If the coefficient of friction between the tires and the road is $\mu$,the maximum safe speed is $10\;m/s$. If the coefficient of friction becomes $\mu' = \frac{\mu}{2}$,what will be the new maximum safe speed?

$A$ block $(P)$ is rotating in contact with the vertical wall of a rotor as shown in figures $A$,$B$,and $C$. Find the relation between angular velocities $\omega_A, \omega_B$,and $\omega_C$ such that the block does not slide down. ($R_A < R_B < R_C$ are the radii).

$A$ motorcyclist of mass $m$ is to negotiate a curve of radius $r$ with a speed $v$. The minimum value of the coefficient of friction so that the negotiation may take place safely is

$A$ stone of mass $1\,kg$ is tied to the end of a massless string of length $1\,m$. If the breaking tension of the string is $400\,N$,then the maximum linear velocity the stone can have without breaking the string,while rotating in a horizontal plane,is $.......\,ms^{-1}$.

$A$ car is moving on a horizontal circular track of radius $0.2 \, km$ with a constant speed. If the coefficient of friction between the tyres of the car and the road is $0.45$,then the maximum safe speed of the car is ........ $m/s$. [Take $g = 10 \, m/s^2$]

$A$ circular path of radius $75 \ m$ is banked at an angle of $\tan^{-1}(0.2)$. If the coefficient of static friction between the tyres of the car and the circular path is $0.1$,then the maximum permissible speed of the car to avoid slipping is (in $m/s$)