The moment of inertia of a body rotating about a given axis is $12.0 \mathrm{kgm}^{2}$ in the S.I. system. The value of the moment of inertia in a system of units in which the units of length is 5 cm and mass is 10g is equal to ______ $\times 10^5$ (mark answer to nearest integer)

Question

Aicharya · Accepted Answer

$n_{2}=n_{1}\left(\frac{M_{1}}{M_{2}}\right)^{a}\left(\frac{L_{1}}{L_{2}}\right)^{b}\left(\frac{T_{1}}{T_{2}}\right)^{0}$ Dimensional formula of moment of inertia $=\left[M L^{2} T^{0}\right]$ $\therefore a=1, b=2, c=0$ $n_{1}=12.0, M_{1}=1 \mathrm{~kg}, M_{2}=10 \mathrm{~g}$ $L_{1}=1 \mathrm{~m}_{i} L_{2}=5 \mathrm{~cm}, T_{1}=1 \mathrm{~s}, T_{2}=1 \mathrm{~s}$ $n_{2}=12.0\left(\frac{1 \mathrm{~kg}}{10 \mathrm{~g}}\right)^{1}\left(\frac{1 \mathrm{~m}}{5 \mathrm{~cm}}\right)^{2}\left(\frac{1 \mathrm{sec}}{1 \mathrm{sec}}\right)^{0}$ $=12 \times\left(\frac{1000 \mathrm{~g}}{10 \mathrm{~g}}\right)^{1}\left(\frac{100 \mathrm{~cm}}{5 \mathrm{~cm}}\right)^{2} \times 1$ $=12 \times 100 \times 400=4.8 \times 10^{5}$

The moment of inertia of a body rotating about a given axis is \(12.0 \mathrm{kgm}^{2}\) in the S.I. system.…

Solution