A) \[3\sqrt{3}:8\]
B) \[16:9\sqrt{3}\]
C) \[4:9\]
D) \[2\sqrt{2}:3\]
Correct Answer: B
Solution :
Time period of suspended magnet \[T=2=\sqrt{\frac{i}{MB\,\cos \delta }}\] \[v=\frac{1}{2\pi }=\sqrt{\frac{MB\,\cos \,\delta }{i}}\] \[v=\sqrt{B\,\cos \delta }\] Or \[B\propto \frac{{{v}^{2}}}{\cos \delta }\] \[\therefore \] \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{400}{\cos {{30}^{o}}}\times \frac{\cos {{60}^{o}}}{225}\] \[=\frac{16\times 2}{9\times \sqrt{3}}\times \frac{1}{2}=16:9\,\sqrt{3}\]You need to login to perform this action.
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