A) \[20m/\sec \]
B) \[10\sqrt{3}m/\sec \]
C) \[5\sqrt{2}\,m/\sec \]
D) \[10m/\sec \]
Correct Answer: D
Solution :
Since the maximum tension \[{{T}_{B}}\] in the string moving in the vertical circle is at the bottom and minimum tension \[{{T}_{T}}\] is at the top. \ \[{{T}_{B}}=\frac{mv_{B}^{2}}{L}+mg\] and \[{{T}_{T}}=\frac{mv_{T}^{2}}{L}-mg\] \ \[\frac{{{T}_{B}}}{{{T}_{T}}}=\frac{\frac{mv_{B}^{2}}{L}+mg}{\frac{mv_{T}^{2}}{L}-mg}=\frac{4}{1}\] or \[\frac{v_{B}^{2}+gL}{v_{T}^{2}-gL}=\frac{4}{1}\] or \[v_{B}^{2}+gL=4v_{T}^{2}-4gL\] but \[v_{B}^{2}=v_{T}^{2}+4gL\] \ \[v_{T}^{2}+4gL+gL=4v_{T}^{2}-4gL\]Þ \[3v_{T}^{2}=9gL\] \ \[v_{T}^{2}=3\times g\times L=3\times 10\times \frac{10}{3}\] or \[{{v}_{T}}=10\,m/\sec \]
You need to login to perform this action.
You will be redirected in
3 sec
