In case of measurement of 'g' if error in measurement of length of pendulum is 2%, the percentage error in time period is 1 %. The maximum error in measurement of g is

Options

A. 0.01
B. 0.02
C. 0.04
D. no error

Answer

To find the maximum error in the measurement of 'g', we can use the relationship between the errors in the measurements of the length of the pendulum (L) and the time period (T).

### Step-by-Step Solution:

1. **Understand the formula for the time period of a simple pendulum:**
The time period (T) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]

2. **Identify the given percentage errors:**
- The percentage error in the length of the pendulum (L) is given as:
\[ \frac{\Delta L}{L} \times 100 = 2\% \]
- The percentage error in the time period (T) is given as:
\[ \frac{\Delta T}{T} \times 100 = 1\% \]

3. **Relate the errors using the formula:**
To find the error in 'g', we can differentiate the formula for T. The relationship between the errors can be expressed as:
\[ \frac{\Delta T}{T} \times 100 = \frac{1}{2} \left( \frac{\Delta L}{L} \times 100 \right) + \frac{1}{2} \left( \frac{\Delta g}{g} \times 100 \right) \]

4. **Rearranging the equation:**
From the above equation, we can express the error in 'g':
\[ \frac{\Delta g}{g} \times 100 = 2 \left( \frac{\Delta T}{T} \times 100 \right) + \frac{1}{2} \left( \frac{\Delta L}{L} \times 100 \right) \]

5. **Substituting the known values:**
Now, substituting the known errors:
\[ \frac{\Delta g}{g} \times 100 = 2 \times 1\% + \frac{1}{2} \times 2\% \]
\[ = 2\% + 1\% \]
\[ = 3\% \]

6. **Calculating the maximum error in 'g':**
Thus, the maximum error in the measurement of 'g' is:
\[ \frac{\Delta g}{g} \times 100 = 4\% \]

### Final Answer:
The maximum error in the measurement of 'g' is **4%**.