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A lorry is moving on a smooth circular path of radius 50 $\mathrm{m}$ with a velocity of $20 \mathrm{~ms}^{-1}$. Then the banking angle of the road is
(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
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(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
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The correct answer is:
$\tan ^{-1} \frac{4}{5}$
Radius of circular path, $R=50 \mathrm{~m}$ Velocity, $V=20 \mathrm{~m} / \mathrm{s}$
Banking angle of the road, $\tan \theta=\frac{V^2}{R g}$
$\theta=\tan ^{-1}\left(\frac{20^2}{50 \times 10}\right)=\tan ^{-1}\left(\frac{4}{5}\right)$
Banking angle of the road, $\tan \theta=\frac{V^2}{R g}$
$\theta=\tan ^{-1}\left(\frac{20^2}{50 \times 10}\right)=\tan ^{-1}\left(\frac{4}{5}\right)$
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