A) 0.2
B) 0.4
C) 0.6
D) 0.8
Correct Answer: D
Solution :
Here, \[m=1500\text{ }kg,\text{ }v=12.5\text{ }m{{s}^{-}}^{1}\] \[r\,\,=\,\,20\,m\] Friction force = centripetal force required \[F=\frac{m{{v}^{2}}}{r}\,\,=\,\,\frac{1500{{(12.5)}^{2}}}{20}\,=\,\,1.172\times {{10}^{4}}N\] \[As\,\,\,F=\mu R=\mu \,mg\] \[\mu \,\,=\,\,\frac{F}{mg}\,\,=\,\,\frac{1.172\times {{10}^{4}}}{1500\times 9.8}\,\,=\,\,0.8\,\]You need to login to perform this action.
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