When an object is made to float in two different liquids of densities `d_1` and `d_2`, the length of the object seen above the liquid surface are 4 cm and 5 cm respectively. Which of the following is the correct alternative?

Options

A. `d_1 gt d_2`
B. `d_1 lt d_2`
C. `d_1 = d_2`
D. None of these

Explanation

To solve the problem, we need to analyze the situation where an object is floating in two different liquids with different densities. The length of the object above the liquid surface gives us information about the relationship between the density of the object and the densities of the liquids.

### Step-by-Step Solution:

1. **Understanding Floating Objects**: When an object floats in a liquid, it displaces a volume of liquid equal to the weight of the object. The part of the object submerged in the liquid is determined by the densities of the object and the liquid.

2. **Given Information**:
- Length of the object above liquid 1 (density \(d_1\)) = 4 cm
- Length of the object above liquid 2 (density \(d_2\)) = 5 cm

3. **Calculating Submerged Length**:
- Let the total length of the object be \(L\).
- The submerged length in liquid 1 = \(L - 4\) cm
- The submerged length in liquid 2 = \(L - 5\) cm

4. **Using Archimedes' Principle**: According to Archimedes' principle, the weight of the liquid displaced is equal to the weight of the object. Therefore, we can set up the following equations based on the densities:
- For liquid 1:
\[ d_1 \times (L - 4) = \text{Weight of the object} \]
- For liquid 2:
\[ d_2 \times (L - 5) = \text{Weight of the object} \]

5. **Equating the Two Expressions**:
Since both expressions equal the weight of the object, we can set them equal to each other:
\[ d_1 \times (L - 4) = d_2 \times (L - 5) \]

6. **Rearranging the Equation**:
Rearranging gives us:
\[ d_1 \times L - 4d_1 = d_2 \times L - 5d_2 \]
\[ d_1 \times L - d_2 \times L = 4d_1 - 5d_2 \]
\[ (d_1 - d_2) \times L = 4d_1 - 5d_2 \]

7. **Analyzing the Result**:
- If \(d_1 > d_2\), then \(4d_1 - 5d_2 > 0\) implies \(L > 0\) which is true.
- If \(d_1 < d_2\), then \(4d_1 - 5d_2 < 0\) implies \(L < 0\) which is not possible.

Therefore, we conclude that \(d_1 < d_2\).

### Final Conclusion:
The correct alternative is that \(d_1\) is smaller than \(d_2\).

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