A metal rod of silver at `0^(@)C` is heated to `100^(@)C` . It's length is increased by `0.19 cm` . Coefficient of cubical 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 cubical expansion (γ) of a silver rod when it is heated from 0°C to 100°C and its length increases by 0.19 cm, we can follow these steps:

### Step 1: Understand the relationship between linear expansion and cubical expansion
The linear expansion of a material is given by the formula:
\[ \Delta L = L_1 \cdot \alpha \cdot \Delta T \]
where:
- \(\Delta L\) = change in length
- \(L_1\) = original length of the rod
- \(\alpha\) = coefficient of linear expansion
- \(\Delta T\) = change in temperature

The coefficient of cubical expansion (γ) is related to the coefficient of linear expansion (α) by the formula:
\[ \gamma = 3\alpha \]

### Step 2: Identify the known values
From the problem, we have:
- \(\Delta L = 0.19 \, \text{cm} = 0.0019 \, \text{m}\) (convert cm to m)
- \(\Delta T = 100°C - 0°C = 100°C\)

### Step 3: Rearrange the linear expansion formula to find α
Rearranging the formula for linear expansion gives:
\[ \alpha = \frac{\Delta L}{L_1 \cdot \Delta T} \]
Since we do not have the original length \(L_1\), we can assume \(L_1 = 1 \, \text{m}\) for simplicity, as it will cancel out later.

### Step 4: Substitute the known values into the equation
Substituting the known values into the equation gives:
\[ \alpha = \frac{0.0019 \, \text{m}}{1 \, \text{m} \cdot 100 \, \text{°C}} = \frac{0.0019}{100} = 0.000019 \, \text{°C}^{-1} = 1.9 \times 10^{-5} \, \text{°C}^{-1} \]

### Step 5: Calculate the coefficient of cubical expansion (γ)
Using the relationship between α and γ:
\[ \gamma = 3\alpha = 3 \times 1.9 \times 10^{-5} \, \text{°C}^{-1} = 5.7 \times 10^{-5} \, \text{°C}^{-1} \]

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

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