TS EAMCET · Physics · Laws of Motion
A body of mass \(M \mathrm{~kg}\) is on the top point of a smooth hemisphere of radius \(5 \mathrm{~m}\). It is released to slide down the surface of the hemisphere. It leaves the surface when velocity is \(5 \mathrm{~m} / \mathrm{s}\). At this instant the angle made by the radius vector of the body with the vertical is (Acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
- A \(30^{\circ}\)
- B \(45^{\circ}\)
- C \(60^{\circ}\)
- D \(90^{\circ}\)
Answer & Solution
Correct Answer
(C) \(60^{\circ}\)
Step-by-step Solution
Detailed explanation
Suppose the body leaves the surface of the hemisphere at point \(P\). Let at point \(P\) radius vector of the body makes an angle \(\theta\). \[ \therefore \quad m g \cos \theta=\frac{m v^2}{r} \] When the body leaves the surface then…
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