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A circular freeway entrance and exit are commonly banked to control a moving car at $14 \mathrm{~m} / \mathrm{s}$. To design similar ramp for $28 \mathrm{~m} / \mathrm{s}$ one should
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The correct answer is:
increase the radius by factor 4
Given, $v_1=14 \mathrm{~m} / \mathrm{s}$ and $v_2=28 \mathrm{~m} / \mathrm{s}$
As we know that, $\tan \theta=\frac{v^2}{r g} \Rightarrow r=\frac{v^2}{g \tan \theta}$
In the first case, $r_1=\frac{v_1^2}{g \tan \theta}=\frac{(14)^2}{g \tan \theta}$ $\ldots$ (i)
In the second case,
$r_2=\frac{(28)^2}{g \tan \theta}$ [for similar ramp remains same]
$r_2=\frac{(14 \times 2)^2}{g \tan \theta}=4 \frac{(14)^2}{g \tan \theta} \Rightarrow r_2=4 r_1 \quad \text { [using Eq. (i)] }$
Hence, radius should be increased by factor 4 .
As we know that, $\tan \theta=\frac{v^2}{r g} \Rightarrow r=\frac{v^2}{g \tan \theta}$
In the first case, $r_1=\frac{v_1^2}{g \tan \theta}=\frac{(14)^2}{g \tan \theta}$ $\ldots$ (i)
In the second case,
$r_2=\frac{(28)^2}{g \tan \theta}$ [for similar ramp remains same]
$r_2=\frac{(14 \times 2)^2}{g \tan \theta}=4 \frac{(14)^2}{g \tan \theta} \Rightarrow r_2=4 r_1 \quad \text { [using Eq. (i)] }$
Hence, radius should be increased by factor 4 .
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