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Q.

If J is angular momentum and E is the Kinetic energy, then $J^{2} / E$ has the dimensions of

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a

moment of inertia

b

Power

c

angular velocity

d

Impulse

answer is A.

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Detailed Solution

$\frac{J^{2}}{E} = \frac{{(m v r)}^{2}}{\frac{1}{2} m v^{2}} = 2 m r^{2} = I$

$\frac{J^{2}}{E}$ has the dimensions of moment of inertia (I)

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