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Standard XII

Physics

Angular Momentum

A particle of...

Question

A particle of mass 6 kg is moving along y = x + 4. Speed of particle is 4 m/s. Areal velocity (in $m^{2} / s$ ) of position vector of particle with respect to the origin is

2.4

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5.6

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Solution

The correct option is C 5.6

$p = m v$
$= 6 (4)$
$= 24$
$L = r_{⊥} p = (4 c o s 45) (24) = \frac{96}{\sqrt{2}}$
Areal velocity = $\frac{d A}{d t}$ $= \frac{L}{2 m} = \frac{96}{\sqrt{2} \times 2 \times 6} = \frac{8}{\sqrt{2}}$
$= \frac{8 \sqrt{2}}{2} = 4 \sqrt{2}$
$= 5.6 m^{2} / s$

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A particle of mass 6 kg is moving along y = x + 4. Speed of particle is 4 m/s. Areal velocity (in m2/s) of position vector of particle with respect to the origin is

The correct option is C 5.6 p=mv =6(4) =24 L=r⊥p=(4 cos 45)(24)=96√2 Areal velocity =dAdt=L2m=96√2×2×6=8√2 =8√22=4√2 =5.6 m2/s

A particle of mass 6 kg is moving along y = x + 4. Speed of particle is 4 m/s. Areal velocity (in $m^{2} / s$ ) of position vector of particle with respect to the origin is

The correct option is C 5.6

$p = m v$
$= 6 (4)$
$= 24$
$L = r_{⊥} p = (4 c o s 45) (24) = \frac{96}{\sqrt{2}}$
Areal velocity = $\frac{d A}{d t}$ $= \frac{L}{2 m} = \frac{96}{\sqrt{2} \times 2 \times 6} = \frac{8}{\sqrt{2}}$
$= \frac{8 \sqrt{2}}{2} = 4 \sqrt{2}$
$= 5.6 m^{2} / s$