Two metallic spheres `S_1 and S_2` are made of the same material and have got identical surface finish. The mass of `S_1` is thrice that of `S_2`. Both the spheres are heated to the same high temperature and placed in the same room having lower temperature but are thermally insulated from each other. the ratio of the initial rate of cooling of `S_1` to that of `S_2` is
`(a)1/3 (b)1/(sqrt3) (c) (sqrt3)/1 (d) (1/3)^(1/3)`

Explanation

To solve the problem, we need to determine the ratio of the initial rate of cooling of two metallic spheres, \( S_1 \) and \( S_2 \), given that the mass of \( S_1 \) is thrice that of \( S_2 \).

### Step-by-step Solution:

1. **Understanding the Problem**:
- We have two spheres, \( S_1 \) and \( S_2 \).
- The mass of \( S_1 \) is \( m_1 = 3m_2 \).
- Both spheres are made of the same material and have identical surface finishes.
- They are heated to the same temperature and placed in the same environment.

2. **Using Stefan-Boltzmann Law**:
- The rate of heat loss due to radiation is given by:
\[ \frac{dQ}{dt} = \sigma E A T^4 \]
where:
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( E \) is the emissivity (same for both since they are identical),
- \( A \) is the surface area,
- \( T \) is the temperature.

3. **Finding the Surface Area**:
- The mass of a sphere is given by:
\[ m = \rho V = \rho \left(\frac{4}{3} \pi R^3\right) \]
- Therefore, for spheres \( S_1 \) and \( S_2 \):
\[ m_1 = \rho \left(\frac{4}{3} \pi R_1^3\right) \quad \text{and} \quad m_2 = \rho \left(\frac{4}{3} \pi R_2^3\right) \]
- Since \( m_1 = 3m_2 \):
\[ \rho \left(\frac{4}{3} \pi R_1^3\right) = 3 \rho \left(\frac{4}{3} \pi R_2^3\right) \]
- This simplifies to:
\[ R_1^3 = 3 R_2^3 \implies R_1 = 3^{1/3} R_2 \]

4. **Calculating Surface Areas**:
- The surface area of a sphere is given by:
\[ A = 4 \pi R^2 \]
- Thus, for \( S_1 \) and \( S_2 \):
\[ A_1 = 4 \pi R_1^2 = 4 \pi (3^{1/3} R_2)^2 = 4 \pi (3^{2/3} R_2^2) = 3^{2/3} A_2 \]

5. **Finding the Rate of Cooling**:
- The rate of cooling for each sphere can be expressed as:
\[ \frac{d\theta}{dt} \propto \frac{dQ}{dt} \propto \frac{A}{m} \]
- Therefore, the ratio of the rates of cooling is:
\[ \frac{\frac{d\theta_1}{dt}}{\frac{d\theta_2}{dt}} = \frac{A_1/m_1}{A_2/m_2} = \frac{A_1}{A_2} \cdot \frac{m_2}{m_1} \]

6. **Substituting Values**:
- We know \( A_1 = 3^{2/3} A_2 \) and \( m_1 = 3m_2 \):
\[ \frac{d\theta_1}{dt} / \frac{d\theta_2}{dt} = \frac{3^{2/3} A_2}{A_2} \cdot \frac{m_2}{3m_2} = 3^{2/3} \cdot \frac{1}{3} = \frac{3^{2/3}}{3} = 3^{-1/3} \]

7. **Final Answer**:
- The ratio of the initial rate of cooling of \( S_1 \) to that of \( S_2 \) is:
\[ \frac{d\theta_1}{dt} : \frac{d\theta_2}{dt} = \left(\frac{1}{3}\right)^{1/3} \]

### Conclusion:
The correct answer is option \( (d) \left(\frac{1}{3}\right)^{1/3} \).