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
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 -
√(3/2)
Option 2 -
√(2/3)
Option 3 -
2/√3
Option 4 -
√3/2
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1 Answer
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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 I with respect to L and set the derivative to zero.
dI/dL = M * [ -M/ (4πρL²) + 2L/12 ] = 0
M/ (4πρL²) = 2L/12 = L/6
M/ (πρ) = (4/6)L³ = (2/3)L³
Since R² = M/ (πρL), we have R²L = M/ (πρ).
Substitute M/ (πρ) int...more
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