A bus turns a slippery road having coefficient of friction of 0.5 with a speed of 10 m / s. The minimum radius of the arc in which bus turns is [. Take .g=10 m / s 2]

Question

A bus turns a slippery road having coefficient of friction of 0.5 with a speed of 10 m / s. The minimum radius of the arc in which bus turns is [. Take .g=10 m / s 2] (A) 4 m (B) 10 m (C) 15 m (D) 20 m

Tardigrade Admin · Accepted Answer

(v2/r)=μ g (10 × 10/r)=0.5 × 10 r=20 m

Q. A bus turns a slippery road having coefficient of friction of $0.5$ with a speed of $10 m / s$ . The minimum radius of the arc in which bus turns is $[$ Take $g = 10 m / s^{2}]$

4 m

10 m

15 m

20 m

$\frac{v ^{2}}{r} = μg$
$\frac{10 \times 10}{r} = 0.5 \times 10$
$r = 20 m$

Q. A bus turns a slippery road having coefficient of friction of 0.5 with a speed of 10m/s. The minimum radius of the arc in which bus turns is [ Take g=10m/s2]

4 m

10 m

15 m

20 m

rv2​=μg r10×10​=0.5×10 r=20m

Q. A bus turns a slippery road having coefficient of friction of $0.5$ with a speed of $10 m / s$ . The minimum radius of the arc in which bus turns is $[$ Take $g = 10 m / s^{2}]$

$\frac{v ^{2}}{r} = μg$
$\frac{10 \times 10}{r} = 0.5 \times 10$
$r = 20 m$