A) 20 m / s
B) \[10\sqrt{3}m/s\]
C) \[50\sqrt{2}m/s\]
D) 10 m / s
Correct Answer: C
Solution :
Here, \[r=L=\frac{10}{3}m\] \[{{T}_{\max }}=mg+\frac{m{{v}_{2}}^{2}}{r}\] and \[{{T}_{\min }}=-mg+\frac{m{{v}_{1}}^{2}}{r}\] Given: \[\frac{{{T}_{\max }}}{{{T}_{\min }}}=4\] \[\Rightarrow \] \[{{T}_{\max }}=4{{T}_{\min }}\] \[\therefore \] \[mg+\frac{m{{v}_{2}}^{2}}{r}=-4mg+\frac{4m{{v}_{1}}^{2}}{r}\] \[\Rightarrow \] \[5g+\frac{{{v}_{2}}^{2}}{r}=\frac{4{{v}_{1}}^{2}}{r}\] \[\Rightarrow \] \[5\,\,g\,\,r+{{v}_{2}}^{2}=4{{v}_{1}}^{2}\] \[\Rightarrow \] \[5g\times \frac{10}{3}+{{v}_{1}}^{2}+2g\times \frac{10}{3}\times 2=4{{v}_{1}}^{2}\] \[(\because \,\,{{v}_{2}}^{2}={{v}_{1}}^{2}-2gh)\] \[\Rightarrow \] \[\frac{50g}{3}+\frac{40g}{3}=3{{v}_{1}}^{2}\] \[\Rightarrow \] \[\frac{90g}{3}=3{{v}_{1}}^{2}\] \[\Rightarrow \] \[10\times 10={{v}_{1}}^{2}\] \[(\because \,\,g=10\,\,m/{{s}^{2}})\] \[\therefore \] \[{{v}_{1}}=\mathbf{10}\,\,\mathbf{m/s}\]You need to login to perform this action.
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