A hole is drilled in a metal sheet . At `27^@C`, the diameter of hole is 5 cm . When the sheet is heated to `177^@C` ,the change in the diameter of hole is `d times 10^(-3)` cm.The value of d will be ____ if coefficient of linearexpansion of the metal is `1.6 times 10^(-5)@C`.

Answer

To solve the problem, we need to calculate the change in diameter of the hole when the metal sheet is heated from 27°C to 177°C. The formula for linear expansion is given by:

\[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \]

where:
- \(\Delta L\) is the change in length (or diameter in this case),
- \(L_0\) is the original length (or diameter),
- \(\alpha\) is the coefficient of linear expansion,
- \(\Delta T\) is the change in temperature.

### Step 1: Identify the given values
- Original diameter \(L_0 = 5 \, \text{cm}\)
- Coefficient of linear expansion \(\alpha = 1.6 \times 10^{-5} \, \text{°C}^{-1}\)
- Initial temperature \(T_1 = 27 \, \text{°C}\)
- Final temperature \(T_2 = 177 \, \text{°C}\)

### Step 2: Calculate the change in temperature
\[ \Delta T = T_2 - T_1 = 177 \, \text{°C} - 27 \, \text{°C} = 150 \, \text{°C} \]

### Step 3: Substitute the values into the linear expansion formula
Using the formula for change in diameter:
\[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \]
Substituting the known values:
\[ \Delta L = 5 \, \text{cm} \cdot (1.6 \times 10^{-5} \, \text{°C}^{-1}) \cdot (150 \, \text{°C}) \]

### Step 4: Calculate \(\Delta L\)
\[ \Delta L = 5 \cdot 1.6 \times 10^{-5} \cdot 150 \]
Calculating this step by step:
1. Calculate \(5 \cdot 150 = 750\)
2. Now multiply by \(1.6 \times 10^{-5}\):
\[ \Delta L = 750 \cdot 1.6 \times 10^{-5} = 1200 \times 10^{-5} \, \text{cm} \]

### Step 5: Express \(\Delta L\) in the required format
\[ \Delta L = 1.2 \times 10^{-2} \, \text{cm} = 12 \times 10^{-3} \, \text{cm} \]
Thus, the change in diameter is \(d \times 10^{-3} \, \text{cm}\) where \(d = 12\).

### Final Answer
The value of \(d\) is:
\[ \boxed{12} \]