Question

Two discs of same mass and different radii are made of different materials such that their thicknesses are 1 cm and 0.5 cm respectively. The densities of materials are in the ratio 3:5. The moment of inertia of these discs respectively about their diameters will be in the ratio of \(\frac{x}{6}\). The value of x is ____.

Hint:

The moment of inertia of a disc depends on its mass and radius, and changes in the density and thickness affect both.

Answer 1

Correct Answer: 5

Let the radii of the two discs be \( R_1 \) and \( R_2 \), and let their densities be \( \rho_1 \) and \( \rho_2 \). According to the problem, \(\frac{\rho_1}{\rho_2} = \frac{3}{5}\) and the thicknesses are 1 cm and 0.5 cm respectively.

The formula for mass (m) of a disc is given by: \(m = \pi R^2 \cdot \text{thickness} \cdot \text{density}\).

Because the masses are equal, we have:

\(\pi R_1^2 \cdot 1 \cdot \rho_1 = \pi R_2^2 \cdot 0.5 \cdot \rho_2 \)

Substituting the density ratio, \(\frac{\rho_1}{\rho_2} = \frac{3}{5}\):

\[\pi R_1^2 \cdot 1 \cdot \left(\frac{3}{5}\rho_2\right) = \pi R_2^2 \cdot 0.5 \cdot \rho_2 \]

Canceling out \(\pi\) and \(\rho_2\) and solving for the radii ratio:

\((\frac{3}{5})R_1^2 = 0.5R_2^2 \)

\(\frac{3}{5}R_1^2 = \frac{1}{2}R_2^2\)

\( R_1^2 = \frac{5}{3} \times \frac{1}{2}R_2^2 \)

\( R_1^2 = \frac{5}{6}R_2^2 \)

The moment of inertia (\(I\)) of a disc about its diameter is \(\frac{1}{4}mR^2\). Therefore:

\(I_1 = \frac{1}{4} \pi R_1^2 \cdot 1 \cdot \rho_1 \cdot R_1^2 \)

\(I_2 = \frac{1}{4} \pi R_2^2 \cdot 0.5 \cdot \rho_2 \cdot R_2^2 \)

Using the relation \(\frac{\rho_1}{\rho_2} = \frac{3}{5}\) and simplifying:

\( I_1 = \frac{1}{4} \pi R_1^4 \cdot \frac{3}{5} \rho_2 \)

\( I_2 = \frac{1}{8} \pi R_2^4 \cdot \rho_2 \)

The ratio of inertia:

\(\frac{I_1}{I_2} = \frac{\frac{1}{4} \pi R_1^4 \cdot \frac{3}{5} \rho_2}{\frac{1}{8} \pi R_2^4 \cdot \rho_2}\)

Canceling terms gives:

\(\frac{I_1}{I_2} = \frac{2 \cdot 3}{5} \cdot \left(\frac{R_1}{R_2}\right)^4\)

\(\frac{I_1}{I_2} = \frac{6}{5} \cdot \left(\frac{5}{6}\right)^2 = \frac{6}{5} \cdot \frac{25}{36}\)

\(\frac{I_1}{I_2} = \frac{5}{6}\)

The ratio is \(\frac{x}{6}\), so \(x = 5\).

Thus, the value of \(x\) is 5, which is within the range [5,5].