A) \[\overrightarrow{r}\]
B) \[\overrightarrow{F}\]
C) \[\therefore \]
D) \[\overrightarrow{\tau }.\overrightarrow{r}=0\]
Correct Answer: D
Solution :
\[=-B\pi 2r\frac{dr}{dt}\] \[e=-0.04\times \pi \times 2\times 2\times {{10}^{-2}}\times 2\times {{10}^{-3}}\] Here,\[=3.2\pi \mu V\] \[=[{{M}^{0}}{{L}^{1}}{{T}^{-1}}]\]\[=[{{M}^{0}}{{L}^{1}}{{T}^{-2}}]\]You need to login to perform this action.
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