The coefficient of linear expansion of crystal in one direction is `alpha_(1)` and that in every direction perpendicular to it is `alpha_(2)`. The coefficient of cubical expansion is

Options

A. `alpha_1+alpha_2`
B. `2alpha_1+alpha_2`
C. `alpha_1+2alpha_2`
D. none of these

Explanation

To find the coefficient of cubical expansion (γ) of a crystal given the coefficients of linear expansion in one direction (α₁) and in every direction perpendicular to it (α₂), we can follow these steps:

### Step 1: Understand the linear expansion in three dimensions
The linear expansion in one direction can be expressed as:
- For the direction with coefficient α₁:
\[ L_1 = L_0(1 + \alpha_1 \Delta t) \]
- For the two perpendicular directions with coefficient α₂:
\[ L_2 = L_0(1 + \alpha_2 \Delta t) \]
\[ L_3 = L_0(1 + \alpha_2 \Delta t) \]

### Step 2: Calculate the final volume
The final volume \( V \) can be expressed as:
\[ V = L_1 \times L_2 \times L_3 \]
Substituting the expressions for \( L_1, L_2, \) and \( L_3 \):
\[ V = L_0(1 + \alpha_1 \Delta t) \times L_0(1 + \alpha_2 \Delta t) \times L_0(1 + \alpha_2 \Delta t) \]
This simplifies to:
\[ V = L_0^3 (1 + \alpha_1 \Delta t)(1 + \alpha_2 \Delta t)^2 \]

### Step 3: Expand the expression
Now, we expand the expression:
\[ (1 + \alpha_2 \Delta t)^2 = 1 + 2\alpha_2 \Delta t + (\alpha_2 \Delta t)^2 \]
Neglecting the higher-order term \((\alpha_2 \Delta t)^2\) since \(\alpha\) is very small, we have:
\[ (1 + \alpha_2 \Delta t)^2 \approx 1 + 2\alpha_2 \Delta t \]
Thus, substituting back:
\[ V \approx L_0^3 (1 + \alpha_1 \Delta t)(1 + 2\alpha_2 \Delta t) \]
Expanding this gives:
\[ V \approx L_0^3 (1 + \alpha_1 \Delta t + 2\alpha_2 \Delta t) \]

### Step 4: Relate to the initial volume and cubical expansion
The initial volume \( V_0 \) is:
\[ V_0 = L_0^3 \]
The expression for volume change due to cubical expansion is:
\[ V = V_0 (1 + \gamma \Delta t) \]
Setting the two expressions for \( V \) equal:
\[ V_0 (1 + \gamma \Delta t) = L_0^3 (1 + \alpha_1 \Delta t + 2\alpha_2 \Delta t) \]
Cancelling \( V_0 \) gives:
\[ 1 + \gamma \Delta t = 1 + \alpha_1 \Delta t + 2\alpha_2 \Delta t \]

### Step 5: Solve for γ
From the above equation, we can equate the coefficients of \( \Delta t \):
\[ \gamma = \alpha_1 + 2\alpha_2 \]

### Final Answer
Thus, the coefficient of cubical expansion \( \gamma \) is:
\[ \gamma = \alpha_1 + 2\alpha_2 \]

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