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Question

A particle having mass m=2 kg is moving with velocity (2^i+3^j) m/s. Find angular momentum of the particle about the origin when it is at (1,1,0) [in kg m2/s].

A
2^k
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B
4^k
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C
6^k
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D
Zero
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Solution

The correct option is A 2^k
Given,
Mass of particle m=2 kg
Velocity of particle v=(2^i+3^j) m/s
Position vector about origin r=(10)^i+(10)^j+(00)^k
r=(^i+^j)m
As we know,
Angular momentum (L)=m(r×v)
L=2[(^i+^j)×(2^i+3^j)]
L=2[3(^i×^j+2(^j×^i))]kg m2s
[^i×^i=^j×^j=0]
L=2×[3^k+2(^k)]
L=2^k kg m2/s

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