If a cyclist moving with a speed of 4.9 m/s on a level road can take a sharp circular turn of radius 4 m, then coefficient of friction between the cycle tyres and road is

0.41

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0.51

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0.61

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0.71

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Solution

The correct option is C
0.61

$μ = \frac{v^{2}}{r g} = \frac{(4.9)^{2}}{4 \times 9.8} = 0.61$

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If a cyclist moving with a speed of 4.9 m/s on a level road can take a sharp circular turn of radius 4 m, then coefficient of friction between the cycle tyres and road is

The correct option is C 0.61 μ=v2rg=(4.9)24×9.8=0.61

The correct option is C
0.61

$μ = \frac{v^{2}}{r g} = \frac{(4.9)^{2}}{4 \times 9.8} = 0.61$