A body P floats in water with half its volume immersed. Another body Q floats in a liquid of density `(3)/(4)`th of the density of water with two - third of the volume immersed. The ratio of density of P to that of Q is

Options

A. `1:2`
B. `1:1`
C. `2:3`
D. `3:4`

Answer

To solve the problem, we need to analyze the conditions given for both bodies P and Q floating in their respective fluids.

### Step-by-Step Solution:

1. **Understanding Body P**:
- Body P floats in water with half its volume immersed.
- Let the total volume of body P be \( V \).
- The volume of body P that is immersed in water is \( \frac{V}{2} \).
- According to Archimedes' principle, the buoyant force \( F_b \) acting on body P is equal to the weight of the water displaced.
- The density of water is denoted as \( \rho_w \).
- The buoyant force can be calculated as:
\[ F_b = \text{Volume immersed} \times \text{Density of water} \times g = \frac{V}{2} \times \rho_w \times g \]
- The weight of body P is given by:
\[ W_P = \text{Volume} \times \text{Density of P} \times g = V \times \rho_P \times g \]
- For equilibrium (floating condition), the buoyant force equals the weight:
\[ \frac{V}{2} \times \rho_w \times g = V \times \rho_P \times g \]
- Dividing both sides by \( Vg \) (assuming \( V \neq 0 \) and \( g \neq 0 \)):
\[ \frac{\rho_w}{2} = \rho_P \]
- Thus, we have:
\[ \rho_P = \frac{\rho_w}{2} \quad \text{(Equation 1)} \]

2. **Understanding Body Q**:
- Body Q floats in a liquid with a density of \( \frac{3}{4} \) of the density of water.
- The total volume of body Q is also \( V \), and the volume immersed is \( \frac{2V}{3} \).
- The density of the liquid is:
\[ \rho_{liquid} = \frac{3}{4} \rho_w \]
- The buoyant force acting on body Q is:
\[ F_b = \text{Volume immersed} \times \text{Density of liquid} \times g = \frac{2V}{3} \times \left(\frac{3}{4} \rho_w\right) \times g \]
- The weight of body Q is given by:
\[ W_Q = V \times \rho_Q \times g \]
- For equilibrium:
\[ \frac{2V}{3} \times \left(\frac{3}{4} \rho_w\right) \times g = V \times \rho_Q \times g \]
- Dividing both sides by \( Vg \):
\[ \frac{2}{3} \times \frac{3}{4} \rho_w = \rho_Q \]
- Simplifying:
\[ \rho_Q = \frac{1}{2} \rho_w \quad \text{(Equation 2)} \]

3. **Finding the Ratio of Densities**:
- Now we have:
- From Equation 1: \( \rho_P = \frac{\rho_w}{2} \)
- From Equation 2: \( \rho_Q = \frac{\rho_w}{2} \)
- Therefore, the ratio of the densities of P to Q is:
\[ \frac{\rho_P}{\rho_Q} = \frac{\frac{\rho_w}{2}}{\frac{\rho_w}{2}} = 1 \]

### Final Answer:
The ratio of the density of P to that of Q is \( 1:1 \).