The mean radius of the earth's orbit around the sun is `1.5 xx 10^(11)m` and that of the orbit of mercury is `6 xx 10^(10)m`. The mercury will revolve around the sun is nearly

Options

A. `sqrt((2)/(5))yr`
B. `(2)/(5)yr`
C. `((2)/(5))^(2)yr`
D. `((2)/(5))^(3//2)yr`

Explanation

To solve the problem of determining the time period of Mercury's revolution around the Sun, we will use Kepler's Third Law of planetary motion, which states that the square of the time period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Mean radius of Earth's orbit, \( R_E = 1.5 \times 10^{11} \) m
- Mean radius of Mercury's orbit, \( R_M = 6 \times 10^{10} \) m
- Time period of Earth, \( T_E = 1 \) year

2. **Apply Kepler's Third Law**:
According to Kepler's Third Law:
\[ \frac{T_M^2}{T_E^2} = \frac{R_M^3}{R_E^3} \]
Where \( T_M \) is the time period of Mercury.

3. **Rearranging the Equation**:
We can rearrange the equation to find \( T_M \):
\[ T_M = T_E \left( \frac{R_M}{R_E} \right)^{3/2} \]

4. **Substituting the Values**:
Substitute the values of \( R_M \) and \( R_E \):
\[ T_M = 1 \left( \frac{6 \times 10^{10}}{1.5 \times 10^{11}} \right)^{3/2} \]

5. **Simplifying the Ratio**:
Calculate the ratio:
\[ \frac{6 \times 10^{10}}{1.5 \times 10^{11}} = \frac{6}{15} = \frac{2}{5} \]

6. **Calculating \( T_M \)**:
Now substitute this ratio back into the equation:
\[ T_M = 1 \left( \frac{2}{5} \right)^{3/2} \]

7. **Calculating \( \left( \frac{2}{5} \right)^{3/2} \)**:
First, calculate \( \left( \frac{2}{5} \right)^{3/2} \):
\[ \left( \frac{2}{5} \right)^{3/2} = \frac{2^{3/2}}{5^{3/2}} = \frac{2 \sqrt{2}}{5 \sqrt{5}} \approx \frac{2 \times 1.414}{5 \times 2.236} \approx \frac{2.828}{11.18} \approx 0.253 \]

8. **Final Calculation of \( T_M \)**:
Thus, the time period of Mercury is:
\[ T_M \approx 0.253 \text{ years} \]

### Conclusion:
The time period of Mercury's revolution around the Sun is approximately 0.253 years.