Density of a liquid 'A' is `0.5 g//c.c` and that of liquid 'B' is `0.6 g//c.c`. Heat capacity of `8` litres of 'A' is equal to that of `10` litres of 'B'. Then the specific heats ratio of `A` and `B` is

Options

A. `4 : 5`
B. `1 : 1`
C. `2 : 3`
D. `3 : 2`

Explanation

To solve the problem, we need to find the specific heat ratio of liquids A and B based on their densities and given heat capacities. Here’s a step-by-step solution:

### Step 1: Understand the relationship between heat capacity, mass, and specific heat.
The heat capacity (H) of a substance can be expressed as:
\[ H = M \cdot S \]
where \( M \) is the mass of the substance and \( S \) is its specific heat.

### Step 2: Express mass in terms of density and volume.
The mass \( M \) can also be expressed in terms of density (\( \rho \)) and volume (\( V \)):
\[ M = \rho \cdot V \]
Thus, we can rewrite the heat capacity as:
\[ H = \rho \cdot V \cdot S \]

### Step 3: Set up the equation for heat capacities of liquids A and B.
According to the problem, the heat capacity of 8 liters of liquid A is equal to that of 10 liters of liquid B:
\[ H_A = H_B \]
Substituting the expression for heat capacity, we have:
\[ \rho_A \cdot V_A \cdot S_A = \rho_B \cdot V_B \cdot S_B \]

### Step 4: Substitute the known values.
Given:
- Density of liquid A, \( \rho_A = 0.5 \, \text{g/cm}^3 \)
- Volume of liquid A, \( V_A = 8 \, \text{liters} = 8000 \, \text{cm}^3 \)
- Density of liquid B, \( \rho_B = 0.6 \, \text{g/cm}^3 \)
- Volume of liquid B, \( V_B = 10 \, \text{liters} = 10000 \, \text{cm}^3 \)

Substituting these values into the equation:
\[ (0.5) \cdot (8000) \cdot S_A = (0.6) \cdot (10000) \cdot S_B \]

### Step 5: Simplify the equation.
Calculating the left and right sides:
\[ 4000 \cdot S_A = 6000 \cdot S_B \]

### Step 6: Rearrange to find the ratio of specific heats.
Dividing both sides by \( S_B \) and rearranging gives:
\[ \frac{S_A}{S_B} = \frac{6000}{4000} = \frac{3}{2} \]

### Conclusion:
Thus, the ratio of the specific heats of liquid A to liquid B is:
\[ \frac{S_A}{S_B} = \frac{3}{2} \]