Sandhya took three beakers A, B and C containing Zinc Sulphate, Silver Sulphate and Iron (II) Sulphate solutions respectively. Copper pieces were added to each beaker. The solution will appear blue in the case of:

Options

A. Beaker A
B. Beaker B
C. Beaker C
D. Beakers-B and C

Explanation

To solve the question, we need to analyze the reactions that occur when copper pieces are added to the solutions in beakers A, B, and C.

### Step-by-Step Solution:

1. **Identify the Solutions in Each Beaker:**
- Beaker A: Zinc Sulfate (ZnSO₄)
- Beaker B: Silver Sulfate (Ag₂SO₄)
- Beaker C: Iron (II) Sulfate (FeSO₄)

2. **Understand the Reactivity Series:**
The reactivity series of metals helps us determine which metal can displace another from its compound. The order of reactivity from most reactive to least reactive is:
- Zinc (Zn)
- Iron (Fe)
- Copper (Cu)
- Silver (Ag)

3. **Analyze the Reactions:**
- **Beaker A (Zinc Sulfate):**
- Copper (Cu) cannot displace Zinc (Zn) from Zinc Sulfate because Zinc is more reactive than Copper. Therefore, no reaction occurs.
- **Beaker B (Silver Sulfate):**
- Copper (Cu) can displace Silver (Ag) from Silver Sulfate because Copper is more reactive than Silver. The reaction will be:
\[ 2Ag^+ + Cu \rightarrow Cu^{2+} + 2Ag \]
- This reaction produces Copper Sulfate (CuSO₄), which is blue in color.
- **Beaker C (Iron (II) Sulfate):**
- Copper (Cu) cannot displace Iron (Fe) from Iron (II) Sulfate because Iron is more reactive than Copper. Therefore, no reaction occurs.

4. **Conclusion:**
The only beaker where a reaction occurs that produces a blue solution is Beaker B, where Copper displaces Silver from Silver Sulfate, forming Copper Sulfate.

### Final Answer:
The solution will appear blue in the case of **Beaker B (Silver Sulfate)**.