A sheet ABCD in the shape of a Rhombus has mass m and length of each side $l$ as shown in figure. Then moment of inertia of the sheet about side AB is

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$\frac{1}{6} m l^{2}$

$\frac{1}{4} m l^{2}$

$\frac{1}{3} m l^{2}$

$\frac{3}{4} m l^{2}$

answer is B.

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Detailed Solution

$M . I . of ABCD about AB = M . I . of the rectangle CDEF about EF + M . I . of the triangle ADE about AE - M . I . of triangle BCF about BF$

$= M . I . of rectangle CDEF about EF$

$DE = AD \sin 60^{0} = \frac{\sqrt{3} l}{2}$

$∴ I = \frac{1}{3} m {(DE)}^{2} = \frac{1}{3} m {(\frac{\sqrt{3} l}{2})}^{2} = \frac{1}{4} m l^{2}$

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