AP EAMCET · PHYSICS · Work Power Energy
A small body slides down a smooth uneven surface from a height \(H\), which eventually emerges into a circular loop of radius \(R( < H)\). What should be the value of \(H\), so that the force on the body at \(A\) is \(\sqrt{2}\) times its weight?
- A \(H=\frac{3 R}{2}\)
- B \(H=5 R\)
- C \(H=\frac{5 R}{2}\)
- D \(H=3 R\)
Answer & Solution
Correct Answer
(A) \(H=\frac{3 R}{2}\)
Step-by-step Solution
Detailed explanation
Given, height through which body slides \(=H\) Radius of circular loop \(=R\) Force on body \(A, F_A=\sqrt{2}\) weight \((w)\) Let, velocity of body at \(A=v_A\) Acceleration due to gravity \(=g\)…
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