The normal reaction $'{N}^{\prime}$ for a vehicle of $800\, {kg}$ mass, negotiating a turn on a $30^{\circ}$ banked road at maximum possible speed without skidding is $...\,\times 10^{3}\, {kg} {m} / {s}^{2}$ [Given $\left.\cos 30^{\circ}=0.87, \mu_{{s}}=0.2\right]$
JEE MAIN 2021, Difficult
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At $V_{\max }, f$ will be limiting in nature.
$\therefore$ Balancing force in vertical direction,
$N \cos 30^{\circ}-{mg}-\mu {N} \cos 60^{\circ}=0$
$\Rightarrow {N}\left[\cos 30^{\circ}-\mu \cos 60^{\circ}\right]={mg}$
$\therefore {N}=\frac{800 \times 10}{(0.87-0.1)} \approx 10.2 \times 10^{3} {kgm} / {s}^{2}$
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