Dear student The length of the loop is l so the radius of the loop will be 2πr=lr=l2πand mass of the ring will bem=ρlthe moment of inertia of the ring about its diameter parallel to XX' will beI'=12mr2Now using the parallel axis theorem of moment of inertia, the M I about the XX'I=I'+mr2I=12mr2+mr2I=32mr2putting the value of r and mI=32ρl·l24π2I=3ρl38π2so correct option is D

13 th ques

Dear student

T h e l e n g t h o f t h e l o o p i s l s o t h e r a d i u s o f t h e l o o p w i l l b e 2 πr = l r = \frac{l}{2 π} and mass of the ring will be m = ρl the moment of inertia of the ring about its diameter parallel to XX' will be I' = \frac{1}{2} {mr}^{2} Now using the parallel axis theorem of moment of inertia, the M . I . about the XX' I = I' + {mr}^{2} I = \frac{1}{2} {mr}^{2} + {mr}^{2} I = \frac{3}{2} {mr}^{2} putting the value of r and m I = \frac{3}{2} \frac{ρl \cdot l^{2}}{4 π^{2}} I = \frac{3 {ρl}^{3}}{8 π^{2}} so correct option is (D)

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13 th ques ​

13 th ques