A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in the figure. The moment of inertia of this lamina about axes passing through O and P is $I_{O}$ and $I_{P}$ respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $I_{P}/I_{O}$ to the nearest integer is

Step 1: Define the masses of the discs.

Let $M$ be the mass of the complete big disc of radius $2R$. Since the mass density is uniform, the mass $m$ of the removed small disc of radius $R$ is proportional to its area.

$$m = M \cdot \frac{\pi R^2}{\pi (2R)^2}$$ $$m = \frac{M}{4}$$

Step 2: Calculate the moments of inertia about the axis through $O$.

The moment of inertia of the complete big disc about its center $O$ is:

$$I_{\text{big},O} = \frac{1}{2}M(2R)^2$$ $$I_{\text{big},O} = 2MR^2$$

The moment of inertia of the removed small disc about its own center $C$ is:

$$I_{\text{small},C} = \frac{1}{2}mR^2$$ $$I_{\text{small},C} = \frac{MR^2}{8}$$

Using the parallel-axis theorem, where the distance between $O$ and $C$ is $R$, the moment of inertia of the small disc about $O$ is:

$$I_{\text{small},O} = I_{\text{small},C} + mR^2$$ $$I_{\text{small},O} = \frac{MR^2}{8} + \frac{M}{4}R^2$$ $$I_{\text{small},O} = \frac{3MR^2}{8}$$

Step 3: Find the moment of inertia of the lamina about $O$.

Subtract the moment of inertia of the removed disc from that of the big disc.

$$I_O = I_{\text{big},O} - I_{\text{small},O}$$ $$I_O = 2MR^2 - \frac{3MR^2}{8}$$ $$I_O = \frac{13MR^2}{8}$$

Step 4: Calculate the moments of inertia about the axis through $P$.

Using the parallel-axis theorem, where the distance between $O$ and $P$ is $2R$, the moment of inertia of the big disc about $P$ is:

$$I_{\text{big},P} = I_{\text{big},O} + M(2R)^2$$ $$I_{\text{big},P} = 2MR^2 + 4MR^2$$ $$I_{\text{big},P} = 6MR^2$$

The distance $CP$ between the center of the small disc $C$ and point $P$ is found using Pythagoras' theorem.

$$CP^2 = (2R)^2 + R^2$$ $$CP^2 = 5R^2$$

Using the parallel-axis theorem, the moment of inertia of the small disc about $P$ is:

$$I_{\text{small},P} = I_{\text{small},C} + m(CP^2)$$ $$I_{\text{small},P} = \frac{MR^2}{8} + \frac{M}{4}(5R^2)$$ $$I_{\text{small},P} = \frac{11MR^2}{8}$$

Step 5: Find the moment of inertia of the lamina about $P$.

Subtract the moment of inertia of the removed disc from that of the big disc.

$$I_P = I_{\text{big},P} - I_{\text{small},P}$$ $$I_P = 6MR^2 - \frac{11MR^2}{8}$$ $$I_P = \frac{37MR^2}{8}$$

Step 6: Calculate the ratio of the moments of inertia.

Divide $I_P$ by $I_O$.

$$\frac{I_P}{I_O} = \frac{\frac{37MR^2}{8}}{\frac{13MR^2}{8}}$$ $$\frac{I_P}{I_O} = \frac{37}{13}$$ $$\frac{I_P}{I_O} \approx 2.85$$

Rounding to the nearest integer gives $3$.

Answer: 3