an elastic spring has a length `l_(1)` when tension in it is 4N. Its length is `l_(2)` when tension in it is 5N. What will be its length when tension in it is 9N?

Options

A. `5l_(1)-4l_(2)`
B. `5l_(2)-4l_(1)`
C. `4l_(1)+5l_(2)`
D. `4l_(2)+5l_(1)`

Answer

To solve the problem, we will use Hooke's Law, which states that the force (tension) in a spring is proportional to the extension (change in length) of the spring.

Let's denote:
- \( l_1 \) = length of the spring when the tension is 4N
- \( l_2 \) = length of the spring when the tension is 5N
- \( l \) = length of the spring when the tension is 9N
- \( k \) = spring constant
- \( x \) = extension of the spring

### Step 1: Establish the relationship using Hooke's Law

From Hooke's Law, we know:
\[ F = k \cdot x \]
Where \( x \) is the extension of the spring from its natural length.

### Step 2: Determine the extensions for the known tensions

1. For the tension of 4N:
\[ 4 = k \cdot (l_1 - l_0) \quad \text{(where \( l_0 \) is the natural length)} \]
Rearranging gives:
\[ l_1 = l_0 + \frac{4}{k} \]

2. For the tension of 5N:
\[ 5 = k \cdot (l_2 - l_0) \]
Rearranging gives:
\[ l_2 = l_0 + \frac{5}{k} \]

### Step 3: Find the change in length between \( l_1 \) and \( l_2 \)

The change in length (extension) between \( l_1 \) and \( l_2 \) is:
\[ l_2 - l_1 = \left( l_0 + \frac{5}{k} \right) - \left( l_0 + \frac{4}{k} \right) = \frac{5}{k} - \frac{4}{k} = \frac{1}{k} \]

### Step 4: Set up the equation for 9N tension

Now, we need to find the length \( l \) when the tension is 9N:
\[ 9 = k \cdot (l - l_0) \]
Rearranging gives:
\[ l = l_0 + \frac{9}{k} \]

### Step 5: Relate \( l \) to \( l_1 \) and \( l_2 \)

We can express \( l \) in terms of \( l_1 \) and \( l_2 \):
Using the previous equations:
\[ l = l_0 + \frac{9}{k} \]
We know:
\[ l_0 = l_1 - \frac{4}{k} \]
Substituting this into the equation for \( l \):
\[ l = \left( l_1 - \frac{4}{k} \right) + \frac{9}{k} = l_1 + \frac{5}{k} \]

### Step 6: Express \( l \) in terms of \( l_2 \)

Similarly, we can express \( l \) in terms of \( l_2 \):
\[ l_0 = l_2 - \frac{5}{k} \]
Substituting this into the equation for \( l \):
\[ l = \left( l_2 - \frac{5}{k} \right) + \frac{9}{k} = l_2 + \frac{4}{k} \]

### Step 7: Final relation

Now we can set up the relationship:
\[ l - l_1 = \frac{5}{k} \]
\[ l - l_2 = \frac{4}{k} \]

Using the above equations, we can find:
\[ l = l_2 + \frac{4}{k} = l_1 + \frac{5}{k} \]

### Step 8: Solve for \( l \)

To find the length \( l \) when the tension is 9N, we can use the values of \( l_1 \) and \( l_2 \):
\[ l = \frac{5l_2 - 4l_1}{1} \]

### Conclusion

Thus, the length of the spring when the tension is 9N is given by:
\[ l = 5l_2 - 4l_1 \]