The volume of a metal sphere increases by `0.15%` when its temperature is raised by `24^@C`. The coefficient of linear expansion of metal is

Options

A. `2.5xx10^-5//^@C`
B. `2.0xx10^-5//^@C`
C. `1.5xx10^-5//^@C`
D. `12xx10^-5//^@C`

Answer

To find the coefficient of linear expansion (α) of the metal sphere, we can follow these steps:

### Step 1: Understand the relationship between volume expansion and linear expansion
The volume expansion (ΔV) of a material is related to its linear expansion (α) by the formula:
\[ \Delta V = V_0 \cdot \gamma \]
where \( \gamma \) is the coefficient of volumetric expansion. For isotropic materials, the relationship between volumetric expansion and linear expansion is given by:
\[ \gamma = 3\alpha \]

### Step 2: Calculate the change in volume
The problem states that the volume of the sphere increases by 0.15%. This can be expressed as:
\[ \Delta V = 0.15\% \times V_0 = \frac{0.15}{100} \times V_0 = 0.0015 V_0 \]

### Step 3: Relate the change in volume to temperature change
The change in volume can also be expressed in terms of temperature change (ΔT):
\[ \Delta V = V_0 \cdot \gamma \cdot \Delta T \]
Substituting the expression for ΔV we found in Step 2:
\[ 0.0015 V_0 = V_0 \cdot \gamma \cdot 24 \]

### Step 4: Simplify the equation
We can cancel \( V_0 \) from both sides (assuming \( V_0 \neq 0 \)):
\[ 0.0015 = \gamma \cdot 24 \]

### Step 5: Solve for γ
Now, we can solve for \( \gamma \):
\[ \gamma = \frac{0.0015}{24} = 0.0000625 \, \text{per degree Celsius} \]

### Step 6: Find the coefficient of linear expansion (α)
Using the relationship \( \gamma = 3\alpha \), we can find α:
\[ \alpha = \frac{\gamma}{3} = \frac{0.0000625}{3} = 0.0000208333 \, \text{per degree Celsius} \]
This can be expressed in scientific notation:
\[ \alpha \approx 2.08 \times 10^{-5} \, \text{per degree Celsius} \]

### Final Answer
The coefficient of linear expansion of the metal is approximately \( 2 \times 10^{-5} \, \text{per degree Celsius} \).

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