JEE PYQ: Laws Of Motion Question 12
Question 12 - 2021 | 24 Feb Shift 1] (Numerical)
An inclined plane is bent in such a way that the vertical cross-section is given by $y = \frac{x^2}{4}$ where $y$ is in vertical and $x$ in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction $\mu = 0.5$, the maximum height in cm at which a stationary block will not slip downward is ____ cm.
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Answer: 25
Solution
$\tan\theta = \mu$. $\frac{dy}{dx} = \frac{x}{2} = 0.5$. $x = 1$. $y = \frac{1}{4} = 0.25$ m $= 25$ cm.
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