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A parabolic bowl with its bottom at origin has the shape $y = x^{2} / 20$ . Here x and y are in metres. The maximum height at which a small mass m can be placed on the bowl without slipping (coefficient of static friction is 0.5) is

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2.5 m

1.25 m

1.0 m

4.0 m

answer is B.

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Detailed Solution

$\frac{dy}{dx} = \frac{x}{10}$

$or \tan θ = \frac{x}{10}$ ……..(1)

This angle should be angle of repose.

$or \tan θ = μ = \frac{1}{2}$ ……(2)

From Eqs. (1) and (2),

$\begin{array}{l} \frac{x}{10} = \frac{1}{2} or x = 5 m \\ ∴ y = \frac{x^{2}}{20} = \frac{25}{20} = 1 .25 m \end{array}$

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