Moment of inertia of a cylinder of mass M, length L and radius R about an axis passing through its centre and perpendicular to the axis of the cylinder is I = M((R²/4) + (L²/12)). If such a cylinder is to be made for a given mass of a material, the ratio L/R for it to have minimum possible I is

Option 1 - <p>√(3/2)</p>
Option 2 - <p>√(2/3)</p>
Option 3 - <p>2/√3<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>√3/2<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
3 Views|Posted 5 months ago
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1 Answer
A
5 months ago
Correct Option - 2
Detailed Solution:

The mass of the cylinder M = ρV = ρ (πR²L), where ρ is the density of the material.
We can express R² in terms of L: R² = M / (πρL).
The moment of inertia is I = M (R²/4 + L²/12).
Substitute R² into the equation for I:
I (L) = M * [ (M / (4πρL) + (L²/12) ]
To find the minimum possible I, we differentiate

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Physics System of Particles and Rotational Motion 2025

Physics System of Particles and Rotational Motion 2025

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