Question

The period of oscillation of a simple pendulum is \(T = 2π(L/g)^{1/2}\). Measured value of 'L' is 10 cm known to 1 mm accuracy and time for 100 oscillations is 50 second, using a watch of 1 second resolution. The percentage error in the measurement of 'g' is

• A. \(2\%\)
• B. \(4\%\)
• C. \(5\%\)
• D. \(8\%\)

Hint:

Write g in terms of L and T, then add the fractional error in L to twice the fractional error in T.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Errors add up when quantities are combined. For a formula \(g = 4\pi^2 L/T^2\), the fractional error in \(g\) is the fractional error in \(L\) plus twice the fractional error in \(T\), because \(T\) is squared.

Step 2: Key Formula or Approach:
\[ T = 2\pi\sqrt{\frac{L}{g}} \Rightarrow g = \frac{4\pi^2 L}{T^2} \Rightarrow \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T} \]

Step 3: Error in L.
\(L = 10\) cm and \(\Delta L = 1\) mm \(= 0.1\) cm. So \(\dfrac{\Delta L}{L} = \dfrac{0.1}{10} = 0.01 = 1\%\).

Step 4: Error in T.
100 oscillations take 50 s, with a watch resolution of 1 s. The error in the total time is 1 s. So \(\dfrac{\Delta T}{T} = \dfrac{1}{50} = 0.02 = 2\%\) (the ratio is the same whether we use the total time or the period).

Step 5: Combine.
\[ \frac{\Delta g}{g}\times 100 = 1\% + 2\times 2\% = 5\% \]

Final Answer:
The percentage error in \(g\) is 5 %, option (C). \[ \boxed{5\%} \]