The normal reaction N for a vehicle of 800 kg mass, negotiating a turn on a 30° banked road at maximum possible speed without skidding is ……$\times {10}^3 \mathrm{kg}-\mathrm{m}/{\mathrm{s}}^2$. $\left[\text{ Take, }{\mathrm{\mu }}_{\mathrm{s}}=0.2\right]$

The given situation can be represented as

Equating forces perpendicular to the inclined plane,

$\mathrm{N}=\mathrm{mgcos}{30}^{\circ }+\frac{{\mathrm{mv}}^2}{\mathrm{R}}\sin {30}^{\circ }$

$\Rightarrow \mathrm{N}-\mathrm{mgcos}{30}^{\circ }=\frac{{\mathrm{mv}}^2}{\mathrm{R}}\sin {30}^{\circ }$         ...(i)

Equating forces along the inclined plane,

$\mathrm{mgsin}{30}^{\circ }+{\mathrm{\mu }}_{\mathrm{s}}\mathrm{N}=\frac{{\mathrm{mv}}^2}{\mathrm{R}}\cos {30}^{\circ }$          ...(ii)

On dividing Eq. (i) by Eq. (ii), we get

$\begin{array}{l}\frac{\mathrm{N}-\mathrm{mgcos}{30}^{\circ }}{\mathrm{mgsin}{30}^{\circ }+{\mathrm{\mu }}_{\mathrm{s}}\mathrm{N}}=\tan {30}^{\circ }\left[\because \cos {30}^{\circ }=\frac{\sqrt{3}}{2}\text{ and }\sin {30}^{\circ }=\frac{1}{2}\right]\\ \Rightarrow \frac{\mathrm{N}-\mathrm{mg}(\sqrt{3}/2)}{\mathrm{mg}(1/2)+(0.2)\mathrm{N}}=\frac{1}{\sqrt{3}}\\ \mathrm{N}\sqrt{3}-\mathrm{mg}\frac{\sqrt{3}}{2}\cdot \sqrt{3}=\frac{\mathrm{mg}}{2}+0.2\mathrm{N}\\ \Rightarrow (\sqrt{3}-0.2)\mathrm{N}=\mathrm{mg}\frac{(1+3)}{2}=2\mathrm{mg}\end{array}$

$\begin{array}{c}\Rightarrow \mathrm{N}=\frac{2\mathrm{mg}}{\sqrt{3}-0.2}=\frac{2\times 800\times 10}{1.532}\\ \simeq 10.2\times {10}^3\mathrm{N}\end{array}$