A) \[40\,\,m{{s}^{-1}}\]
B) \[20\,\,m{{s}^{-1}}\]
C) \[15\,\,m{{s}^{-1}}\]
D) \[10\,\,m{{s}^{-1}}\]
Correct Answer: D
Solution :
Key Idea: Force of friction provides the necessary centripetal force. For turning-around a curve, centripetal force is provided by friction, so \[\frac{m{{v}^{2}}}{r}<f\] \[\frac{m{{v}^{2}}}{r}<\mu mg\] \[\Rightarrow \] \[v<\sqrt{\mu gr}\], So that \[{{v}_{\max }}=\sqrt{\mu gr}\] Substituting, \[\mu =0.25\], \[r=40\,m\], \[g=10\,\,m{{s}^{-2}}\], we get \[{{v}_{\max }}=10\,\,m{{s}^{-1}}\]
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