JEE Mains · Physics · STD 11 - 3.2 motion in plane
A particle of mass \(100\,g\) is projected at time \(t =0\) with a speed \(20\,ms ^{-1}\) at an angle \(45^{\circ}\) to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time \(t=2\,s\) is found to be \(\sqrt{ K }\,kg\,m ^2 / s\). The value of \(K\) is \(............\) \(\left(\right.\) Take \(\left.g =10\,ms ^{-2}\right)\)

- A \(80\)
- B \(800\)
- C \(8\)
- D \(0.8\)
Answer & Solution
Correct Answer
(B) \(800\)
Step-by-step Solution
Detailed explanation
\(\text { Use } \Delta L =\int \limits_0^{ t } \tau dt\) \(L _0=\int \limits_0^2 mg \left( v _{ x } t \right) dt\) \(= mg _{ x } \frac{ t ^2}{2}=(0.1)(10)(10 \sqrt{2}) \frac{2^2}{2}\) \(=20 \sqrt{2}\) \(=\sqrt{800}\,kgm ^2 / s\)
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