Rotational Motion

JEE Main level, numerical answer, medium

204 of 280

Question 204 of 280

A body of mass 5kg5kg moving with a uniform speed 32ms−13\sqrt{2}m{s}^{-1} in X−YX-Y plane along the line y=x+4y=x+4. The angular momentum of the particle about the origin will be _______kgm2s−1kg{m}^{2}{s}^{-1}.

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Answer: 60

Equation of line given: y−x−4=0y-x-4=0.
Let d1{d}_{1} be the perpendicular distance of given line from origin.
d1=∣−412+12∣=22{d}_{1}=\left|\dfrac{-4}{\sqrt{{1}^{2}+{1}^{2}}}\right|=2\sqrt{2}
So, ∣L⃗∣=mvd1=5×32×22kgm2s−1|\vec{L}|=mv{d}_{1}=5\times 3\sqrt{2}\times 2\sqrt{2}kg{m}^{2}{s}^{-1}
=60kgm2s−1=60kg{m}^{2}{s}^{-1}

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