A spring balance reads 10 kg when a bucket of water is suspended from it. What is the reading on the spring balance when (i) an ice cube of mass 1.5 kg is put into the bucket (ii) an iron piece of mass 7.8 kg suspended by another spring is immersed with half its volume inside the water n the bucket. Relative density of iron = 7.8

Explanation

To solve the problem step by step, we will analyze both parts of the question regarding the spring balance readings.

### Part (i): Ice Cube in the Bucket

1. **Initial Reading of the Spring Balance**: The spring balance reads 10 kg when the bucket of water is suspended from it. This means the weight of the bucket and water combined is 10 kg.

2. **Adding the Ice Cube**: When we add an ice cube with a mass of 1.5 kg to the bucket, we need to consider the total weight that the spring balance will now measure.

3. **Calculating the New Reading**:
- The new weight = Initial weight + Weight of the ice cube
- New weight = 10 kg + 1.5 kg = 11.5 kg

4. **Final Reading**: Therefore, the reading on the spring balance after adding the ice cube will be **11.5 kg**.

### Part (ii): Iron Piece Immersed in Water

1. **Understanding the Iron Piece**: We have an iron piece with a mass of 7.8 kg and a relative density of 7.8. This means the density of iron is 7.8 times that of water.

2. **Calculating the Volume of the Iron Piece**:
- Density of iron = 7.8 x 10³ kg/m³ (since the density of water is 1 x 10³ kg/m³).
- Volume of the iron piece = Mass / Density = 7.8 kg / (7.8 x 10³ kg/m³) = 1 x 10^-3 m³.

3. **Determining the Volume Immersed**: Since half of the iron piece is immersed in water, the volume submerged (V') is:
- V' = (1 x 10^-3 m³) / 2 = 0.5 x 10^-3 m³.

4. **Calculating the Upward Thrust (Buoyant Force)**:
- Buoyant force (thrust) = Volume submerged x Density of water x g
- Buoyant force = (0.5 x 10^-3 m³) x (1 x 10³ kg/m³) x (9.8 m/s²) = 4.9 N.

5. **Converting Buoyant Force to kgf**:
- 1 kgf = 9.8 N, so the buoyant force in kgf = 4.9 N / 9.8 N/kgf = 0.5 kgf.

6. **Calculating the Total Reading on the Spring Balance**:
- Initial reading (bucket + water) = 10 kg.
- The upward thrust (buoyant force) = 0.5 kgf.
- Total reading = Initial weight + Buoyant force = 10 kg + 0.5 kg = 10.5 kg.

7. **Final Reading**: Therefore, the reading on the spring balance when the iron piece is immersed will be **10.5 kg**.

### Summary of Answers:
- (i) The reading on the spring balance after adding the ice cube is **11.5 kg**.
- (ii) The reading on the spring balance after immersing the iron piece is **10.5 kg**.