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# The kinetic energy of rotation k depends on angular momentum J and moment of inertia I.The expression for kinetic energy is ( where c is a constant )

$(a)\;k=\frac{cI}{I}\quad (b)\;k=\frac{cI}{J} \quad (c)\;k=\frac{cI^2}{J}\quad (d)\;k=\frac{cJ^2}{I}$

$k=cJ^aI^b$
$[ML^2T^{-2}]=[ML^2T^{-1}]^a[ML^2]^b$
$-a=-2=>a=2$
$a+b=1$
$2+b=1$
$b=-1$
Applying all the values we get
$k=\large\frac{cJ^2}{I}$
Hence d is the correct answer.
edited Jan 9, 2014 by meena.p