A car of 800 kg is taking turn on a banked road of radius 300 m and angle of banking 30 ° . If coefficient of static friction is 0.2 then the maximum speed with which car can negotiate the turn safely: ( g =10 m / s 2 , √(3)=1.73 )

Physics · Horizontal Circular Motion

JEE Main 2024 — 6 April, Shift 2 — Question 37

A car of 800 kg is taking turn on a banked road of radius 300 m and angle of banking $30^{\circ}$ . If coefficient of static friction is

0.2 then the maximum speed with which car can negotiate the turn safely: $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}, \sqrt{3}=1.73\right)$

$\mathrm{m}=800 \mathrm{~kg}$

$\mathrm{r}=300 \mathrm{~m}$

$\theta=30^{\circ}$

$\mu_{\mathrm{s}}=0.2$

$V_{\text {max }}=\sqrt{\operatorname{Rg}\left[\frac{\tan \theta+\mu}{1-\mu \tan \theta}\right]}$

$=\sqrt{300 \times \mathrm{g} \times\left[\frac{\tan 30^{\circ}+0.2}{1-0.2 \times \tan 30^{\circ}}\right]}$

$=\sqrt{300 \times 10 \times\left[\frac{0.57+0.2}{1-0.2 \times 0.57}\right]}$

$\mathrm{V}_{\max }=51.4 \mathrm{~m} / \mathrm{s}$

Answer key and solution verified before publishing.

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