A) \[\frac{\rho {{L}^{3}}}{8{{\pi }^{2}}}\]
B) \[\frac{\rho {{L}^{3}}}{16{{\pi }^{2}}}\]
C) \[\frac{5\rho {{L}^{3}}}{16{{\pi }^{2}}}\]
D) \[\frac{3\rho {{L}^{3}}}{8{{\pi }^{2}}}\]
Correct Answer: D
Solution :
Mass per unit length of the wire \[=\rho \] Mass of L length, \[M=\rho L\] and since the wire of length L is bent in a form of circular loop therefore \[2\pi R=L\Rightarrow R=\frac{L}{2\pi }\] Moment of inertia of loop about given axis \[=\frac{3}{2}M{{R}^{2}}\] \[=\frac{3}{2}\rho L{{\left( \frac{L}{2\pi } \right)}^{2}}=\frac{3\rho {{L}^{3}}}{8{{\pi }^{2}}}\]You need to login to perform this action.
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