In an experiment for estimating the value of focal length of converging mirror,image of an object placed at `40cm` from the pole of the mirror is formed at distance `120cm` from the pole of the mirror.These distances are measured with a modified scale in which there are `20` small divisions in `1cm`.The value of error in measurement of focal length of the mirror is `(1)/(K)cm`.The value of `K` is

Answer

To solve the problem, we need to find the value of \( K \) in the context of the error in the measurement of the focal length of a converging mirror.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The object distance \( u = -40 \, \text{cm} \) (negative as per the sign convention).
- The image distance \( v = -120 \, \text{cm} \) (also negative).
- The distances are measured with a scale that has 20 small divisions in 1 cm, which gives us the least count \( \Delta u = \Delta v = \frac{1}{20} \, \text{cm} \).

2. **Use the Mirror Formula**:
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the values of \( u \) and \( v \):
\[ \frac{1}{f} = \frac{1}{-120} + \frac{1}{-40} \]

3. **Calculate \( \frac{1}{f} \)**:
\[ \frac{1}{f} = -\frac{1}{120} - \frac{1}{40} \]
To combine these fractions, find a common denominator (which is 120):
\[ \frac{1}{f} = -\frac{1}{120} - \frac{3}{120} = -\frac{4}{120} = -\frac{1}{30} \]
Thus, \( f = -30 \, \text{cm} \).

4. **Calculate the Error in Focal Length**:
The formula for the error in focal length is given by:
\[ \frac{\Delta f}{f^2} = \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} \]
Since \( \Delta u = \Delta v = \frac{1}{20} \), we can substitute:
\[ \frac{\Delta f}{(-30)^2} = \frac{\frac{1}{20}}{(-120)^2} + \frac{\frac{1}{20}}{(-40)^2} \]

5. **Calculate Each Term**:
- For \( v = -120 \):
\[ \frac{\Delta v}{(-120)^2} = \frac{\frac{1}{20}}{14400} = \frac{1}{288000} \]
- For \( u = -40 \):
\[ \frac{\Delta u}{(-40)^2} = \frac{\frac{1}{20}}{1600} = \frac{1}{32000} \]

6. **Combine the Errors**:
\[ \frac{\Delta f}{900} = \frac{1}{288000} + \frac{1}{32000} \]
Finding a common denominator (which is 288000):
\[ \frac{1}{32000} = \frac{9}{288000} \]
Thus,
\[ \frac{\Delta f}{900} = \frac{1 + 9}{288000} = \frac{10}{288000} \]

7. **Solve for \( \Delta f \)**:
\[ \Delta f = \frac{10 \times 900}{288000} = \frac{9000}{288000} = \frac{1}{32} \, \text{cm} \]

8. **Relate to \( K \)**:
The problem states that the error in measurement of focal length is given as \( \frac{1}{K} \, \text{cm} \). Therefore, we have:
\[ \frac{1}{K} = \frac{1}{32} \implies K = 32 \]

### Final Answer:
Thus, the value of \( K \) is \( \boxed{32} \).