A) \[=\omega \sqrt{{{A}^{2}}-{{y}^{2}}}\]
B) \[y=\frac{a}{2}\]
C) \[\therefore \]
D) \[v=\omega \sqrt{{{a}^{2}}-\frac{{{a}^{2}}}{4}}=\omega \sqrt{\frac{3{{a}^{2}}}{4}}=\frac{2\pi }{T}\frac{a\sqrt{3}}{2}=\frac{\pi a\sqrt{3}}{T}\]
Correct Answer: A
Solution :
Hence, \[\to \] \[\to \] \[\to \]\[\to \] \[{{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\,-\,\frac{\sigma 4\pi {{\alpha }^{2}}}{a}\,-\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{\sigma 4\pi {{b}^{2}}}{b}\] \[\to \] \[=\frac{\sigma }{{{\varepsilon }_{0}}}\,\left( \frac{{{a}^{2}}}{c}-b+c \right)\,=\frac{\sigma }{{{\varepsilon }_{0}}}(2a)\,\,(\because \,c=a+b)\]and\[\to \]\[\to \] \[\to \] Hence,\[\to \]You need to login to perform this action.
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