The coefficient of friction between the tyres and the road is $0.25$. The maximum speed with which a car can be driven round a curve of radius $40 \,m$ without skidding is ........ $ms^{-1}$ (assume $g = 10 \,ms^{-2}$)

$A$ circular road of radius $1000\, m$ has a banking angle of $45^o$. The maximum safe speed of a car having a mass of $2000\, kg$ will be,if the coefficient of friction between the tyre and the road is $0.5$.

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 $\cos 30^{\circ} = 0.87, \mu_{s} = 0.2$]

$A$ 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 $^{\circ}$)

$A$ modern grand-prix racing car of mass $m$ is travelling on a flat track in a circular arc of radius $R$ with a speed $v$. If the coefficient of static friction between the tyres and the track is $\mu_{s},$ then the magnitude of negative lift $F_{L}$ acting downwards on the car is (Assume forces on the four tyres are identical and $g =$ acceleration due to gravity)

$A$ motor car has a width of $1.1 \ m$ between its wheels. Its center of gravity is $0.62 \ m$ above the ground and the coefficient of friction between the wheels and the road is $0.8$. What is the maximum possible speed in $m/s$ if the center of gravity moves in a circle of radius $15 \ m$ on a horizontal road surface?