Taking that earth revolves round the sun in circular orbit of radius `15 xx 10^(10)` m with a time period of lyear. The time taken by another planet which is at a distance of `540 xx 10^(10)m` , to revolve round the sun in circular orbit once will be

Options

A. 216 years
B. 144 years
C. 72 years
D. 36 years

Answer

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

### Step-by-Step Solution

1. **Identify the given values:**
- Radius of Earth's orbit, \( r_1 = 15 \times 10^{10} \) m
- Time period of Earth's revolution, \( T_1 = 1 \) year
- Radius of the other planet's orbit, \( r_2 = 540 \times 10^{10} \) m
- Time period of the other planet's revolution, \( T_2 = ? \)

2. **Apply Kepler's Third Law:**
According to Kepler's Third Law:
\[ \frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3} \]

3. **Substitute the known values:**
\[ \frac{T_2^2}{(1 \text{ year})^2} = \frac{(540 \times 10^{10})^3}{(15 \times 10^{10})^3} \]

4. **Simplify the equation:**
\[ T_2^2 = (1 \text{ year})^2 \cdot \frac{(540)^3}{(15)^3} \]

5. **Calculate the ratio:**
First, calculate \( \frac{540^3}{15^3} \):
\[ \frac{540^3}{15^3} = \left(\frac{540}{15}\right)^3 = (36)^3 = 46656 \]

6. **Substitute back to find \( T_2^2 \):**
\[ T_2^2 = 1^2 \cdot 46656 = 46656 \text{ years}^2 \]

7. **Take the square root to find \( T_2 \):**
\[ T_2 = \sqrt{46656} = 216 \text{ years} \]

### Final Answer:
The time taken by the other planet to revolve around the sun once is **216 years**.