A spring balance reads `10kg` when a bucket of water is suspened from it . What will be the reading of the balance when an iron piece of mass `7.2 kg` suspended by a string is immersed with half its volume inside the water in the bucket? Relative density of iron is `7.2` a) 10 kg b) 10.5 kg c) 13.6 kg d) 17.2 kg

Options

A. `10kg`
B. `10.5 kg`
C. `13.6kg`
D. `17.2kg`

Explanation

To solve the problem, we need to determine the reading of the spring balance when a bucket of water is suspended from it, and an iron piece is immersed in the water. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
The spring balance reads `10 kg` when the bucket of water is suspended. This means the weight of the bucket of water is equivalent to `10 kg` under the influence of gravity.

### Step 2: Calculate the Weight of the Iron Piece
The mass of the iron piece is `7.2 kg`. The weight (W) of the iron piece can be calculated using the formula:
\[ W = m \cdot g \]
Where:
- \( m = 7.2 \, \text{kg} \)
- \( g \approx 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Thus, the weight of the iron piece is:
\[ W = 7.2 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 70.56 \, \text{N} \]

### Step 3: Calculate the Buoyant Force on the Iron Piece
The relative density of iron is given as `7.2`, which means the density of iron is `7.2` times the density of water. The density of water (\( \rho_w \)) is approximately \( 1000 \, \text{kg/m}^3 \), so:
\[ \text{Density of iron} = 7.2 \cdot 1000 \, \text{kg/m}^3 = 7200 \, \text{kg/m}^3 \]

The volume (V) of the iron piece can be calculated using its mass and density:
\[ V = \frac{m}{\text{Density}} = \frac{7.2 \, \text{kg}}{7200 \, \text{kg/m}^3} = 0.001 \, \text{m}^3 \]

Since half of the iron piece is immersed in water, the volume of the submerged part is:
\[ V_{\text{submerged}} = \frac{1}{2} \cdot 0.001 \, \text{m}^3 = 0.0005 \, \text{m}^3 \]

The buoyant force (F_b) can be calculated using Archimedes' principle:
\[ F_b = V_{\text{submerged}} \cdot \rho_w \cdot g \]
Substituting the values:
\[ F_b = 0.0005 \, \text{m}^3 \cdot 1000 \, \text{kg/m}^3 \cdot 9.8 \, \text{m/s}^2 = 4.9 \, \text{N} \]

### Step 4: Calculate the Effective Weight of the Iron Piece
The effective weight of the iron piece when it is submerged is given by:
\[ \text{Effective Weight} = \text{Weight of Iron} - \text{Buoyant Force} \]
\[ \text{Effective Weight} = 70.56 \, \text{N} - 4.9 \, \text{N} = 65.66 \, \text{N} \]

### Step 5: Calculate the Total Weight on the Spring Balance
The total weight on the spring balance is the weight of the bucket of water plus the effective weight of the iron piece:
\[ \text{Total Weight} = \text{Weight of Bucket} + \text{Effective Weight} \]
Converting the weight of the bucket from kg to N (using \( g \)):
\[ \text{Weight of Bucket} = 10 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 98 \, \text{N} \]

Thus:
\[ \text{Total Weight} = 98 \, \text{N} + 65.66 \, \text{N} = 163.66 \, \text{N} \]

### Step 6: Convert Total Weight Back to kg
To convert the total weight back to kg, we divide by \( g \):
\[ \text{Total Weight in kg} = \frac{163.66 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 16.67 \, \text{kg} \]

### Conclusion
The reading of the balance will be approximately `16.67 kg`, which is closest to option (d) `17.2 kg`.