A cubical solid aluminium (bulk modulus `= -V (dp)/(dV) = 70 Gpa`) block has adge length of 1m on the surface of the earth. It is kept on the floor of a 5 km deep ocean. Taking the average density of water and the acceleration due to gravity to be `10^3 kg m^(-3) and 10 ms^(-2)`, respectively, the change in the adge length of the block in mm is ___________ .

Explanation

To solve the problem of the change in edge length of a cubical solid aluminum block submerged in a 5 km deep ocean, we will follow these steps:

### Step 1: Understand the given data
- Bulk modulus of aluminum, \( K = 70 \, \text{GPa} = 70 \times 10^9 \, \text{Pa} \)
- Initial edge length of the cube, \( a = 1 \, \text{m} \)
- Depth of the ocean, \( h = 5000 \, \text{m} \)
- Density of water, \( \rho = 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the change in pressure (\( \Delta P \))
The change in pressure due to the depth of the ocean can be calculated using the hydrostatic pressure formula:
\[ \Delta P = \rho g h \]
Substituting the values:
\[ \Delta P = (10^3 \, \text{kg/m}^3)(10 \, \text{m/s}^2)(5000 \, \text{m}) = 5 \times 10^7 \, \text{Pa} \]

### Step 3: Relate change in pressure to change in volume
The bulk modulus is defined as:
\[ K = -\frac{V \Delta P}{\Delta V} \]
Rearranging gives:
\[ \Delta V = -\frac{V \Delta P}{K} \]

### Step 4: Calculate the initial volume (\( V \))
The volume of the cube is given by:
\[ V = a^3 = (1 \, \text{m})^3 = 1 \, \text{m}^3 \]

### Step 5: Substitute values to find \( \Delta V \)
Substituting \( V \), \( \Delta P \), and \( K \) into the equation:
\[ \Delta V = -\frac{(1 \, \text{m}^3)(5 \times 10^7 \, \text{Pa})}{70 \times 10^9 \, \text{Pa}} \]
Calculating this gives:
\[ \Delta V = -\frac{5 \times 10^7}{70 \times 10^9} = -\frac{5}{70} \times 10^{-2} = -\frac{1}{14} \times 10^{-2} \approx -0.0007142857 \, \text{m}^3 \]

### Step 6: Relate change in volume to change in edge length
The change in volume can also be expressed in terms of the change in edge length (\( \Delta a \)):
\[ \Delta V = 3a^2 \Delta a \]
Substituting \( a = 1 \, \text{m} \):
\[ \Delta V = 3(1^2) \Delta a = 3 \Delta a \]

### Step 7: Solve for \( \Delta a \)
Equating the two expressions for \( \Delta V \):
\[ 3 \Delta a = -0.0007142857 \]
Thus:
\[ \Delta a = -\frac{0.0007142857}{3} \approx -0.0002380952 \, \text{m} \]

### Step 8: Convert to mm
To convert from meters to millimeters:
\[ \Delta a \approx -0.2380952 \, \text{mm} \approx -0.24 \, \text{mm} \]

### Final Answer
The change in the edge length of the block is approximately:
\[ \Delta a \approx -0.24 \, \text{mm} \]