The temperature of an isotropic cubical solid of length `l_(0)`, density `rho_(0)` and coefficient of linear expansion `alpha` is increased by `20^(@)C`. Then at higher temperature , to a good approximation:-

Options

A. Length is `l_(0) (1+20alpha)`
B. Total surface area is `l_(0)^(2) (1 +40alpha)`
C. Total volume is `l_(0)^(3)(1+60alpha)`
D. Density is `(rho_(0))/(1+ 60 alpha)`

Explanation

To solve the problem of how the dimensions of an isotropic cubical solid change when the temperature is increased by \(20^\circ C\), we will consider the effects of linear expansion on length, surface area, volume, and density.

### Step 1: Calculate the new length of the cube

1. **Initial Length**: Let the initial length of the cube be \(l_0\).
2. **Coefficient of Linear Expansion**: The coefficient of linear expansion is given as \(\alpha\).
3. **Temperature Change**: The temperature increases by \(\Delta T = 20^\circ C\).
4. **Formula for Linear Expansion**: The new length \(l\) after expansion can be calculated using the formula:
\[ l = l_0 (1 + \alpha \Delta T) \]
5. **Substituting Values**: Plugging in the values:
\[ l = l_0 (1 + \alpha \cdot 20) \]

### Step 2: Calculate the new surface area of the cube

1. **Initial Surface Area**: The initial surface area \(A_0\) of a cube is given by:
\[ A_0 = 6l_0^2 \]
2. **Surface Area Expansion**: The surface area expands according to:
\[ A = A_0 (1 + 2\alpha \Delta T) \]
3. **Substituting Values**: Plugging in the values:
\[ A = 6l_0^2 (1 + 2\alpha \cdot 20) = 6l_0^2 (1 + 40\alpha) \]

### Step 3: Calculate the new volume of the cube

1. **Initial Volume**: The initial volume \(V_0\) of the cube is given by:
\[ V_0 = l_0^3 \]
2. **Volume Expansion**: The volume expands according to:
\[ V = V_0 (1 + 3\alpha \Delta T) \]
3. **Substituting Values**: Plugging in the values:
\[ V = l_0^3 (1 + 3\alpha \cdot 20) = l_0^3 (1 + 60\alpha) \]

### Step 4: Calculate the new density of the cube

1. **Initial Density**: The initial density \(\rho_0\) is given.
2. **Density Change**: As the volume increases, the density changes according to:
\[ \rho = \frac{\text{mass}}{\text{new volume}} = \frac{\rho_0 V_0}{V} \]
3. **Using the Volume Expansion**: Since the mass remains constant:
\[ \rho = \frac{\rho_0 l_0^3}{l_0^3 (1 + 60\alpha)} = \frac{\rho_0}{1 + 60\alpha} \]

### Summary of Results

- New Length: \( l = l_0 (1 + 20\alpha) \)
- New Surface Area: \( A = 6l_0^2 (1 + 40\alpha) \)
- New Volume: \( V = l_0^3 (1 + 60\alpha) \)
- New Density: \( \rho = \frac{\rho_0}{1 + 60\alpha} \)