A) At the top of the circle
B) At the bottom of the circle
C) Half-way down
D) None of these
Correct Answer: B
Solution :
At the top, \[T+mg=\frac{m{{v}^{2}}}{r}\] At the bottom, \[T'-mg=\frac{mv{{'}^{2}}}{r}\] \[\therefore \] \[T'-2\times 9.8=\frac{2\times {{(4)}^{2}}}{1}=32\] \[\Rightarrow \] \[T'=32+19.6=\mathbf{51}\mathbf{.6}\,\,\mathbf{N}\] Thus, stone is at the bottom of the circleYou need to login to perform this action.
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