A particle having mass m=2kg 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 A2^k Given,
Mass of particle m=2kg
Velocity of particle →v=(2^i+3^j)m/s
Position vector about origin →r=(1−0)^i+(1−0)^j+(0−0)^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^kkg m2/s