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  4. The ratio of the dimensions of Plancks constant and that of the moment of inertia is the dimension of

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Q. The ratio of the dimensions of Plancks constant and that of the moment of inertia is the dimension of

JIPMERJIPMER 2007Physical World, Units and Measurements

Solution:

$E=h v $
$\Rightarrow h=$ Plancks constant $=\frac{E}{V} $
$\therefore [h]=\frac{[E]}{[v]}=\frac{\left[M L^{2} T^{-2}\right]}{\left[T^{-1}\right]}$
$=\left[M L^{2} T^{-1}\right] $
and $ I=$ moment of inertia $=M R^{2} $
$\Rightarrow [I]=[M]\left[L^{2}\right]=\left[M L^{2}\right] $
Hence, $ \frac{[h]}{[I]}=\frac{\left[M L^{2} T^{-1}\right]}{\left[M L^{2}\right]}=\left[T^{-1}\right] $
$=\frac{1}{[T]}=$ dimensions of frequency
Alternative: $\frac{h}{I}=\frac{E / v}{I} $
$=\frac{E \times T}{I}=\frac{\left(k g-m^{2} / s^{2}\right) \times s}{\left(k g-m^{2}\right)} $
$=\frac{1}{s}=\frac{1}{\text { time }}=$ frequency
Thus, dimensions of $\frac{h}{I}$ is same as of frequency.