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JEE Main 202429 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Evening Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2024 · 29 January · Evening Shift · Physics · Rotational Motion

A body of mass 5kg moving with a uniform speed 3 2 m s ^ - 1 in X -...

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}.

Correct answer

0

Step-by-step solution

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, ∣L→∣=mvd1=5×32×22kgm2s−1| \overset{\to}{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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