is perpendicular to the plane of the Loop. The magnetic force acting on the Loop is (a) \(\mathrm{B}^{-} \operatorname{Ir}\) (b) \(\mathrm{B}^{-} \mathrm{I} \pi \mathrm{r}^{2}\) (c) Zero (d) BI \((2 \pi \mathrm{r})\)

Two straight long conductors \(\mathrm{AOB}\) and \(\mathrm{COD}\) are perpendicular to each other and carry currents \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\). The magnitude of the mag. field at a point " \(\mathrm{P}^{n}\) at a distance " \(\mathrm{a}^{\prime \prime}\) from the point "O" in a direction perpendicular to the plane \(\mathrm{ABCD}\) is (a) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(I_{1}+I_{2}\right)\) (b) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(I_{1}-I_{2}\right)\) (c) $\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(\mathrm{I}_{1}^{2}+\mathrm{I}_{2}^{2}\right)^{1 / 2}$ (d) $\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left[\left(I_{1} I_{2}\right) /\left(I_{1}-I_{2}\right)\right]$

An element \(\mathrm{d} \ell^{-}=\mathrm{dx} \uparrow\) (where $\mathrm{dx}=1 \mathrm{~cm}$ ) is placed at the origin and carries a large current \(\mathrm{I}=10 \mathrm{Amp}\). What is the mag. field on the Y-axis at a distance of \(0.5\) meter ? (a) \(2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (b) \(4 \times 10^{8} \mathrm{k} \wedge \mathrm{T}\) (c) \(-2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (d) \(-4 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\)

The direction of mag. field lines close to a straight conductor carrying current will be (a) Along the length of the conductor (b) Radially outward (c) Circular in a plane perpendicular to the conductor (d) Helical
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