Consider a uniform wire of mass M and length L. It is bent into a semi-circle. Its moment of inertia about a line perpendicular to the plane of the wire passing through centre is :

Option 1 - <p>ML²/π²<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>2ML²/5π²<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>ML²/2π²<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>ML²/4π²</p>
4 Views|Posted 5 months ago
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I = I 1 + I 2 + I 3

I 1 = m l 2 3

I 2 = m l 2 1 2 + 5 m l 2 4 = I 3

I = m l 2 3 + 2 [ m l 2 1 2 + 5 m l 2 4 ]

= m l 2 3 + 2 × 1 1 2 ( 1 6 ) m l 2

= ( 1 3 + 8 3 ) m l 2

= 9 3 m l 2 = 3 m l 2

According to question, we can write

l = 2 π r r = l 2 π

I 1 = m l 2 3 , a n d I 2 = m r 2 2 = m l 2 8 π 2

I 1 I 2 = m l 2 3 × 8 π 2 m l 2 = 8 π 2 3

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Physics System of Particles and Rotational Motion 2025

Physics System of Particles and Rotational Motion 2025

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