A) \[\frac{m{{v}^{2}}}{r}+mg\]
B) \[\frac{m{{v}^{2}}}{r}-mg\]
C) \[\frac{mv}{r}\]
D) mg
Correct Answer: A
Solution :
Key Idea: Tension will be more than the centripetal force at the lowest point for circular motion. For a stone of mass m and length of string r at the point B, weight mg acts vertically downwards, while tension \[{{T}_{B}}\] acts vertically upwards. Their resultant provides the necessary centripetal force, that is \[{{T}_{B}}-mg=\frac{m{{v}^{2}}}{r}\] \[\Rightarrow \] \[{{T}_{B}}=\frac{m{{v}^{2}}}{r}+mg\] This is the tension of string at lowest point. Note: Tension of string at highest point A will be less than that at B. \[{{T}_{A}}+mg\frac{mv_{A}^{2}}{r}\] \[\Rightarrow \] \[{{T}_{A}}=\frac{mv_{A}^{2}}{r}-mg\]You need to login to perform this action.
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