A) \[\frac{R\alpha }{4{{\pi }^{2}}}(2\pi -\alpha )\]
B) \[\frac{R}{2\pi }(2\pi -\alpha )\]
C) R (2p ? a)
D) \[\frac{4\pi }{R\alpha }(2\pi -\alpha )\]
Correct Answer: A
Solution :
Here \[{{R}_{XWY}}=\frac{R}{2\pi r}\times (r\alpha )=\frac{R\alpha }{2\pi }\] \[\left( \because \,\alpha =\frac{l}{r} \right)\] and \[{{R}_{XZY}}=\frac{R}{2\pi r}\times r(2\pi -\alpha )=\frac{R}{2\pi }(2\pi -\alpha )\] \[{{R}_{eq}}=\frac{{{R}_{XWY}}{{R}_{XZY}}}{{{R}_{XWY}}+{{R}_{XZY}}}=\frac{\frac{R\alpha }{2\pi }\times \frac{R}{2\pi }(2\pi -\alpha )}{\frac{R\alpha }{2\pi }+\frac{R(2\pi -\alpha )}{2\pi }}\]\[=\frac{R\alpha }{4{{\pi }^{2}}}(2\pi -\alpha )\]
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