The radius of a curved road is $R$,and the width of the road is $b$. The outer edge of the road is raised by $h$ with respect to the inner edge so that a car with velocity $V$ can pass safely over it. The value of $h$ is ($g =$ acceleration due to gravity).

The maximum velocity (in $m/s$) with which a car driver must traverse a flat curve of radius $150 \, m$ and coefficient of friction $0.6$ to avoid skidding is

$A$ motor vehicle of mass $1000 \,kg$ is moving on a circular road having a banking angle of $30^{\circ}$ and a coefficient of friction of $0.2$. Then the normal reaction force on the motor vehicle is about (Acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,N$)

To mop-clean a floor,a cleaning machine presses a circular mop of radius $R$ vertically down with a total force $F$ and rotates it with a constant angular speed about its axis. If the force $F$ is distributed uniformly over the mop and the coefficient of friction between the mop and the floor is $\mu$,the torque applied by the machine on the mop is

The normal reaction $N$ for a vehicle of $800 \, kg$ mass,negotiating a turn on a $30^{\circ}$ banked road at maximum possible speed without skidding is $... \times 10^{3} \, kg \cdot m/s^{2}$. [Given $\cos 30^{\circ} = 0.87, \mu_{s} = 0.2$]

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?