JEE Mains · Physics · STD 11 - Friction
- JEE Mains 2021
- Hard
- 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............\(cm\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(25\)
Step-by-step Solution
Detailed explanation
At maximum ht. block will experience maximum friction force. Therefore if at this height slope of the tangent is \(\tan \theta,\) then \(\theta=\) Angle of repose.
\(\therefore \tan \theta=\frac{d y}{d x}=\frac{2 x}{4}=\frac{x}{2}=0.5\)
\(\Rightarrow x=1\) and therefore \(y=\frac{x^{2}}{4}=0.25 m\)
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