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A particle of mass 1kg is subjected to a force which depends on the position as F=-k(xi+yj)kg-m/s^2. At time t=0, the particle’s position r={[1/SQRT(2)]i+SQRT(2)j}m and its velocity v={SQRT(2)i+SQRT(2)j+[2/(pi)]k}m/s. Let v(x) and v(y) denote the x and y components of the particle’s velocity, respectively. Ignore gravity. When z=0.5m. The value of [xv(y)-yv(x)]m is _______ (m^2)/2.

Previous video (JEE Advance 2021 Paper 1) in this series can be seen at: https://youtu.be/McoKufY7ARw
Title Search on Other Platforms: JEE Advanced Physics 2022 Paper 1: #18 Multi-Lens Problems Part 4

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Title Search on Other Platforms: JEE Advanced Physics 2022 Paper 2: #2 Radioactive Deay