A) 8 mg
B) 7 mg
C) 1 mg
D) 3 mg
Correct Answer: A
Solution :
Apply energy conservation b/w highest point & pt C, \[mg\times 4r=\frac{1}{2}m{{v}^{2}}\] \[v=\sqrt{8gr}\] \[N\,\,=\,\frac{m{{v}^{2}}}{R}\,\,at\,\,po\operatorname{int}\,\,C\] \[N\,\,=\,\,8mg\]You need to login to perform this action.
You will be redirected in
3 sec