The moment of inertia of a particle rotating about a given axis is 12.0 kg m² in the SI system. What is the value of the moment of inertia in a system of units in which the unit of length is 5 cm and the unit of mass is 10 g?

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$6.0 \times 10^{3}$

$5.4 \times 10^{5}$

$2.4 \times 10^{3}$

$4.8 \times 10^{5}$

answer is D.

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Detailed Solution

$n_{2} = n_{1} {(\frac{M_{1}}{M_{2}})}^{a} {(\frac{L_{1}}{L_{2}})}^{b} {(\frac{T_{1}}{T_{2}})}^{c}$

$Dimensional formula of moment of inertia = [{ML}^{2} T^{0}]$

$\begin{array}{l} ∴ a = 1, b = 2, c = 0 \\ n_{1} = 12 .0, M_{1} = 1 kg, M_{2} = 10 g \\ L_{1} = 1 m, L_{2} = 5 cm, T_{1} = 1 s, T_{2} = 1 s \end{array}$

$\begin{matrix} n_{2} = 12 .0 {(\frac{1 kg}{10 g})}^{1} {(\frac{1 m}{5 cm})}^{2} {(\frac{1 s}{1 s})}^{0} \\ = 12 \times {(\frac{1000 g}{10 g})}^{1} {(\frac{100 cm}{5 cm})}^{2} \times 1 \\ = 12 \times 100 \times 400 = 4 .8 \times 10^{5} \end{matrix}$