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A cyclist is moving in a circular track of radius $80 \mathrm{~m}$ with a velocity of $36 \mathrm{~km} \mathrm{~h}^{-1}$. He has to lean from the vertical approximately through an angle (Take $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ )
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The correct answer is:
$\tan ^{-1}(1 / 8)$
$\begin{array}{r}\text { Here, } r=80 \mathrm{~m}, v=36 \mathrm{kmh}^{-1}=36 \times(5 / 18) \mathrm{ms}^{-1} \\ =10 \mathrm{~m} \mathrm{~s}^{-1}\end{array}$
$\therefore \quad \tan \theta=\frac{v^2}{r g}=\frac{(10)^2}{80 \times 10}=\frac{1}{8}$
or $\theta=\tan ^{-1}\left(\frac{1}{8}\right)$
$\therefore \quad \tan \theta=\frac{v^2}{r g}=\frac{(10)^2}{80 \times 10}=\frac{1}{8}$
or $\theta=\tan ^{-1}\left(\frac{1}{8}\right)$
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