Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm) about its axis be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is

Concept:

Moment of inertia of hollow cylinder about its axis is given as:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Where,

R₁ = Inner radius and R₂ = Outer radius

Moment of inertia of thin hollow cylinder of radius ‘R’ about its axis is given as:

I₂ = MR²

Calculation:

From question, we know that,

⇒ I₁ = I₂

_{\(\Rightarrow \frac{{\rm{M}}}{2}\left( {{\rm{R}}_1^2 + {\rm{R}}_2^2} \right) = {\rm{M}}{{\rm{R}}^2}\)}

Both cylinders have same mass (M)
\(\Rightarrow \frac{{\left( {{\rm{R}}_1^2 + {\rm{R}}_2^2} \right)}}{2} = {{\rm{R}}^2}\)

\(\Rightarrow \frac{{\left( {{{10}^2} + {{20}^2}} \right)}}{2} = {{\rm{R}}^2}\)

⇒ R² = 250 = 15.8