The density of a material A is 1500`kg//m^(3)` and that of another material B is 2000 `kg//m^(3)`. It is found that the heat capacity of 8 volumes of A is equal to heat capacity of 12 volumes of B. The ratio of specific heats of A and B will be

Options

A. `1:2`
B. `3:1`
C. `3:2`
D. `2:1`

Answer

To find the ratio of specific heats of materials A and B, we can follow these steps:

### Step 1: Define the given quantities
- Density of material A, \( \rho_A = 1500 \, \text{kg/m}^3 \)
- Density of material B, \( \rho_B = 2000 \, \text{kg/m}^3 \)
- Heat capacity of 8 volumes of A is equal to the heat capacity of 12 volumes of B.

### Step 2: Write the formula for heat capacity
The heat capacity \( C \) of a material can be expressed as:
\[ C = m \cdot s \]
where \( m \) is the mass and \( s \) is the specific heat.

### Step 3: Calculate the mass for each material
For material A (8 volumes):
\[ m_A = \rho_A \cdot V_A = \rho_A \cdot 8V \]
For material B (12 volumes):
\[ m_B = \rho_B \cdot V_B = \rho_B \cdot 12V \]

### Step 4: Write the heat capacity equations
The heat capacity for 8 volumes of A:
\[ C_A = m_A \cdot s_A = (\rho_A \cdot 8V) \cdot s_A \]
The heat capacity for 12 volumes of B:
\[ C_B = m_B \cdot s_B = (\rho_B \cdot 12V) \cdot s_B \]

### Step 5: Set the heat capacities equal to each other
According to the problem statement:
\[ C_A = C_B \]
Thus,
\[ (\rho_A \cdot 8V) \cdot s_A = (\rho_B \cdot 12V) \cdot s_B \]

### Step 6: Cancel out the common terms
Since \( V \) is common in both sides, we can cancel it out:
\[ \rho_A \cdot 8 \cdot s_A = \rho_B \cdot 12 \cdot s_B \]

### Step 7: Rearrange to find the ratio of specific heats
Rearranging gives us:
\[ \frac{s_A}{s_B} = \frac{\rho_B \cdot 12}{\rho_A \cdot 8} \]

### Step 8: Substitute the known values
Substituting the values of \( \rho_A \) and \( \rho_B \):
\[ \frac{s_A}{s_B} = \frac{2000 \cdot 12}{1500 \cdot 8} \]

### Step 9: Simplify the expression
Calculating the right-hand side:
\[ \frac{s_A}{s_B} = \frac{24000}{12000} = 2 \]

### Step 10: Write the final ratio
Thus, the ratio of specific heats \( s_A : s_B \) is:
\[ s_A : s_B = 1 : 2 \]