Two forces `F_(1)` and `F_(2)` are acting at a point, whose resultant is `F`.If `F_(2)` is doubled `F` is also doubled. If `F_(2)` is reversed then also `F` is doubled. Then `F_(1):F_(2):F` is

Options

A. `sqrt2:sqrt2:sqrt3`
B. `sqrt3:sqrt3:sqrt2`
C. `sqrt3:sqrt2:sqrt3`
D. `sqrt2:sqrt3:sqrt2`

Explanation

To solve the problem, we need to analyze the conditions given and derive the relationship between the forces \( F_1 \), \( F_2 \), and the resultant force \( F \).

### Step-by-Step Solution:

1. **Understanding the Forces**:
- Let \( F_1 \) and \( F_2 \) be two forces acting at a point, and their resultant is \( F \).
- We can represent the resultant force using the parallelogram law of vector addition.

2. **First Condition**: When \( F_2 \) is doubled:
- The new force \( F_2' = 2F_2 \).
- The resultant becomes \( F' = F_1 + 2F_2 \).
- According to the problem, this resultant is also double the original resultant: \( F' = 2F \).
- Therefore, we can write the equation:
\[ F_1 + 2F_2 = 2F \tag{1} \]

3. **Second Condition**: When \( F_2 \) is reversed:
- The new force \( F_2'' = -F_2 \).
- The resultant becomes \( F'' = F_1 - F_2 \).
- Again, this resultant is double the original resultant: \( F'' = 2F \).
- Thus, we can write the equation:
\[ F_1 - F_2 = 2F \tag{2} \]

4. **Setting Up the Equations**:
- From equation (1):
\[ F_1 + 2F_2 = 2F \tag{1} \]
- From equation (2):
\[ F_1 - F_2 = 2F \tag{2} \]

5. **Solving the Equations**:
- We can solve these two equations simultaneously.
- From equation (2), we can express \( F_1 \):
\[ F_1 = 2F + F_2 \tag{3} \]
- Substitute equation (3) into equation (1):
\[ (2F + F_2) + 2F_2 = 2F \]
\[ 2F + 3F_2 = 2F \]
\[ 3F_2 = 0 \implies F_2 = 0 \]
- Substitute \( F_2 = 0 \) back into equation (3):
\[ F_1 = 2F + 0 = 2F \]

6. **Finding the Ratios**:
- Now we have \( F_1 = 2F \) and \( F_2 = 0 \).
- The ratio \( F_1 : F_2 : F \) becomes:
\[ 2F : 0 : F \]
- This can be simplified to:
\[ 2 : 0 : 1 \]

### Final Result:
The ratio \( F_1 : F_2 : F \) is \( 2 : 0 : 1 \).