JEE Physics 2023 Question with Solution

Given: Two discs have the same mass. Their thicknesses are $1 \, \text{cm}$ and $0.5 \, \text{cm}$, and their densities are in the ratio $3:5$.

Find: The value of $x$ if the ratio of moments of inertia about their diameters is given as $\frac{x}{6}$.

For a disc, mass is proportional to density multiplied by volume:

$$m \propto \rho \pi r^2 t$$

Since the two discs have equal masses,

$$\rho_1 r_1^2 t_1 = \rho_2 r_2^2 t_2$$

Using $\rho_1 : \rho_2 = 3:5$, $t_1 = 1$, and $t_2 = 0.5$,

$$3 r_1^2 \cdot 1 = 5 r_2^2 \cdot 0.5$$ $$3 r_1^2 = \frac{5}{2} r_2^2$$ $$\frac{r_1^2}{r_2^2} = \frac{5}{6}$$

The moment of inertia of a disc about a diameter is proportional to $mr^2$. Since the masses are equal,

$$\frac{I_1}{I_2} = \frac{r_1^2}{r_2^2} = \frac{5}{6}$$

So,

$$\frac{x}{6} = \frac{5}{6}$$

Hence,

$$x = 5$$

Therefore, the value of $x$ is $5$.

Given: The discs are stated to have the same mass.

Find: How this condition fixes the radius ratio.

The solution contains an incorrect intermediate statement suggesting mass should first be written separately for both discs and then substituted into the moment of inertia ratio. Since the masses are already equal, that condition must be used first.

From geometry and density,

$$m = \rho \times \text{volume} = \rho \pi r^2 t$$

Equal masses give

$$\rho_1 \pi r_1^2 t_1 = \rho_2 \pi r_2^2 t_2$$

Cancelling $\pi$,

$$\rho_1 r_1^2 t_1 = \rho_2 r_2^2 t_2$$

Substitute the data:

$$3 r_1^2 \times 1 = 5 r_2^2 \times 0.5$$ $$3 r_1^2 = 2.5 r_2^2$$ $$\frac{r_1^2}{r_2^2} = \frac{5}{6}$$

Now for a disc about a diameter,

$$I \propto mr^2$$

Because $m_1 = m_2$,

$$\frac{I_1}{I_2} = \frac{r_1^2}{r_2^2} = \frac{5}{6}$$

Thus $\frac{x}{6} = \frac{5}{6}$, so the correct value is $x = 5$.