Question

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

Answer 1

Answer: 5

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
$=\left[M^1{L}^2{T}^0\right]$
$\therefore a=1,b=2$
$M_1=1kg,M_2=10g,L_1=1m,L_2=5cm,$
$\therefore N=9{\left(\frac{1kg}{10g}\right)}^1{\left(\frac{1m}{5cm}\right)}^2$
$=9\times {\left(\frac{1000g}{10g}\right)}^1{\left(\frac{100cm}{5cm}\right)}^2$
$=9\times 100\times 400$
$=3.6\times 10^5$
$\therefore x=5$