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 step by step, we will analyze the effects of temperature change on the dimensions, surface area, volume, and density of an isotropic cubical solid.

### Step 1: Calculate the new length of the cube
The linear expansion of a solid can be expressed as:
\[ L = L_0 (1 + \alpha \Delta T) \]
where:
- \( L_0 \) is the original length,
- \( \alpha \) is the coefficient of linear expansion,
- \( \Delta T \) is the change in temperature.

Given that the temperature is increased by \( 20^\circ C \):
\[ L = L_0 (1 + \alpha \cdot 20) \]

### Step 2: Calculate the new total surface area
The total surface area \( A \) of a cube is given by:
\[ A = 6L^2 \]
Using the new length from Step 1:
\[ A = 6[L_0 (1 + \alpha \cdot 20)]^2 \]
Expanding this, we have:
\[ A = 6L_0^2 (1 + 20\alpha)^2 \]
Using the approximation for small expansions:
\[ A \approx A_0 (1 + 40\alpha) \]
where \( A_0 = 6L_0^2 \).

### Step 3: Calculate the new total volume
The volume \( V \) of a cube is given by:
\[ V = L^3 \]
Using the new length from Step 1:
\[ V = [L_0 (1 + \alpha \cdot 20)]^3 \]
Expanding this, we have:
\[ V = L_0^3 (1 + 20\alpha)^3 \]
Using the approximation for small expansions:
\[ V \approx V_0 (1 + 60\alpha) \]
where \( V_0 = L_0^3 \).

### Step 4: Calculate the new density
Density \( \rho \) is defined as mass per unit volume:
\[ \rho = \frac{m}{V} \]
Since the mass \( m \) remains constant, the new density \( \rho' \) can be expressed as:
\[ \rho' = \frac{\rho_0 V_0}{V} \]
Substituting the volume from Step 3:
\[ \rho' = \frac{\rho_0 V_0}{V_0 (1 + 60\alpha)} \]
Thus:
\[ \rho' = \frac{\rho_0}{1 + 60\alpha} \]

### Summary of Results
1. **New Length**: \( L = L_0 (1 + 20\alpha) \)
2. **New Total Surface Area**: \( A = 6L_0^2 (1 + 40\alpha) \)
3. **New Total Volume**: \( V = L_0^3 (1 + 60\alpha) \)
4. **New Density**: \( \rho' = \frac{\rho_0}{1 + 60\alpha} \)