The coefficient of friction between the tyres and the road is 0.1. The maximum speed with which a cyclist can take a circular turn of radius 3 m without skidding is

(Take g = 10 m $s^{- 2}$ )

Question

The coefficient of friction between the tyres and the road is 0.1. The maximum speed with which a cyclist can take a circular turn of radius 3 m without skidding is

(Take g = 10 m $s^{- 2}$ )

Smart Achievers · Accepted Answer

(2)

Here, $r = 3 m, μ = 0.1, g = 10 {m s}^{- 2}$

The maximum speed with which a cyclist can take a turn without skidding is

$= \sqrt{μ r g} = \sqrt{0.1 \times (3 m) (10 {m s}^{- 2})} = \sqrt{3} {m s}^{- 1}$

Solution

Q.
The coefficient of friction between the tyres and the road is 0.1. The maximum speed with which a cyclist can take a circular turn of radius 3 m without skidding is

(Take g = 10 m $s^{- 2}$ )

1 $\sqrt{15} {m s}^{- 1}$

2 $\sqrt{3} {m s}^{- 1}$

3 $\sqrt{30} {m s}^{- 1}$

4 $\sqrt{10} {m s}^{- 1}$

Ans.
(2)

Here, $r = 3 m, μ = 0.1, g = 10 {m s}^{- 2}$

The maximum speed with which a cyclist can take a turn without skidding is

$= \sqrt{μ r g} = \sqrt{0.1 \times (3 m) (10 {m s}^{- 2})} = \sqrt{3} {m s}^{- 1}$

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Solution

Q. The coefficient of friction between the tyres and the road is 0.1. The maximum speed with which a cyclist can take a circular turn of radius 3 m without skidding is (Take g = 10 m s-2)

1 15 m s-1

2 3 m s-1

3 30 m s-1

4 10 m s-1

Ans. (2) Here, r=3 m, μ=0.1, g=10 m s-2 The maximum speed with which a cyclist can take a turn without skidding is =μrg=0.1×(3 m)(10 m s-2)=3 m s-1