
Two discs of the same mass and same thickness have densities as 17 g / $\text{c}{{\text{m}}^{\text{3}}}$ and 51 g / $\text{c}{{\text{m}}^{\text{3}}}$. The ratio of their moment of inertia are in the ratio:
A). 1:3
B). 3:1
C). $\dfrac{1}{\sqrt[3]{9}}$
D). $\sqrt[3]{9}$
Answer
586.2k+ views
Hint: We can easily define the moment of inertia and the measure of the resistance of a particular body to the angular acceleration about a given axis which is equivalent to the sum of the product of every element of mass present in the body and also the square of the distance of the elements on the axis. Keeping this in mind we can easily solve the given question.
Complete step by step answer:
We know that the momentum of inertia of the disc about its centre will be given by I = $\dfrac{1}{2}M{{r}^{2}}$.
In the above formula, r represents the radius of the disc.
Now the volume of the disc will be equal to base area $\times$thickness.
So, we can say that V or volume = $\pi {{r}^{2}}\times L$.
So, the mass of the disc is = density $\times$volume.
So, the mass of m = $d\times V=d\pi {{r}^{2}}L$
So, ${{r}^{2}}=\dfrac{M}{d\pi L}$
Now, I = $M(\dfrac{M}{d\pi L})=\dfrac{{{M}^{2}}}{d\pi L}$.
In this case , $\pi$and L are constant.
Therefore, the moment of inertia is inversely proportional to density of disc.
So, if mass and thickness of disc are same then, ratio of their moment of inertia should be $\dfrac{{{I}_{1}}}{{{I}_{2}}}=\dfrac{{{d}_{2}}}{{{d}_{1}}}=\dfrac{51}{17}$ = $\dfrac{3}{1}$
Hence, the correct answer is Option (B).
Note: For ease of understanding, we can consider the moment of inertia to be basically the quantitative measure of the rotational inertia of a body, which means that it is the measure of the opposition that the particular body exhibits when the application of a turning force or torque alters the speed of rotation of the body about an axis. This axis might be internal or external and it is not necessarily fixed, but however the moment of inertia is always specified with respect to that axis and is defined as the sum of products obtained by multiplication of the mass of each particle of matter in a given body with the square of the distance of the particles from the axis.
Complete step by step answer:
We know that the momentum of inertia of the disc about its centre will be given by I = $\dfrac{1}{2}M{{r}^{2}}$.
In the above formula, r represents the radius of the disc.
Now the volume of the disc will be equal to base area $\times$thickness.
So, we can say that V or volume = $\pi {{r}^{2}}\times L$.
So, the mass of the disc is = density $\times$volume.
So, the mass of m = $d\times V=d\pi {{r}^{2}}L$
So, ${{r}^{2}}=\dfrac{M}{d\pi L}$
Now, I = $M(\dfrac{M}{d\pi L})=\dfrac{{{M}^{2}}}{d\pi L}$.
In this case , $\pi$and L are constant.
Therefore, the moment of inertia is inversely proportional to density of disc.
So, if mass and thickness of disc are same then, ratio of their moment of inertia should be $\dfrac{{{I}_{1}}}{{{I}_{2}}}=\dfrac{{{d}_{2}}}{{{d}_{1}}}=\dfrac{51}{17}$ = $\dfrac{3}{1}$
Hence, the correct answer is Option (B).
Note: For ease of understanding, we can consider the moment of inertia to be basically the quantitative measure of the rotational inertia of a body, which means that it is the measure of the opposition that the particular body exhibits when the application of a turning force or torque alters the speed of rotation of the body about an axis. This axis might be internal or external and it is not necessarily fixed, but however the moment of inertia is always specified with respect to that axis and is defined as the sum of products obtained by multiplication of the mass of each particle of matter in a given body with the square of the distance of the particles from the axis.
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