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The slope of the smooth banked horizontal road is p. If the radius of curve be r, then the maximum velocity with which a car can negotiate the curve is given by

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$\sqrt{(p / rg)}$

$\sqrt{(prg)}$

p r g

p / rg

answer is B.

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Detailed Solution

If $θ$ be the angle of banking, then slope $= \tan θ = p \dots (1)$

For smooth negotiation of the curve

$\tan θ = v^{2} / rg$ ….. (2)

$∴ p = \frac{V^{2}}{rg} or v = \sqrt{(prg)}$

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