Question

The moment of inertia of a body rotating about a given axis is $9\, kg\, m ^{2}$ in the $SI$ system. If the value of the moment of inertia in a new system of units in which the unit of length is $5\, cm$ and the unit of mass is $10\, g$ is $3.6 \times 10^{ x }$, then find the value of $x$.

Answer 1

Value of moment of inertia in new system in terms of dimensions
$(N)=9\left(\frac{M_{1}}{M_{2}}\right)^{a}\left(\frac{L_{1}}{L_{2}}\right)^{b}$
Dimensional formula of moment of inertia
$=\left[ M ^{1} L ^{2} T ^{0}\right]$
$\therefore a=1, b=2$
$M_{1}=1\, kg, M_{2}=10\, g, L_{1}=1\, m, L_{2}=5\, cm , $
$\therefore N=9\left(\frac{1\, kg }{10\, g }\right)^{1}\left(\frac{1\, m }{5\, cm }\right)^{2}$
$=9 \times\left(\frac{1000\, g }{10\, g }\right)^{1}\left(\frac{100 \,cm }{5\, cm }\right)^{2}$
$=9 \times 100 \times 400$
$=3.6 \times 10^{5}$
$\therefore x=5$