A car is moving on a circular road of radius of curvature 500m. If the coefficient of friction is 0.8 and g=10m/s2, the maximum speed of the car can have without slipping is in (km/hr).
A
5√10
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
20√10
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
72√10
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
54√10
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C72√10 Given, radius of curvature (r)=500m Coefficient of friction (μ)=0.8
Condition for max speed of the car for not slipping is,
μmg=mv2maxr, here the frictional force will provide the necessary centripetal force ⇒(vmax)=√μrg=√0.8×500×10=√45×500×10 =√4×100×10 =20√10m/s or 72√10km/hr (As to convert (m/s) to (km/hr) we multiply it by 185)