For a banked curved road,if the velocity of the vehicle $v < v_0$ (where $v_0$ is the optimum speed),what is the direction of the frictional force?

$A$ body of mass $200 \text{ g}$ is tied to a spring of spring constant $12.5 \text{ N/m}$,while the other end of the spring is fixed at point '$O$'. If the body moves about '$O$' in a circular path on a smooth horizontal surface with a constant angular speed of $5 \text{ rad/s}$,then the ratio of the extension in the spring to its natural length will be:

$A$ car is driven on a banked road of radius of curvature $20 \ m$ with maximum safe speed. In order to increase its safe speed by $20 \%$,without changing the angle of banking,the increase in the radius of curvature will be (Assume friction is same on the road). (in $m$)

$A$ body of mass $M \text{ kg}$ is at the top point of a smooth hemisphere of radius $5 \text{ m}$. It is released to slide down the surface of the hemisphere. It leaves the surface when its velocity is $5 \text{ m/s}$. At this instant, the angle made by the radius vector of the body with the vertical is (Acceleration due to gravity $g = 10 \text{ m/s}^2$) (in $^{\circ}$)

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?

$A$ 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 $\,N$)