A) \[\frac{3{{L}^{2}}D}{16\,\pi }\]
B) \[\frac{2{{L}^{2}}D}{3\,\pi }\]
C) \[\frac{3{{L}^{2}}D}{8{{\pi }^{2}}}\]
D) \[\frac{8{{L}^{2}}D}{13{{\pi }^{2}}}\]
Correct Answer: C
Solution :
The moment of inertia of a circular loop about diameter is \[=\,\,\frac{M{{R}^{2}}}{2}\] [R is radius, M is mass] \[\therefore \] Moment of inertia about AB will be \[{{I}_{AB}}=\frac{M{{R}^{2}}}{2}+M{{R}^{2}}=\frac{3}{2}M{{R}^{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{I}_{AB}}=\frac{3}{2}M{{R}^{2}}\] ... (i) As linear mass density is D Mass of loop \[\operatorname{M}=LD\] ... (ii) and length of wire is L. \[\therefore \] Radius of circular loop will be \[R=\frac{L}{2\pi }\] ? (iii) From Eqs. (i), (ii) and (iii), we get \[{{I}_{AB}}=\frac{3}{2}(LD).{{\left( \frac{L}{2\pi } \right)}^{2}}=\frac{3{{L}^{2}}D}{8{{\pi }^{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
