A) \[3.4\times {{10}^{-5}}\,kg\,{{m}^{2}}\]
B) \[1.2\times {{10}^{-4}}\,kg\,{{m}^{2}}\]
C) \[2.6\times {{10}^{-4}}\,kg\,{{m}^{2}}\]
D) \[4.7\times {{10}^{-5}}\,kg\,{{m}^{2}}\]
Correct Answer: B
Solution :
[b] The magnetic moment of the coil is \[M=NIA=16\times 0.75\times \pi \times {{(0.1)}^{2}}=0.377\,A{{m}^{2}}\] If K be the moment of inertia of the coil about its axis of rotation, then its period of oscillation in a magnetic field B is given by \[T=2\pi \sqrt{\frac{K}{MB}}\] or its frequency v is\[=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{MB}{K}}\] This gives \[K=\frac{MB}{4{{\pi }^{2}}{{v}^{2}}}\] Given that \[B=5.0\times {{10}^{-2}}T\], \[M=0.377\,A-{{m}^{2}}\]and \[v=2{{s}^{-1}}\] \[\therefore \,\,\,\,\,\,\,K=\frac{0.377\times 5.0\times {{10}^{-2}}}{4\times {{(3.14)}^{2}}\times {{(2)}^{2}}}=1.2\times {{10}^{-4}}\,kg{{m}^{2}}\]
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