Question

Two discs of the same material and thickness have radii \(0.2 \mathrm{~m}\) and \(0.6 \mathrm{~m}\) their moment of inertia about their axes will be in the ratio \(\\{\mathrm{A}\\} 1: 81\) \(\\{\mathrm{B}\\} 1: 27\) \(\\{C\\} 1: 9\) \(\\{\mathrm{D}\\} 1: 3\)

Short answer

The ratio of the moments of inertia for the two discs is 1:81. Since both discs have the same material and thickness, their masses are proportional to the square of their radii. Using the formula for the moment of inertia of a disc (\(I = \frac{1}{2}mr^2\)), we calculate the moments of inertia for both discs and determine the ratio \(\frac{I_1}{I_2} = \frac{1}{81}\). The correct answer is \(\mathrm{A} \: 1: 81\).

Step-by-step solution

Formula for the moment of inertia of a disc

Recall the formula for the moment of inertia (I) of a solid disc about its central axis: \[I = \frac{1}{2}mr^2\] where m is the mass and r is the radius of the disc.

Calculate the moments of inertia for both discs

Since both discs have the same material and thickness, their masses will be proportional to the square of their radii, i.e., \(m_2 = k(0.2)^2\) and \(m_1 = k(0.6)^2\) , where k is a constant of proportionality. Now we can calculate the moments of inertia for both discs using the given radii and the proportional masses: \(I_1 = \frac{1}{2}(k(0.2)^2)(0.2)^2 = \frac{k}{50}\) \(I_2 = \frac{1}{2}(k(0.6)^2)(0.6)^2 = \frac{9k}{10}\)

Calculate the ratio of the moments of inertia

We are interested in finding the ratio \(\frac{I_1}{I_2}\). Using the values we found in the previous step, we can write: \[\frac{I_1}{I_2} = \frac{\frac{k}{50}}{\frac{9k}{10}} = \frac{1}{81}\]

Find the matching option

The ratio we obtained is 1:81, which matches option A. Therefore, the correct answer is: \(\boxed{\mathrm{A}\: 1: 81}\).