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Q. A particle of mass $15\, kg$ is moving with a uniform speed $8\, ms ^{-1}$ in $x y$ -plane along the line $3 y=4\,x+10$, then the magnitude of its angular momentum about the origin in
$kg - m ^{2} s ^{-1}$ is ... $\left(\sin 53^{\circ}=\frac{4}{5}\right)$

AP EAMCETAP EAMCET 2018

Solution:

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Angular momentum $=m v r_{1}$
$=15 \times 8 \times\left(\frac{10}{\sqrt{3^{2}+4^{2}}}\right)$
$=15 \times 8 \times 2=240\, kg - m ^{2} s ^{-1}$