Question

Three identical rods, each of length I, are joined to form a rigid equilateral triangle. Its radius of gyration about an axis passing through a corner and perpendicular to the plane of the triangle is

• A. $\frac{\ell }{\sqrt{3}}$
• B. $\frac{\ell }{\sqrt{2}}$
• C. $\frac{\ell }{\sqrt{5}}$
• D. $\frac{\ell }{\sqrt{7}}$

Answer 1

The correct option is B $\frac{\ell }{\sqrt{2}}$

$\begin{array}{c}\text{we know}\\ \text{Moment of Inertia of a Rod about its axis}\\ \text{and perpendicular is -}\\ I=\frac{M{L}^2}{3}\\ \Rightarrow I_{0A}=I_{0B}=\frac{M{L}^2}{3}\\ \text{* Moment of Inertia of a Rod about its centre}\\ I_c=\frac{M{L}^2}{12}\end{array}$

$\begin{array}{c}\Rightarrow \text{By parallel axis theorem-}\\ \begin{array}{cc}{I}_{AB}=I_c+M{d}^2& d=\text{distanu from the Axis}\\ I_{AB}=\frac{M{L}^2}{12}+M{\left(\frac{\sqrt{3}L}{2}L\right)}^2=\frac{5M{L}^2}{6}& \end{array}\\ \Rightarrow \text{Total moment of Inertia I}=I_{OA}+I_{DB}+I_{AB}=\frac{3M{L}^2}{2}\\ \Rightarrow \text{Radius of gyration (R)-}\end{array}$

$3M{R}^2=\frac{3}{2}M{L}^2\Rightarrow [R=\frac{L}{\sqrt{2}}]\text{option(B)}$