A) \[\text{6}.\text{6 N, 35m}{{\text{s}}^{\text{-1}}}\]
B) \[\text{6N},\text{37m}{{\text{s}}^{\text{-1}}}\]
C) \[~\text{7}.\text{5 N, 46 m}{{\text{s}}^{\text{-1}}}\]
D) \[\text{8 N},\text{ 38 m}{{\text{s}}^{\text{-1}}}\]
Correct Answer: A
Solution :
Frequency of revolution of stone, \[f=40rev/\min =\frac{40}{60}rev/s\] Angular speed of the stone, \[\omega =2\pi f\] \[\text{= 2 }\!\!\pi\!\!\text{ }\!\!\times\!\!\text{ }\frac{\text{40}}{\text{60}}\text{=}\frac{\text{4 }\!\!\pi\!\!\text{ }}{\text{3}}\text{rad }{{\text{s}}^{\text{-1}}}\] The centripetal force is provided by the tension (T) in the string i.e. \[\text{T =}\frac{m{{v}^{2}}}{r}=mr{{\omega }^{2}}\] \[=0.25\times 1.5\times {{\left( \frac{4\pi }{3} \right)}^{2}}N=6.48N\approx 6.6N\] As the string can withstand a maximum tension of 200N. \[\therefore \] \[{{T}_{\max }}=\frac{mv_{\max }^{2}}{r}\] \[\Rightarrow \] \[{{v}_{\max }}=\sqrt{\frac{r{{T}_{\max }}}{m}}=\sqrt{\frac{1.5\times 200}{0.25}}\] \[\text{= 34}\text{.64 m}{{\text{s}}^{\text{-1}}}\text{k}\approx \text{35 m}{{\text{s}}^{\text{-1}}}\] \[T=6.6N,{{v}_{\max }}=35m{{s}^{-1}}\]
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