$A$ car turns a corner on a slippery road at a constant speed of $10\,m/s$. If the coefficient of friction is $0.5$,the minimum radius of the arc in meters in which the car turns is:

$A$ car of $800 \,kg$ is taking a turn on a banked road of radius $300 \,m$ and angle of banking $30^{\circ}$. If the coefficient of static friction is $0.2$, then the maximum speed with which the car can negotiate the turn safely is: $(g=10 \,m/s^2, \sqrt{3}=1.73)$ (in $\,m/s$)

$A$ 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:

$A$ coin placed on a rotating table just slips if it is placed at a distance $4r$ from the centre. On doubling the angular velocity of the table,the coin will just slip when at a distance from the centre equal to

$A$ vehicle is moving with speed $v$ on a curved road of radius $r$. The coefficient of friction between the vehicle and the road is $\mu$. The angle $\theta$ of banking needed is given by

$A$ 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 $m$)