A thin wire of length L and linear mass densi | vedclass

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(C) Total mass $M = m 	imes L$.The radius $R$ is found from $L = 2\pi R$,so $R = \frac{L}{2\pi}$.The moment of inertia of a ring about its diameter i

Vedclass · Accepted Answer

$\frac{3mL^3}{8\pi^2}$

Vedclass · Answer

(C) Total mass $M = m 	imes L$.The radius $R$ is found from $L = 2\pi R$,so $R = \frac{L}{2\pi}$.The moment of inertia of a ring about its diameter is $I_{diam} = \frac{1}{2}MR^2$.Using the parallel axis theorem,the moment of inertia about the tangent $yy'$ is $I = I_{cm} + MR^2$,where $I_{cm} = I_{diam} = \frac{1}{2}MR^2$.Thus,$I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.Substituting $M = mL$ and $R = \frac{L}{2\pi}$:$I = \frac{3}{2} (mL) \left(\frac{L}{2\pi}ight)^2 = \frac{3}{2} mL \left(\frac{L^2}{4\pi^2}ight) = \frac{3mL^3}{8\pi^2}$.Therefore,the correct option is $C$.

Column-$I$	Column-$II$
$(1)$ Perpendicular Axis Theorem	$(a)$ $I = I_C + Md^2$
$(2)$ Parallel Axis Theorem	$(b)$ $I_z = I_x + I_y$

$A$ thin wire of length $L$ and linear mass density $m$ is bent into a circular ring (in $x-y$ plane) with centre $C$ as shown in the figure. The moment of inertia of the ring about an axis $yy'$ will be:

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Match Column-$I$ with Column-$II$:
Column-$I$ Column-$II$
$(1)$ Perpendicular Axis Theorem $(a)$ $I = I_C + Md^2$
$(2)$ Parallel Axis Theorem $(b)$ $I_z = I_x + I_y$

Where, $d =$ distance between two parallel axes.

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