A hot body is placed in cold surroundings It's rate of cooling is `3^(@)C` per minute when its temperature is `70^(@)C` and `1.5^(@)C` per minute when its temperature `50^(@)C` it s rate of cooling when its temperature is `40^(@)C` .

Options

A. `0.25^(@)C//min`
B. `0.5^(@)C//min`
C. `0.75^(@)C//min`
D. `1^(@)C//min`

Explanation

To solve the problem, we will use Newton's Law of Cooling, which states that the rate of cooling of a body is directly proportional to the difference in temperature between the body and its surroundings. Let's denote:

- \( \theta \) = temperature of the body
- \( \theta_0 \) = temperature of the surroundings
- \( k \) = constant of proportionality

According to Newton's Law of Cooling, we can express the rate of cooling as:

\[ \frac{d\theta}{dt} = k(\theta - \theta_0) \]

### Step 1: Set up the equations based on the given information

1. When \( \theta = 70^\circ C \), the rate of cooling is \( \frac{d\theta}{dt} = -3 \) degrees per minute:
\[ -3 = k(70 - \theta_0) \quad \text{(1)} \]

2. When \( \theta = 50^\circ C \), the rate of cooling is \( \frac{d\theta}{dt} = -1.5 \) degrees per minute:
\[ -1.5 = k(50 - \theta_0) \quad \text{(2)} \]

### Step 2: Solve the equations for \( \theta_0 \) and \( k \)

From equation (1):
\[ k = \frac{-3}{70 - \theta_0} \quad \text{(3)} \]

From equation (2):
\[ k = \frac{-1.5}{50 - \theta_0} \quad \text{(4)} \]

### Step 3: Equate the two expressions for \( k \)

Set equations (3) and (4) equal to each other:
\[ \frac{-3}{70 - \theta_0} = \frac{-1.5}{50 - \theta_0} \]

### Step 4: Cross-multiply to eliminate the fractions

\[ -3(50 - \theta_0) = -1.5(70 - \theta_0) \]

### Step 5: Simplify and solve for \( \theta_0 \)

Expanding both sides:
\[ -150 + 3\theta_0 = -105 + 1.5\theta_0 \]

Rearranging gives:
\[ 3\theta_0 - 1.5\theta_0 = 150 - 105 \]
\[ 1.5\theta_0 = 45 \]

Dividing by 1.5:
\[ \theta_0 = 30^\circ C \]

### Step 6: Substitute \( \theta_0 \) back to find \( k \)

Using equation (3):
\[ k = \frac{-3}{70 - 30} = \frac{-3}{40} = -\frac{3}{40} \]

### Step 7: Find the rate of cooling when \( \theta = 40^\circ C \)

Using the formula:
\[ \frac{d\theta}{dt} = k(\theta - \theta_0) \]
Substituting \( \theta = 40^\circ C \):
\[ \frac{d\theta}{dt} = -\frac{3}{40}(40 - 30) \]
\[ = -\frac{3}{40} \times 10 = -\frac{30}{40} = -\frac{3}{4} \]

Thus, the rate of cooling when the temperature of the body is \( 40^\circ C \) is:
\[ \frac{d\theta}{dt} = -0.75^\circ C \text{ per minute} \]

### Final Answer:
The rate of cooling when the temperature is \( 40^\circ C \) is \( 0.75^\circ C \) per minute.