For a particle of a mass 100 gm, position and velocity at any instant are given as 10i^+6j^ cm and v→ = 5i^ cm/s. Calculate the angular momentum about the point (1, 1) cm.

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25 × 10-5(-k^) kgm2s-1

30× 10-5(-k^) kgm2s-1

3 × 10-3(-k^)kgm2s-1

25× 10-4(-k^) kgm2s-1

answer is A.

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Detailed Solution

The position vector of particle w.r.t. (1,1) is r→=10-1i^+6-1j^ = 9i^+5j^L→ = m(r→ × v→) = 1001000[9i^+5j^100]×5i^100= 25 ×10-5(-k^) kgm2s-1

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