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A particle of mass 100 \mathrm{~g} is projected at timet=0 with a speed 20 \mathrm{~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 \mathrm{~s} is found to be \sqrt{\mathrm{K}} \mathrm{kgm}^2 / \mathrm{s}. The value of \mathrm{K} is ____________.\left(\right.$ Take $\left.g=10 \mathrm{~ms}^{-2}\right)

Option: 1

800


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \text { Use } \Delta \mathrm{L}=\int_0^t \tau \mathrm{dt} \\ & \mathrm{L}_0=\int_0^2(\mathrm{mg})\left(\mathrm{v}_{\mathrm{x}} \mathrm{t}\right) \mathrm{dt} \\ & =\left(\mathrm{mgv}_{\mathrm{x}}\right) \frac{\mathrm{t}^2}{2} \\ & =(0.1)(10)(10)(\sqrt{2}) \times \frac{2^2}{2} \\ & =20 \sqrt{2} \\ & =\sqrt{800} \end{aligned}

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rishi.raj

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