ExamBro
ExamBro
JEE Mains · Physics · STD 11 - 4.2 friction

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............\(cm\)

  1. A \(20\)
  2. B \(25\)
  3. C \(16\)
  4. D \(30\)
Verified Solution

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…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app