JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A particle of mass \(20\,g\) is released with an initial velocity \(5\,m/s\) along the curve from the point \(A,\) as shown in the figure. The point \(A\) is at height \(h\) from point \(B.\) The particle slides along the frictionless surface. When the particle reaches point \(B,\) its angular momentum about \(O\) will be ......... \(kg - m^2/s\). [Take \(g = 10\,m/s^2\) ]
- A \(2\)
- B \(8\)
- C \(6\)
- D \(3\)
Answer & Solution
Correct Answer
(C) \(6\)
Step-by-step Solution
Detailed explanation
\(v = \sqrt {{5^2} + 2gh} = \sqrt {{5^2} + 2 \times 10 \times 10} = \sqrt {225} \) \( = 15\,m/s\) \(h = rmv\, = 20 \times \left( {20 \times {{10}^{ - 3}}kg} \right) \times \left( {15} \right)\) \( = 6\,kg\,{m^2}/\sec \)
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