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

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$\frac{49 M R^{2}}{4}$

$\frac{39 M R^{2}}{4}$

$\frac{39 M R^{2}}{8}$

$\frac{49 M R^{2}}{8}$

answer is D.

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Detailed Solution

Given: l = 6R. From perpendicular axes theorem, the
moment of inertia about the given axis is given by

$I = M (\frac{R^{2}}{4} + \frac{l^{2}}{3}) = M [\frac{R^{2}}{4} + \frac{{(6 R)}^{2}}{3}] = M [\frac{R^{2}}{4} + \frac{36 R^{2}}{3}] = \frac{49 M R^{2}}{4}$

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