A cyclist moves in a circular track of radius $100 m$ . If the coefficient of friction is $0.2$ , then the maximum speed with which the cyclist can take a turn without leaning inwards, in $m s^{- 1}$ , is:

9.8

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1.4

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14

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1.0

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Solution

The correct option is B $14$

$N = m g$

$μ N = \frac{m v^{2}}{R}$

$0.2 \times 9.8 = \frac{v^{2}}{100}$

$V = 14 m / s .$

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A cyclist moves in a circular track of radius 100 m. If the coefficient of friction is 0.2, then the maximum speed with which the cyclist can take a turn without leaning inwards, in ms−1, is:

The correct option is B 14 N=mg μN=mv2R 0.2×9.8=v2100 V=14 m/s.

A cyclist moves in a circular track of radius $100 m$ . If the coefficient of friction is $0.2$ , then the maximum speed with which the cyclist can take a turn without leaning inwards, in $m s^{- 1}$ , is:

The correct option is B $14$

$N = m g$

$μ N = \frac{m v^{2}}{R}$

$0.2 \times 9.8 = \frac{v^{2}}{100}$

$V = 14 m / s .$