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(A) The moment of inertia of a square plate of side '$a$' and mass '$m$' about an axis passing through its center of mass and perpendicular to its pla

Vedclass · Accepted Answer

$\frac{2}{3}ma^2$

Vedclass · Answer

(A) The moment of inertia of a square plate of side '$a$' and mass '$m$' about an axis passing through its center of mass and perpendicular to its plane is given by:$I_{cm} = \frac{1}{12}m(a^2 + a^2) = \frac{ma^2}{6}$According to the parallel axis theorem,the moment of inertia about an axis passing through a corner and perpendicular to the plane is:$I = I_{cm} + md^2$Here,'$d$' is the distance from the center of the square to the corner. For a square of side '$a$',the diagonal is '$a\sqrt{2}$',so the distance from the center to a corner is:$d = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$Substituting the values:$I = \frac{ma^2}{6} + m\left(\frac{a}{\sqrt{2}}\right)^2$$I = \frac{ma^2}{6} + \frac{ma^2}{2}$$I = \frac{ma^2 + 3ma^2}{6} = \frac{4ma^2}{6} = \frac{2}{3}ma^2$

Column-$I$	Column-$II$
$(1)$ Moment of inertia of a solid sphere about its diameter	$(a)$ $\frac{2}{3}MR^2$
$(2)$ Moment of inertia of a solid sphere about a tangent	$(b)$ $\frac{2}{5}MR^2$
	$(c)$ $\frac{7}{5}MR^2$

Consider a uniform square plate of side '$a$' and mass '$m$'. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

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Column-$I$ Column-$II$
$(1)$ Moment of inertia of a solid sphere about its diameter $(a)$ $\frac{2}{3}MR^2$
$(2)$ Moment of inertia of a solid sphere about a tangent $(b)$ $\frac{2}{5}MR^2$
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