Question

The period of oscillating simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$ where length ' $l$ ' is $100\text{ cm}$ with error $1\text{ mm}$ . Period is $2\text{ second}$. The time of $100$ oscillations is measured by a stopwatch of least count $0.1\text{s}$. The percentage error in gravitational acceleration ' $g$ ' is

• A. $0.2%$
• B. $0.1%$
• C. $1%$
• D. $2%$

Hint:

For quantities like \[ g\propto \frac{l}{T^2}, \] add percentage errors according to powers: \[ \frac{\Delta g}{g}=\frac{\Delta l}{l}+2\frac{\Delta T}{T} \]

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The fractional error in a quantity calculated from products/powers is the sum of fractional errors of the base quantities multiplied by their respective powers.
Step 2: Key Formula or Approach:
1. Formula for $g$: $g = \frac{4\pi^2 l}{T^2}$
2. Error Relation: $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \frac{\Delta T}{T}$
3. For $N$ oscillations: $\frac{\Delta T}{T} = \frac{\Delta t}{t}$, where $t = N \times T$.
Step 3: Detailed Explanation:
Given:
$l = 100\text{ cm}, \Delta l = 1\text{ mm} = 0.1\text{ cm} \implies \frac{\Delta l}{l} = \frac{0.1}{100} = 0.001$.
Number of oscillations $N = 100$, Time Period $T = 2\text{ s}$.
Total time measured $t = 100 \times 2 = 200\text{ s}$.
Least count of stopwatch $\Delta t = 0.1\text{ s}$.
Fractional error in $T$: $\frac{\Delta T}{T} = \frac{0.1}{200} = 0.0005$.
Total percentage error in $g$:
\[ % \text{ Error in } g = \left( \frac{\Delta l}{l} + 2 \frac{\Delta T}{T} \right) \times 100 \]
\[ % \text{ Error} = (0.001 + 2 \times 0.0005) \times 100 = (0.001 + 0.001) \times 100 = 0.2% \]
Step 4: Final Answer:
The percentage error in $g$ is $0.2%$.