1. For maximum speed on a banked road

At the maximum possible speed without skidding, the vehicle tends to slip up the bank, so static friction acts down the slope.

Forces on the vehicle:

• Weight: $mg$ downward
• Normal reaction: $N$ perpendicular to road
• Friction: $f=\mu N$ down the plane

Given:

• $m=800\,\text{kg}$
• $\theta=30^\circ$
• $\mu=0.2$
• $\cos 30^\circ=0.87$
• $\sin 30^\circ=0.5$
• Take $g=10\,\text{m/s}^2$

2. Vertical equilibrium

Since there is no vertical acceleration, $N\cos\theta - f\sin\theta = mg$ Because $f=\mu N$, $N\cos\theta - \mu N\sin\theta = mg$ $N(\cos\theta - \mu\sin\theta)=mg$

Substitute values: $N(0.87-0.2\times 0.5)=800\times 10$ $N(0.87-0.1)=8000$ $N(0.77)=8000$ $N=\frac{8000}{0.77}\approx 10389.6\,\text{N}$

3. Express in the asked form

They ask for $\underline{\hspace{2cm}}\times 10^3\,\text{kg m/s}^2$

So, $N\approx 10.39\times 10^3\,\text{N}$

Closest option is: $\boxed{10.2}$

4. Option check

• A: $12.4$ ❌
• B: $7.2$ ❌
• C: $6.96$ ❌
• D: $10.2$ ✅

Hence, the correct option is D.