If a cyclist moving with a speed of $4.9$ m/s on leveled road can take a sharp circular turn of radius $4$ m, then the coefficient of friction between cycle tyre and road will be

0.81

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0.41

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0.71

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0.61

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Solution

The correct option is D 0.61

For movement on circular path,centripetal force = force of friction

$\frac{{m v}^{2}}{r} = μ m g$

$v^{2} = μ r g \Rightarrow μ = \frac{v^{2}}{r g} = \frac{4.9 \times 4.9}{10 \times 4} = 0.61$

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

The correct option is D 0.61 For movement on circular path,centripetal force = force of friction mv2r=μmg v2=μrg⇒μ=v2rg=4.9×4.910×4=0.61

If a cyclist moving with a speed of $4.9$ m/s on leveled road can take a sharp circular turn of radius $4$ m, then the coefficient of friction between cycle tyre and road will be

The correct option is D 0.61

For movement on circular path,centripetal force = force of friction

$\frac{{m v}^{2}}{r} = μ m g$

$v^{2} = μ r g \Rightarrow μ = \frac{v^{2}}{r g} = \frac{4.9 \times 4.9}{10 \times 4} = 0.61$