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(B) The condition for a block to remain stationary on an inclined plane is that the angle of inclination $	heta$ must be less than or equal to the an

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$25$

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(B) The condition for a block to remain stationary on an inclined plane is that the angle of inclination $	heta$ must be less than or equal to the angle of repose $\alpha$,where $	an \alpha = \mu$.At the maximum height,the block is at the verge of slipping,so the slope of the tangent to the curve at that point is equal to the coefficient of friction.$	an 	heta = \frac{dy}{dx} = \mu$Given $y = \frac{x^2}{4}$,we have $\frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2}$.Setting $\frac{x}{2} = \mu = 0.5$,we get $x = 1 \ m$.Now,find the corresponding height $y$ at $x = 1 \ m$:$y = \frac{x^2}{4} = \frac{(1)^2}{4} = 0.25 \ m$.Since $1 \ m = 100 \ cm$,the height is $0.25 	imes 100 = 25 \ cm$.

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