A body floats in water with one third of its volume above the surface of water if it is placed in oil it floats with half of its volume above the surface of the oil the specific gravity of the oil is

Options

A. (5/3)
B. (4/3)
C. (3/2)
D. 1

Answer

To solve the problem, we need to determine the specific gravity of the oil based on the given conditions of the body floating in water and oil.

### Step-by-Step Solution:

1. **Understand the Problem**:
- A body floats in water with one-third of its volume above the surface. This means two-thirds of the body is submerged in water.
- When placed in oil, half of its volume is above the surface, meaning half of the body is submerged in oil.

2. **Use Archimedes' Principle**:
- According to Archimedes' principle, the buoyant force acting on the body is equal to the weight of the fluid displaced by the submerged part of the body.
- Let the volume of the body be \( V \).
- The density of water is denoted as \( \rho_W \) and the density of oil as \( \rho_O \).

3. **Calculate the Buoyant Force in Water**:
- The volume submerged in water is \( \frac{2}{3}V \).
- The weight of the water displaced is given by:
\[ F_{b, water} = \text{Weight of displaced water} = \rho_W g \left(\frac{2}{3}V\right) \]
- The weight of the body is equal to the buoyant force when floating:
\[ W = \rho_B g V = \rho_W g \left(\frac{2}{3}V\right) \]
- Simplifying gives:
\[ \rho_B = \frac{2}{3} \rho_W \quad \text{(Equation 1)} \]

4. **Calculate the Buoyant Force in Oil**:
- The volume submerged in oil is \( \frac{1}{2}V \).
- The weight of the oil displaced is given by:
\[ F_{b, oil} = \rho_O g \left(\frac{1}{2}V\right) \]
- Again, the weight of the body is equal to the buoyant force:
\[ W = \rho_B g V = \rho_O g \left(\frac{1}{2}V\right) \]
- Simplifying gives:
\[ \rho_B = \frac{1}{2} \rho_O \quad \text{(Equation 2)} \]

5. **Equate the Two Equations**:
- From Equation 1 and Equation 2, we have:
\[ \frac{2}{3} \rho_W = \frac{1}{2} \rho_O \]
- Rearranging gives:
\[ \rho_O = \frac{4}{3} \rho_W \]

6. **Determine the Specific Gravity of Oil**:
- The specific gravity (SG) of a substance is defined as the ratio of its density to the density of water:
\[ SG = \frac{\rho_O}{\rho_W} \]
- Substituting the value of \( \rho_O \):
\[ SG = \frac{\frac{4}{3} \rho_W}{\rho_W} = \frac{4}{3} \]

### Final Answer:
The specific gravity of the oil is \( \frac{4}{3} \).