Question

# Moment of inertia of a uniform horizontal solid cylinder of mass M about an axis passing through its edge and perpendicular to the axis of the cylinder when its length is 6 times its radius $$'R'$$ is:

A
MR2/4
B
39MR/4
C
49MR/4
D
49MR2/4

Solution

## The correct option is D $${49MR^2}/{4}$$From parallel axix theorem $$I = M[\frac{L^2}{12} + \frac{L^2}{4}] + M [\frac{R}{2}]^2$$$$I = \dfrac{ML^2}{3} + \dfrac{MR^2}{4}$$given L = 6R$$L = \dfrac{M}{3} (6R)^2 + \dfrac{MR^2}{4} = \dfrac{49}{4}MR^2$$Hence option (D) is correctPhysics

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