A metre rod of silver at `0^(@)C` is heated to `100^(@)C`. It's length is increased by 0.19 cm. coefficient of volume expansion of the silver rod is

Options

A. `5.7xx10^(-5)//.^(@)C`
B. `0.63xx10^(-5)//.^(@)C`
C. `1.9xx10^(-5)//.^(@)C`
D. `16.1xx10^(-5)//.^(@)C`

Answer

To find the coefficient of volume expansion (γ) of a silver rod that is heated from 0°C to 100°C and experiences an increase in length of 0.19 cm, we can follow these steps:

### Step 1: Understand the relationship between linear expansion and volume expansion.
The coefficient of volume expansion (γ) is related to the coefficient of linear expansion (α) by the equation:
\[ \gamma = 3\alpha \]

### Step 2: Identify the given values.
- Initial length of the rod (L₀) = 1 meter = 100 cm
- Change in length (ΔL) = 0.19 cm
- Change in temperature (Δθ) = 100°C - 0°C = 100°C

### Step 3: Use the formula for linear expansion to find α.
The formula for linear expansion is:
\[ \Delta L = L_0 \cdot \alpha \cdot \Delta \theta \]

Rearranging this to solve for α gives:
\[ \alpha = \frac{\Delta L}{L_0 \cdot \Delta \theta} \]

### Step 4: Substitute the known values into the equation.
Substituting the values we have:
\[ \alpha = \frac{0.19 \, \text{cm}}{100 \, \text{cm} \cdot 100 \, \text{°C}} \]

### Step 5: Calculate α.
\[ \alpha = \frac{0.19}{10000} = 0.000019 = 1.9 \times 10^{-5} \, \text{°C}^{-1} \]

### Step 6: Calculate γ using the relationship between γ and α.
Now, substituting α into the equation for γ:
\[ \gamma = 3 \cdot \alpha = 3 \cdot (1.9 \times 10^{-5}) \]

### Step 7: Perform the multiplication.
\[ \gamma = 5.7 \times 10^{-5} \, \text{°C}^{-1} \]

### Conclusion:
The coefficient of volume expansion of the silver rod is:
\[ \gamma = 5.7 \times 10^{-5} \, \text{°C}^{-1} \]

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