For the given uniform square lamina ABCD, whose center is O, (a) IAC = √(2) IEF (b) √(2)IAC = IEF (c) IAD = 3IEF (d) IAC = IEF

Answer: (d). Perpendicular axis theorem: IEF + I⊥ EF = Iz and IAC + IBD = Iz. By symmetry of the square, IEF = I⊥ EF and IAC = IBD, so both equal Iz/2. Hence IAC = IEF. (Also IAD = (Ma²)/(12) + Ma²/4 = 4IEF, so (c) is false.) So the answer is (d) IAC = IEF.

For the given uniform square lamina ABCD, whose center is O, | JEE Main 2007 (AIEEE)

Q189JEE Main 2007 · AIEEEMedium

Moment of Inertia & Radius of Gyration

For the given uniform square lamina $ABCD$ , whose center is $O,$

Show answer and solution

Answer: (d)

Perpendicular axis theorem: $I_{EF} + I_{\perp EF} = I_z$ and $I_{AC} + I_{BD} = I_z$ .
By symmetry of the square, $I_{EF} = I_{\perp EF}$ and $I_{AC} = I_{BD}$ , so both equal $\frac{I_z}{2}$ .
Hence $I_{AC} = I_{EF}$ . (Also $I_{AD} = \frac{Ma^2}{12} + \frac{Ma^2}{4} = 4I_{EF}$ , so (c) is false.)
So the answer is (d) $I_{AC} = I_{EF}$ .

Report a mistake

Moment of Inertia & Radius of Gyration question from JEE Main 2007 (AIEEE)

More questions on Moment of Inertia & Radius of Gyration