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Moment of Inertia Physics System of Particles and Rotational Motion Physics Class 11th

Two circular rings (of same material) are of same thickness. The diameter of first ring is twice that of second. The moment of inertia of first ring as compared to that of second is

Option 1 -

4 times

Option 2 -

8 times

Option 3 -

16 times

Option 4 -

2 times

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 weeks ago

Correct Option - 2

Detailed Solution:

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According to question, we can write

$l = 2 π r \Rightarrow r = \frac{l}{2 π}$

$I_{1} = \frac{m l^{2}}{3}, a n d I_{2} = \frac{m r^{2}}{2} = \frac{m l^{2}}{8 π^{2}}$

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The momentum of inertia of a uniform thin rod about a perpendicular axis passing through on end is I₁. The same rod is into a ring and its moment of inertia about a diameter is I₂. If $\frac{I_{1}}{I_{2}} i s \frac{X π^{2}}{3},$ then the value of x will be__________.

alok kumar singh

According to question, we can write

$l = 2 π r \Rightarrow r = \frac{l}{2 π}$

$I_{1} = \frac{m l^{2}}{3}, a n d I_{2} = \frac{m r^{2}}{2} = \frac{m l^{2}}{8 π^{2}}$

$\Rightarrow \frac{I_{1}}{I_{2}} = \frac{m l^{2}}{3} \times \frac{8 π^{2}}{m l^{2}} = \frac{8 π^{2}}{3}$

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Two circular rings (of same material) are of same thickness. The diameter of first ring is twice that of second. The moment of inertia of first ring as compared to that of second is

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