$A$ 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 $10 \%$,the increase in the radius of curvature will be (Angle of banking and friction is unchanged in both the cases.) (in $m$)

$A$ car is moving on a circular track banked at an angle of $45^{\circ}$. If the maximum permissible speed of the car to avoid slipping is twice the optimum speed of the car to avoid the wear and tear of the tyres,then the coefficient of static friction between the wheels of the car and the road is

$A$ motorcyclist moving with a velocity of $72\, km/h$ on a flat road takes a turn at a point where the radius of curvature of the road is $20\, m$. The acceleration due to gravity is $10\, m/s^2$. In order to avoid skidding,he must not bend with respect to the vertical plane by an angle greater than

$A$ car moves on a horizontal circular road of radius $16 \ m$ with an increasing speed at a constant rate of $3 \ m \ s^{-2}$. If the coefficient of friction between the road and the tyres is $0.5$,then the speed at which the car will skid is (assume $g = 10 \ m \ s^{-2}$): (in $m \ s^{-1}$)

$A$ small mass $m$ rests at the edge of a horizontal disc of radius $R$. The coefficient of static friction between the mass and the disc is $\mu$. The disc is rotated about its axis at an angular velocity such that the mass slides off the disc and lands on the floor $h$ meters below. What was its horizontal distance of travel from the point it left the disc?

$A$ 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 $^{\circ}$)