JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A particle of mass \('{m}'\) is moving in time \('t'\) on a trajectory given by \(\overrightarrow{{r}}=10 \alpha {t}^{2}\, \hat{{i}}+5 \beta({t}-5)\, \hat{{j}}\) Where \(\alpha\) and \(\beta\) are dimensional constants. The angular momentum of the particle becomes the same as it was for \({t}=0\) at time \({t}=\) .....\(seconds.\)
- A \(15\)
- B \(10\)
- C \(20\)
- D \(25\)
Answer & Solution
Correct Answer
(B) \(10\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ r }=10 \alpha t ^{2} \hat{ i }+5 \beta( t -5) \hat{ j }\) \(\overrightarrow{ v }=20 \alpha t \hat{ i }+5 \beta \hat{ j }\) \(\overrightarrow{ L }= m (\overrightarrow{ r } \times \overrightarrow{ v })\)…
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