Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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

0.51

0.61

0.71

Step-by-Step Solution

Principle: For a cyclist to negotiate a circular turn on a level road without skidding, the necessary centripetal force is provided by the force of static friction between the tyres and the road [Source 72].
Formula: The limiting condition where the maximum static friction balances the centripetal force is: $\mu m g = \frac{m v^2}{R}$ Rearranging for the coefficient of friction ( $\mu$ ): $\mu = \frac{v^2}{Rg}$
Calculation:

Speed ( $v$ ) = $4.9 \text{ m/s}$
Radius ( $R$ ) = $4 \text{ m}$
Acceleration due to gravity ( $g$ ) $\approx 9.8 \text{ m/s}^2$ [Source 73]. Substitute the values: $\mu = \frac{(4.9)^2}{4 \times 9.8}$ $\mu = \frac{4.9 \times 4.9}{4 \times 2 \times 4.9}$ $\mu = \frac{4.9}{8} = 0.6125$ Rounding to two decimal places, $\mu \approx 0.61$ .

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