The moment of inertia of a uniform thin rod of mass m and length about two axis PQ and RS passing through centre of rod C and in the plane of the rod are IPQ and IRS respectively. Then IPQ + IRS is equal to

Engineering

Physics

Moment of Inertia

Question

The moment of inertia of a uniform thin rod of mass m and length $l$ about two axis PQ and RS passing through centre of rod C and in the plane of the rod are I_PQ and I_RS respectively. Then I_PQ + I_RS is equal to

$\frac{{mℓ}^{2}}{2}$

$\frac{{mℓ}^{2}}{3}$

$\frac{{mℓ}^{2}}{4}$

$\frac{{mℓ}^{2}}{12}$

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The MI of rod about axis PQ figure(a) and MI of rod about axis P'Q' figure (b) are same by symmetry.

$∴ I_{PQ} + I_{RS} = I_{P' Q'} + I_{RS} = \frac{{mℓ}^{2}}{12}$

by perpendicular axis theorem.