A) \[\sqrt{2\pi dgT}\]
B) \[\sqrt{\frac{2T}{\pi dg}}\]
C) \[\sqrt{\frac{\pi dg}{2T}}\]
D) \[\sqrt{\frac{2Tg}{\pi d}}\]
Correct Answer: B
Solution :
The wire does not sink, so net force on it will be zero \[\therefore \] \[mg=T.2l\] \[\Rightarrow \] \[{{\pi }^{2}}ldg=T.2l\] \[\therefore \] \[r=\sqrt{\frac{2R}{\pi dg}}\]
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