Find the ratio of the weights, as measured by a spring balance, of a 1 kg block of iron and a 1kg block of wood. Density of iron `=7800 kg m^-3` density of wood `=800 kg m^-3` and density of air `=1.293 kgm^-3`.

Explanation

To find the ratio of the weights measured by a spring balance of a 1 kg block of iron and a 1 kg block of wood, we can follow these steps:

### Step 1: Understand the Concept of Weight Measurement
The weight measured by a spring balance in air is affected by the buoyancy force exerted by the air. The apparent weight (W') measured by the spring balance is given by:
\[ W' = W - F_b \]
where \( W \) is the actual weight and \( F_b \) is the buoyancy force.

### Step 2: Calculate the Actual Weight
The actual weight of both blocks (iron and wood) is given by:
\[ W = mg \]
For both blocks, since they have a mass of 1 kg:
\[ W = 1 \, \text{kg} \cdot g \]

### Step 3: Calculate the Buoyancy Force
The buoyancy force can be calculated using Archimedes' principle:
\[ F_b = \text{density of fluid} \cdot g \cdot V \]
where \( V \) is the volume of the object submerged in the fluid. The volume \( V \) can be calculated using the mass and density of the material:
\[ V = \frac{m}{\text{density}} \]

### Step 4: Calculate the Volume Displaced
For the iron block:
- Density of iron = 7800 kg/m³
\[ V_{\text{iron}} = \frac{1 \, \text{kg}}{7800 \, \text{kg/m}^3} = \frac{1}{7800} \, \text{m}^3 \]

For the wood block:
- Density of wood = 800 kg/m³
\[ V_{\text{wood}} = \frac{1 \, \text{kg}}{800 \, \text{kg/m}^3} = \frac{1}{800} \, \text{m}^3 \]

### Step 5: Calculate the Buoyancy Forces
For the iron block:
\[ F_{b,\text{iron}} = \text{density of air} \cdot g \cdot V_{\text{iron}} \]
\[ F_{b,\text{iron}} = 1.293 \, \text{kg/m}^3 \cdot g \cdot \frac{1}{7800} \]

For the wood block:
\[ F_{b,\text{wood}} = \text{density of air} \cdot g \cdot V_{\text{wood}} \]
\[ F_{b,\text{wood}} = 1.293 \, \text{kg/m}^3 \cdot g \cdot \frac{1}{800} \]

### Step 6: Substitute into the Apparent Weight Formula
Now substituting the buoyancy forces back into the apparent weight formula:
For the iron block:
\[ W'_{\text{iron}} = mg - F_{b,\text{iron}} \]
\[ W'_{\text{iron}} = 1g - 1.293g \cdot \frac{1}{7800} \]

For the wood block:
\[ W'_{\text{wood}} = mg - F_{b,\text{wood}} \]
\[ W'_{\text{wood}} = 1g - 1.293g \cdot \frac{1}{800} \]

### Step 7: Calculate the Ratio of Apparent Weights
The ratio of the weights measured by the spring balance is:
\[ \text{Ratio} = \frac{W'_{\text{iron}}}{W'_{\text{wood}}} \]

Substituting the expressions we derived:
\[ \text{Ratio} = \frac{1 - \frac{1.293}{7800}}{1 - \frac{1.293}{800}} \]

### Step 8: Simplify and Calculate
Now we can simplify the expression:
- Calculate \( \frac{1.293}{7800} \) and \( \frac{1.293}{800} \)
- Substitute these values back into the ratio and simplify.

After performing the calculations, we find:
\[ \text{Ratio} \approx 1.0015 \]

### Final Answer
The ratio of the weights measured by a spring balance of a 1 kg block of iron to a 1 kg block of wood is approximately **1.0015**.

---