A sphere and a cube of same material and same volume are heated up to same temperature and allowed to cool in the same sorroundings. The radio of the amounts of radiations emitted in equal time intervals will be

Options

A. `1:1`
B. `(4pi)/(3):(1)/(86400)//^@C 1`
C. `((pi)/(6))^(1//3):1`
D. `(1)/(2)((4pi)/(3))^(2//3):1`

Explanation

To solve the problem, we need to determine the ratio of the amounts of radiation emitted by a sphere and a cube of the same material and volume, heated to the same temperature, and allowed to cool in the same surroundings. We will use the principles of Stefan's law of radiation and the geometry of the objects involved.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a sphere and a cube made of the same material and having the same volume.
- Both are heated to the same temperature and allowed to cool in the same surroundings.

2. **Using Stefan's Law:**
- According to Stefan's law, the emissive power (Q) of a body is given by:
\[ Q = A \cdot e \cdot \sigma (T^4 - T_0^4) \]
where:
- \(A\) = surface area of the body,
- \(e\) = emissivity (same for both since they are made of the same material),
- \(\sigma\) = Stefan-Boltzmann constant (same for both),
- \(T\) = absolute temperature of the body,
- \(T_0\) = temperature of the surroundings.

3. **Setting Up the Equations:**
- Let \(Q_1\) be the heat emitted by the sphere and \(Q_2\) be the heat emitted by the cube.
- Since both bodies are at the same temperature and made of the same material, we can express:
\[ Q_1 = A_1 \cdot e \cdot \sigma (T^4 - T_0^4) \]
\[ Q_2 = A_2 \cdot e \cdot \sigma (T^4 - T_0^4) \]

4. **Finding the Ratio of Heat Emitted:**
- The ratio of heat emitted by the sphere to that emitted by the cube is:
\[ \frac{Q_1}{Q_2} = \frac{A_1}{A_2} \]

5. **Calculating Surface Areas:**
- The surface area of the sphere \(A_1\) is given by:
\[ A_1 = 4\pi r^2 \]
- The volume of the sphere is:
\[ V = \frac{4}{3}\pi r^3 \]
- The surface area of the cube \(A_2\) is given by:
\[ A_2 = 6a^2 \]
- The volume of the cube is:
\[ V = a^3 \]

6. **Equating Volumes:**
- Since the volumes are equal:
\[ \frac{4}{3}\pi r^3 = a^3 \implies a = \left(\frac{4}{3}\pi\right)^{1/3} r \]

7. **Substituting for Surface Areas:**
- Substitute \(a\) into the area of the cube:
\[ A_2 = 6\left(\left(\frac{4}{3}\pi\right)^{1/3} r\right)^2 = 6\left(\frac{4}{3}\pi\right)^{2/3} r^2 \]

8. **Finding the Ratio of Areas:**
- Now, we can find the ratio of the areas:
\[ \frac{A_1}{A_2} = \frac{4\pi r^2}{6\left(\frac{4}{3}\pi\right)^{2/3} r^2} = \frac{4\pi}{6\left(\frac{4}{3}\right)^{2/3}} = \frac{4\pi}{6 \cdot \frac{16}{9^{2/3}}} \]

9. **Simplifying the Ratio:**
- This simplifies to:
\[ \frac{4\pi}{\frac{96}{9^{2/3}}} = \frac{4\pi \cdot 9^{2/3}}{96} = \frac{\pi \cdot 9^{2/3}}{24} \]

10. **Final Ratio:**
- The final ratio of the amounts of radiation emitted by the sphere and the cube is:
\[ \frac{Q_1}{Q_2} = \frac{4\pi}{6\left(\frac{4}{3}\right)^{2/3}} = \frac{2\pi}{3\left(\frac{4}{3}\right)^{2/3}} = \frac{2\pi \cdot 3^{2/3}}{4^{2/3}} = \frac{2\pi \cdot 3^{2/3}}{16^{1/3}} = \frac{2\pi}{8} = \frac{\pi}{4} \]

### Conclusion:
The ratio of the amounts of radiation emitted by the sphere and the cube is:
\[ \frac{Q_1}{Q_2} = \frac{1}{3} \]