AP EAMCET · PHYSICS · Work Power Energy
A uniform chain of length \(l\) and mass \(m\) lies on the surface of a smooth hemisphere of radius \(R(R>l)\) with one end tied to the top the hemisphere as shown in the figure.
Gravitational potential energy of the chain with respect to the base of the hemisphere is
- A \(\frac{m g l}{2}\)
- B \(\frac{m g R^2}{l} \sin \left(\frac{I}{R}\right)\)
- C \(\frac{m g R^2}{l} \sin \left(\frac{R}{l}\right)\)
- D \(\frac{m g l^2}{R} \sin \left(\frac{l}{R}\right)\)
Answer & Solution
Correct Answer
(B) \(\frac{m g R^2}{l} \sin \left(\frac{I}{R}\right)\)
Step-by-step Solution
Detailed explanation
We have, \(\quad \frac{h}{R}=\sin \theta\) or \(h=R \sin \theta\) Also, \(d l=R d \theta\) Mass of \(d l\) length of chain \(=d m=\frac{m}{l} \cdot d l\) PE of \(d m\) mass…
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