A sphere and a cube of same material and same total surface area are placed in the same evaculated space turn by turn after they are heated to the same temperature. Find the ratio of their initial rates of cooling in the enclosure.

Options

A. `sqrt(pi/6) : 1`
B. `sqrt(pi/3) : 1`
C. `(pi)/sqrt(6) : 1`
D. `(pi)/sqrt(3) : 1`

Explanation

To find the ratio of the initial rates of cooling of a sphere and a cube made of the same material and having the same total surface area, we can follow these steps:

### Step 1: Understand the cooling process
The rate of cooling of an object is determined by Newton's law of cooling, which states that the rate of heat loss of a body is proportional to the difference in temperature between the body and its surroundings.

### Step 2: Define the parameters
Let:
- \( R_s \) = rate of cooling of the sphere
- \( R_c \) = rate of cooling of the cube
- \( A_s \) = surface area of the sphere
- \( A_c \) = surface area of the cube
- \( V_s \) = volume of the sphere
- \( V_c \) = volume of the cube
- \( m_s \) = mass of the sphere
- \( m_c \) = mass of the cube
- \( C \) = specific heat capacity of the material

### Step 3: Relate surface area and volume
For a sphere of radius \( r \):
- Surface area: \( A_s = 4\pi r^2 \)
- Volume: \( V_s = \frac{4}{3}\pi r^3 \)

For a cube with side length \( a \):
- Surface area: \( A_c = 6a^2 \)
- Volume: \( V_c = a^3 \)

### Step 4: Set the surface areas equal
Since the surface areas are equal:
\[ 4\pi r^2 = 6a^2 \]
From this, we can express \( a \) in terms of \( r \):
\[ a^2 = \frac{2\pi}{3} r^2 \implies a = \sqrt{\frac{2\pi}{3}} r \]

### Step 5: Calculate the masses
Since both objects are made of the same material, their masses can be expressed as:
\[ m_s = \rho V_s = \rho \left(\frac{4}{3}\pi r^3\right) \]
\[ m_c = \rho V_c = \rho a^3 = \rho \left(\sqrt{\frac{2\pi}{3}} r\right)^3 = \rho \left(\frac{2\pi}{3}\right)^{3/2} r^3 \]

### Step 6: Find the ratio of the rates of cooling
Using Newton's law of cooling, the rate of cooling can be expressed as:
\[ R_s = \frac{A_s}{m_s} \cdot C \cdot \Delta T \]
\[ R_c = \frac{A_c}{m_c} \cdot C \cdot \Delta T \]

The ratio of the rates of cooling is:
\[ \frac{R_s}{R_c} = \frac{A_s / m_s}{A_c / m_c} = \frac{A_s \cdot m_c}{A_c \cdot m_s} \]

### Step 7: Substitute the values
Substituting the expressions for surface areas and masses:
\[ \frac{R_s}{R_c} = \frac{4\pi r^2 \cdot \left(\frac{2\pi}{3}\right)^{3/2} r^3}{6a^2 \cdot \left(\frac{4}{3}\pi r^3\right)} \]

### Step 8: Simplify the expression
After substituting \( a^2 \) from step 4 and simplifying, we find:
\[ \frac{R_s}{R_c} = \frac{4\pi r^2 \cdot \frac{2\pi}{3\sqrt{3}} r^3}{6 \cdot \frac{2\pi}{3} r^2 \cdot \frac{4}{3}\pi r^3} \]

### Step 9: Final ratio
After simplification, we find:
\[ \frac{R_s}{R_c} = \frac{3}{2} \]

### Conclusion
The ratio of the initial rates of cooling of the sphere to the cube is \( \frac{3}{2} \).