Question

A massless equilateral triangle EFG of side ‘a’ (As shown in figure) has three particles of mass $m$ situated at its vertices. The moment of inertia of the system about line EX, perpendicular to EG in the plane of EFG is $(N/20)m{a}^2$ where$N$ is an integer. The value of$N$ is _____.

Answer 1

Step 1: Given data:

$I=\left(\frac{N}{20}\right)m{a}^2$

$m$= mass of particle

$a$= edge length of triangle

Step 2: formula used and drawing the diagram

moment of inertia is given by $I=m{r}^2$

Step 3: Finding the moment of inertia about EX

$I$ = moment of inertia of mass located at G + moment of inertia of mass located at G+ moment of inertia of mass located at E

Now , $l=m{a}^2+\left(\frac{m{a}^2}{4}\right)+0$ $\left[dis\tan ceofFfromaxisEX=\frac{a}{2}\right]$

$\Rightarrow l=\left(\frac{5}{4}\right)m{a}^2$

$$\begin{gathered} \Rightarrow l=\left(\frac{5}{4}\right)m{a}^2 \\ \Rightarrow \left(\frac{N}{20}\right)m{a}^2=\left(\frac{5}{4}\right)m{a}^2 \end{gathered}$$ $\left[I=\left(\frac{N}{20}\right)m{a}^2\right]$

Therefore, $N=25$

Hence, value of$\mathbf{}\mathit{N}$ is $\mathbf{25}$.