A) \[mg+\frac{m{{v}^{2}}}{r}\]
B) \[mg-\frac{m{{v}^{2}}}{r}\]
C) \[\frac{{{m}^{2}}{{v}^{2}}g}{r}\]
D) \[\frac{{{v}^{2}}g}{r}\]
Correct Answer: A
Solution :
Thrust at the lowest point of concave bridge \[=mg+\frac{m{{v}^{2}}}{r}\]
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