A block of steel of size `5 cm xx 5cm xx 5cm` is weighed in water. If the relative density of steel is 7. its apparent weight is:

Options

A. `6 xx 5 xx 5 xx 5 gf`
B. `4 xx 4 xx 4 xx 7 gf`
C. `5 xx 5 xx 5 xx 7 gf`
D. `4 xx4xx 4 xx 6 gf`

Explanation

To find the apparent weight of a block of steel when submerged in water, we can follow these steps:

### Step 1: Calculate the Volume of the Steel Block
The volume \( V \) of the steel block can be calculated using the formula for the volume of a cube:
\[ V = \text{side}^3 \]
Given that the size of the block is \( 5 \, \text{cm} \times 5 \, \text{cm} \times 5 \, \text{cm} \):
\[ V = 5 \, \text{cm} \times 5 \, \text{cm} \times 5 \, \text{cm} = 125 \, \text{cm}^3 \]

### Step 2: Calculate the Weight of the Steel Block in Air
The weight \( W \) of the steel block in air is given by:
\[ W = m \cdot g \]
where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity. The mass can be calculated using the density of steel and the volume:
\[ m = \rho \cdot V \]
Given that the relative density of steel is 7, we can express the density of steel \( \rho \) in terms of the density of water \( \rho_w \):
\[ \rho = 7 \cdot \rho_w \]
Substituting this into the mass equation:
\[ m = 7 \cdot \rho_w \cdot V \]
Thus, the weight in air becomes:
\[ W = 7 \cdot \rho_w \cdot V \cdot g \]

### Step 3: Calculate the Buoyant Force
The buoyant force \( F_b \) acting on the block when submerged in water is given by Archimedes' principle:
\[ F_b = \rho_w \cdot V \cdot g \]

### Step 4: Calculate the Apparent Weight
The apparent weight \( W' \) of the block when submerged in water can be calculated using the formula:
\[ W' = W - F_b \]
Substituting the expressions for \( W \) and \( F_b \):
\[ W' = (7 \cdot \rho_w \cdot V \cdot g) - (\rho_w \cdot V \cdot g) \]
Factoring out \( \rho_w \cdot V \cdot g \):
\[ W' = (\rho_w \cdot V \cdot g) \cdot (7 - 1) = 6 \cdot \rho_w \cdot V \cdot g \]

### Step 5: Substitute the Volume and Density of Water
Now, substituting the volume \( V = 125 \, \text{cm}^3 \) and using \( \rho_w \) (density of water) which is approximately \( 1 \, \text{g/cm}^3 \):
\[ W' = 6 \cdot (1 \, \text{g/cm}^3) \cdot (125 \, \text{cm}^3) \cdot g \]
\[ W' = 750 \, \text{g} \cdot g \]

### Final Result
Thus, the apparent weight of the steel block when submerged in water is:
\[ W' = 750 \, \text{g}f \]