A) At the top of the circle
B) At the bottom of the circle
C) Halfway down
D) None of the above
Correct Answer: B
Solution :
\[mg=20N\] and \[\frac{m{{v}^{2}}}{r}=\frac{2\times {{(4)}^{2}}}{1}=32N\] It is clear that 52 N tension will be at the bottom of the circle. Because we know that \[{{T}_{\text{Bottom}}}=mg+\frac{m{{v}^{2}}}{r}\]
You need to login to perform this action.
You will be redirected in
3 sec
