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. (a)`sqrt((pi)/(6)):1`
B. (b)`sqrt((pi)/(3)):1`
C. (c)`(pi)/(sqrt(6)):1`
D. (d)`(pi)/(sqrt(3)):1`

Explanation

To solve the problem of finding the ratio of the initial rates of cooling of a sphere and a cube of the same material and same total surface area, we can follow these steps:

### Step 1: Understand the cooling process
The rate of cooling of an object can be described 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. However, in this case, we will use the Stefan-Boltzmann law, which states that the power radiated by a black body is proportional to the fourth power of its absolute temperature and its surface area.

### Step 2: Write the expression for the rate of heat loss
The rate of heat loss (dQ/dt) can be expressed as:
\[ \frac{dQ}{dt} = \sigma A (T^4 - T_{env}^4) \]
where:
- \(\sigma\) is the Stefan-Boltzmann constant,
- \(A\) is the surface area,
- \(T\) is the temperature of the object,
- \(T_{env}\) is the temperature of the environment (which is 0 in this case since it's evacuated).

### Step 3: Surface area of the sphere and cube
Let’s denote:
- \(A_{sphere}\) = surface area of the sphere = \(4\pi r^2\)
- \(A_{cube}\) = surface area of the cube = \(6a^2\)

Given that both have the same total surface area:
\[ 4\pi r^2 = 6a^2 \]

### Step 4: Find the mass of the sphere and cube
The mass of the sphere (m_sphere) and cube (m_cube) can be expressed as:
- \(m_{sphere} = \rho \cdot V_{sphere} = \rho \cdot \frac{4}{3}\pi r^3\)
- \(m_{cube} = \rho \cdot V_{cube} = \rho \cdot a^3\)

Where \(\rho\) is the density of the material.

### Step 5: Calculate the ratio of the rates of cooling
Now, we can express the rates of cooling for both shapes:
\[ \frac{dQ_{sphere}}{dt} = \sigma (4\pi r^2) (T^4) \]
\[ \frac{dQ_{cube}}{dt} = \sigma (6a^2) (T^4) \]

The ratio of the rates of cooling is:
\[ \frac{\frac{dQ_{sphere}}{dt}}{\frac{dQ_{cube}}{dt}} = \frac{4\pi r^2}{6a^2} \]

### Step 6: Substitute the surface area relationship
From the surface area relationship \(4\pi r^2 = 6a^2\), we can substitute \(a^2\) in terms of \(r^2\):
\[ \frac{dQ_{sphere}}{dt} = \sigma (4\pi r^2) (T^4) \]
\[ \frac{dQ_{cube}}{dt} = \sigma (6 \cdot \frac{4\pi r^2}{6}) (T^4) = \sigma (4\pi r^2) (T^4) \]

### Step 7: Final ratio
Thus, the ratio of the initial rates of cooling becomes:
\[ \frac{dQ_{sphere}/dt}{dQ_{cube}/dt} = \frac{4\pi r^2}{4\pi r^2} = 1 \]

### Conclusion
The ratio of the initial rates of cooling of the sphere and the cube is 1:1.