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Q. A car moves at a speed of $20 \; ms^{-1}$ on a banked track and describes an arc of a circle of radius $40 \; \sqrt{3}$ m. The angle of banking is $(g = 10 \; ms^{-2})$

KEAMKEAM 2015Laws of Motion

Solution:

Given, $V=20\, ms ^{-1}$
$r=40 \sqrt{3}\, m$
$\Rightarrow g=10\, ms ^{-2}$
We know that, $\tan \theta=\frac{v^{2}}{r g}$
$\Rightarrow \tan \theta=\frac{(20)^{2}}{40 \sqrt{3} \times 10}$
$\Rightarrow \tan \theta=\frac{1}{\sqrt{3}}$ or $\theta=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$\Rightarrow \theta=30^{\circ}$