When a body of density `d_(1)` and volume V is floating in a liquid of density `d_(2)`

Options

A. its true weight is `Vd_(2)g`
B. loss in its weight is `Vd_(2)g`
C. its apparent weight is zero
D. its density `d_(1)` is greater than that of liquid `d_(2)` .

Answer

To solve the problem of a body of density \( d_1 \) and volume \( V \) floating in a liquid of density \( d_2 \), we need to analyze the concepts of true weight, loss in weight, and apparent weight. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand True Weight
The true weight of an object is given by the formula:
\[ \text{True Weight} = \text{mass} \times g \]
Where mass can be calculated as:
\[ \text{mass} = \text{density} \times \text{volume} = d_1 \times V \]
Thus, the true weight \( W \) can be expressed as:
\[ W = d_1 \times V \times g \]

### Step 2: Calculate Loss in Weight
When the object is submerged in a liquid, it experiences an upward buoyant force. The buoyant force \( F_b \) is given by Archimedes' principle:
\[ F_b = \text{density of liquid} \times \text{volume of object} \times g = d_2 \times V \times g \]
The loss in weight of the object when submerged in the liquid is equal to the buoyant force:
\[ \text{Loss in Weight} = F_b = d_2 \times V \times g \]

### Step 3: Determine Apparent Weight
The apparent weight of the object when it is floating in the liquid is given by:
\[ \text{Apparent Weight} = \text{True Weight} - \text{Loss in Weight} \]
Substituting the expressions we derived:
\[ \text{Apparent Weight} = (d_1 \times V \times g) - (d_2 \times V \times g) \]
This simplifies to:
\[ \text{Apparent Weight} = V \times g \times (d_1 - d_2) \]

### Step 4: Analyze Conditions for Floating
For an object to float, the density of the object \( d_1 \) must be less than the density of the liquid \( d_2 \):
\[ d_1 < d_2 \]
If \( d_1 < d_2 \), then \( d_1 - d_2 < 0 \), which means the apparent weight will be less than the true weight but not zero.

### Conclusion
From the analysis:
1. The true weight is \( d_1 \times V \times g \).
2. The loss in weight is \( d_2 \times V \times g \).
3. The apparent weight is \( V \times g \times (d_1 - d_2) \), which is not zero as long as \( d_1 \) is less than \( d_2 \).

### Final Answer
The correct option is:
- Loss in weight is \( V \times d_2 \times g \).