A van is moving with a speed of 72 km/h on a level road. Where the coefficient of friction between its tyres and road is 0.5. The minimum radius of curvature, that the road must have, for safe driving of van is (g=10ms-2) :

Question

A van is moving with a speed of 72 km/h on a level road. Where the coefficient of friction between its tyres and road is 0.5. The minimum radius of curvature, that the road must have, for safe driving of van is (g=10ms-2) : (A) 80 m (B) 40 m (C) 20 m (D) 4 m

Tardigrade Admin · Accepted Answer

From the relation, μ mg=(mv2/r) ⇒ r=(v2/μ g) (Given: v=72 km/h=72× (5/18)=20 m/s, μ =0.5,g=10 m/s2 ) =((20)2/0.5× 10)=(400/5)=80m

Q. A van is moving with a speed of 72 km/h on a level road. Where the coefficient of friction between its tyres and road is 0.5. The minimum radius of curvature, that the road must have, for safe driving of van is $(g = 10 m s^{- 2})$ :

80 m

40 m

20 m

4 m

From the relation, $μ m g = \frac{m v ^{2}}{r}$ $\Rightarrow$ $r = \frac{v ^{2}}{μg}$ (Given: $v = 72 km / h = 72 \times \frac{5}{18} = 20 m / s,$ $μ = 0.5, g = 10 m / s^{2}$ ) $= \frac{( 20 ) ^{2}}{0.5 \times 10} = \frac{400}{5} = 80 m$

Q. A van is moving with a speed of 72 km/h on a level road. Where the coefficient of friction between its tyres and road is 0.5. The minimum radius of curvature, that the road must have, for safe driving of van is (g=10ms−2) :

80 m

40 m

20 m

4 m

From the relation, μmg=rmv2​ ⇒ r=μgv2​ (Given: v=72km/h=72×185​=20m/s, μ=0.5,g=10m/s2 ) =0.5×10(20)2​=5400​=80m

Q. A van is moving with a speed of 72 km/h on a level road. Where the coefficient of friction between its tyres and road is 0.5. The minimum radius of curvature, that the road must have, for safe driving of van is $(g = 10 m s^{- 2})$ :

From the relation, $μ m g = \frac{m v ^{2}}{r}$ $\Rightarrow$ $r = \frac{v ^{2}}{μg}$ (Given: $v = 72 km / h = 72 \times \frac{5}{18} = 20 m / s,$ $μ = 0.5, g = 10 m / s^{2}$ ) $= \frac{( 20 ) ^{2}}{0.5 \times 10} = \frac{400}{5} = 80 m$