Question

If \( T = 2\pi \sqrt{\frac{L}{g}} \), \( g \) is a constant and the relative error in \( T \) is \( k \) times to the percentage error in \( L \), then \( \frac{1}{k} = \) ? 

• A. \( 200 \)
• B. \( \frac{1}{200} \)
• C. \( 2 \)
• D. \( \frac{1}{2} \)

Hint:

For error propagation in functions involving square roots, use logarithmic differentiation and identify the relative error coefficients.

Answer 1

The Correct Option is C

Step 1: Expressing the Relative Error 
The given equation for the time period of a simple pendulum is: \[ T = 2\pi \sqrt{\frac{L}{g}}. \] Taking the natural logarithm on both sides: \[ \ln T = \ln \left( 2\pi \right) + \frac{1}{2} \ln L - \frac{1}{2} \ln g. \] Differentiating both sides: \[ \frac{dT}{T} = \frac{1}{2} \frac{dL}{L}. \] This implies the relative error in \( T \): \[ \frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L}. \] Step 2: Finding the Value of \( k \) 
The problem states that the relative error in \( T \) is \( k \) times the percentage error in \( L \): \[ \frac{\Delta T}{T} = k \times \frac{\Delta L}{L}. \] Comparing with the earlier result: \[ k = \frac{1}{2}. \] Thus, \[ \frac{1}{k} = 2. \] Final Answer: \( \boxed{2} \).

Answer 2

Step 1: Take logarithm and differentiate 

Given: \[ T = 2\pi \sqrt{\frac{L}{g}} \Rightarrow T \propto \sqrt{L} \] Taking logarithm: \[ \ln T = \ln (2\pi) + \frac{1}{2} \ln L - \frac{1}{2} \ln g \] Differentiating: \[ \frac{dT}{T} = \frac{1}{2} \cdot \frac{dL}{L} \quad \text{(since \( g \) is constant)} \]

Step 2: Interpret in terms of relative and percentage errors

Relative error in \( T \): \( \frac{\Delta T}{T} \)
Relative error in \( L \): \( \frac{\Delta L}{L} \) From above: \[ \frac{\Delta T}{T} = \frac{1}{2} \cdot \frac{\Delta L}{L} \Rightarrow \text{Relative error in } T = \frac{1}{2} \text{ (Relative error in } L) \] So, \[ \boxed{k = \frac{1}{2} \Rightarrow \frac{1}{k} = 2} \]

Answer:

\( \boxed{2} \)