Answer
64.8k+ views
Hint We will use the general formula for the moment of inertia of a circular ring, which is equal to $m{r^2}$, and the moment of inertia of a rod, which is equal to $\dfrac{{m{l^2}}}{{12}}$. We will relate the length of the rod with the circumference of the ring since both are the same physical length.
Complete step by step answer
The formula for the moment of inertia of the ring about its own axis is $m{r^2}$, where $r$ is the radius and $m$ is the mass of the ring.
Therefore we have ${I_1} = m{r^2}$.
When this same ring is cut into a rod, the length of the rod becomes equal to the perimeter of the ring.
This means that
$2\pi r = l$,
where $l$ is the length of the rod.
Thus the new moment of inertia, the moment of inertia of the rod about an axis perpendicular to the length of the rod and passing through its center is given by the general formula
$\dfrac{{m{l^2}}}{{12}}$,
where $m$ is the mass of the rod and
$l$ is the length of the rod.
Here we will substitute the value of the length of the rod with $l = 2\pi r$.
Therefore the moment of inertia becomes
${I_2} = \dfrac{{m{{(2\pi r)}^2}}}{{12}} = \dfrac{{m{{(\pi r)}^2}}}{3}$.
We need to calculate the ratio between the moment of inertias of the ring and the rod, i.e. $\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{m{r^2}}}{{\dfrac{{m{{(\pi r)}^2}}}{3}}}$.
$ \Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{3}{{{\pi ^2}}}$
This is the required ratio of the moments of inertias, i.e. the answer is $\dfrac{3}{{{\pi ^2}}}$.
Hence option the correct answer is option (B).
Note Here we have considered the mass moment of inertia and not the area moment of inertia. The mass moment of inertia is easier and required to be calculated here since nothing is given about the area and we will consider the thin rod and ring to be effectively a thin straight line.
Complete step by step answer
The formula for the moment of inertia of the ring about its own axis is $m{r^2}$, where $r$ is the radius and $m$ is the mass of the ring.
Therefore we have ${I_1} = m{r^2}$.
When this same ring is cut into a rod, the length of the rod becomes equal to the perimeter of the ring.
This means that
$2\pi r = l$,
where $l$ is the length of the rod.
Thus the new moment of inertia, the moment of inertia of the rod about an axis perpendicular to the length of the rod and passing through its center is given by the general formula
$\dfrac{{m{l^2}}}{{12}}$,
where $m$ is the mass of the rod and
$l$ is the length of the rod.
Here we will substitute the value of the length of the rod with $l = 2\pi r$.
Therefore the moment of inertia becomes
${I_2} = \dfrac{{m{{(2\pi r)}^2}}}{{12}} = \dfrac{{m{{(\pi r)}^2}}}{3}$.
We need to calculate the ratio between the moment of inertias of the ring and the rod, i.e. $\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{m{r^2}}}{{\dfrac{{m{{(\pi r)}^2}}}{3}}}$.
$ \Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{3}{{{\pi ^2}}}$
This is the required ratio of the moments of inertias, i.e. the answer is $\dfrac{3}{{{\pi ^2}}}$.
Hence option the correct answer is option (B).
Note Here we have considered the mass moment of inertia and not the area moment of inertia. The mass moment of inertia is easier and required to be calculated here since nothing is given about the area and we will consider the thin rod and ring to be effectively a thin straight line.
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
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)
In the ground state an element has 13 electrons in class 11 chemistry 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)
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)
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)