Answer
64.8k+ views
Hint First establish the magnetic moments and the moment of inertia of each of the magnet pieces cut from the original piece. Then substitute these values in the equation for the time period of a magnet.
Formula used
${T_0} = 2\pi \sqrt {\dfrac{I}{{MB}}} $ where $I$ is the moment of inertia of the magnet and $M$ is its magnetic moment, $B$is the magnetic field and ${T_0}$is the time period of the magnet.
Complete step by step answer
For the original magnet, the time period is given by the formula,
${T_0} = 2\pi \sqrt {\dfrac{I}{{MB}}} $ where $I$ is the moment of inertia of the magnet and $M$ is its magnetic moment.
Now when the magnet is cut into three equal parts, the magnetic moments of each of the parts become one-third of its original value $M$ such that $M' = \dfrac{M}{3}$ where $M'$is the magnetic moment of each of the parts.
Now moment of inertia is essentially defined as a quantity expressing a body’s tendency to resist angular acceleration.
Moment of inertia of the magnet about an axis perpendicular to its length through its one end is given as $\dfrac{{m{l^2}}}{3}$ where $m$is its mass and $l$is its length.
Now, when the magnet is cut into three equal parts, mass of each part becomes $\dfrac{m}{3}$ and length of each part becomes $\dfrac{l}{3}$
So, moment of inertia of each of such part is $I' = \dfrac{m}{3}\dfrac{{{{\left( {\dfrac{l}{3}} \right)}^2}}}{3} = \dfrac{1}{{27}}\dfrac{{m{l^2}}}{3} = \dfrac{I}{{27}}$
So, the new time period of each of the pieces is
$\begin{gathered}
T = 2\pi \sqrt {\dfrac{{I'}}{{M'B}}} = 2\pi \sqrt {\dfrac{3}{{27}}\dfrac{I}{{MB}}} \\
\Rightarrow T = \sqrt {\dfrac{1}{9}} 2\pi \sqrt {\dfrac{I}{{MB}}} = \dfrac{{{T_0}}}{3} \\
\end{gathered} $
Therefore, the time period of each of the magnetic pieces is $\dfrac{{{T_0}}}{3}$
So, the correct answer is B.
Note Magnetic lines of force are always closed as magnetic monopoles do not exist. Which means that a magnet will always have a north pole and a south pole no matter how many times it is cut into smaller pieces.
Formula used
${T_0} = 2\pi \sqrt {\dfrac{I}{{MB}}} $ where $I$ is the moment of inertia of the magnet and $M$ is its magnetic moment, $B$is the magnetic field and ${T_0}$is the time period of the magnet.
Complete step by step answer
For the original magnet, the time period is given by the formula,
${T_0} = 2\pi \sqrt {\dfrac{I}{{MB}}} $ where $I$ is the moment of inertia of the magnet and $M$ is its magnetic moment.
Now when the magnet is cut into three equal parts, the magnetic moments of each of the parts become one-third of its original value $M$ such that $M' = \dfrac{M}{3}$ where $M'$is the magnetic moment of each of the parts.
Now moment of inertia is essentially defined as a quantity expressing a body’s tendency to resist angular acceleration.
Moment of inertia of the magnet about an axis perpendicular to its length through its one end is given as $\dfrac{{m{l^2}}}{3}$ where $m$is its mass and $l$is its length.
Now, when the magnet is cut into three equal parts, mass of each part becomes $\dfrac{m}{3}$ and length of each part becomes $\dfrac{l}{3}$
So, moment of inertia of each of such part is $I' = \dfrac{m}{3}\dfrac{{{{\left( {\dfrac{l}{3}} \right)}^2}}}{3} = \dfrac{1}{{27}}\dfrac{{m{l^2}}}{3} = \dfrac{I}{{27}}$
So, the new time period of each of the pieces is
$\begin{gathered}
T = 2\pi \sqrt {\dfrac{{I'}}{{M'B}}} = 2\pi \sqrt {\dfrac{3}{{27}}\dfrac{I}{{MB}}} \\
\Rightarrow T = \sqrt {\dfrac{1}{9}} 2\pi \sqrt {\dfrac{I}{{MB}}} = \dfrac{{{T_0}}}{3} \\
\end{gathered} $
Therefore, the time period of each of the magnetic pieces is $\dfrac{{{T_0}}}{3}$
So, the correct answer is B.
Note Magnetic lines of force are always closed as magnetic monopoles do not exist. Which means that a magnet will always have a north pole and a south pole no matter how many times it is cut into smaller pieces.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)