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' based on the given errors in the length of the pendulum (L) and the time period (T), we can follow these steps:

### Step 1: Understand the relationship between g, L, and T
The formula for the time period (T) of a simple pendulum is given by:

\[ T = 2\pi \sqrt{\frac{L}{g}} \]

From this formula, we can express g in terms of L and T:

\[ g = \frac{4\pi^2 L}{T^2} \]

### Step 2: Determine the percentage errors
We are given:
- The percentage error in the measurement of length (L) is 2%, which can be expressed as:

\[ \frac{\Delta L}{L} = 0.02 \]

- The percentage error in the measurement of time period (T) is 1%, which can be expressed as:

\[ \frac{\Delta T}{T} = 0.01 \]

### Step 3: Use the formula for error propagation
To find the maximum error in g, we can use the formula for error propagation. For a quantity \( g \) that depends on \( L \) and \( T \), the relative error in g can be expressed as:

\[ \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} \]

### Step 4: Substitute the known values
Now, substituting the known values into the error propagation formula:

\[ \frac{\Delta g}{g} = 0.02 + 2 \times 0.01 \]

Calculating this gives:

\[ \frac{\Delta g}{g} = 0.02 + 0.02 = 0.04 \]

### Step 5: Convert to percentage
To express this as a percentage, we multiply by 100:

\[ \text{Percentage error in } g = 0.04 \times 100 = 4\% \]

### Conclusion
The maximum error in the measurement of g is **4%**.

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