A) \[{{l}^{1/2}}\]
B) The lower end will be at a lower potential
C) \[\sin \theta \]
D) \[{{\left( \sin \theta \right)}^{1/2}}\]
Correct Answer: D
Solution :
[d] WET from A to B \[{{\omega }_{mg}}=\Delta K.E.\] \[mg\frac{L}{2}\sin \theta =\frac{1}{2}I{{\omega }^{2}}\] \[mg\frac{L}{2}\sin \theta =\frac{1}{2}\times \frac{m{{L}^{2}}}{3}{{\omega }^{2}}\] \[\frac{3g}{L}\sin \theta ={{\omega }^{2}}\] Induced emf produced in rod is \[\varepsilon =\frac{1}{2}(B\omega {{L}^{2}})=\frac{1}{2}B\times \sqrt{\frac{3g\,\sin \theta }{L}}\times {{L}^{2}}\] \[\varepsilon =\frac{1}{2}\times B\times \sqrt{\frac{3g\,\sin \,\theta \times {{L}^{4}}}{L}}=\frac{1}{2}B\sqrt{3g\,\sin \,\theta {{L}^{3}}}\] \[\varepsilon \propto {{L}^{3/2}}\] \[\varepsilon \propto \sqrt{\sin \theta }\]You need to login to perform this action.
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