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

$A$ car is moving on a horizontal circular track of radius $0.2 \, km$ with a constant speed. If the coefficient of friction between the tyres of the car and the road is $0.45$,then the maximum safe speed of the car is ........ $m/s$. [Take $g = 10 \, m/s^2$]

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

$A$ 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 $^{\circ}$)

$A$ vehicle is moving with a velocity $v$ on a curved road of width $b$ and radius of curvature $R$. For counteracting the centrifugal force on the vehicle,the difference in elevation required between the outer and inner edges of the road is

On a dry road,the maximum speed of a vehicle along a circular path is $V$. When the road becomes wet,the maximum speed becomes $\frac{V}{2}$. If the coefficient of friction of the dry road is $\mu$,then the coefficient of friction of the wet road is: