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

Concept:
From \[ T=2\pi\sqrt{\frac{l}{g}} \] we get \[ g\propto \frac{l}{T^2} \] So fractional error is: \[ \frac{\Delta g}{g}=\frac{\Delta l}{l}+2\frac{\Delta T}{T} \] ip

Step 1: Find percentage error in length.
\[ l=100\text{ cm}=1\text{ m} \] Error in length: \[ \Delta l=1\text{ mm}=0.001\text{ m} \] So, \[ \frac{\Delta l}{l}=\frac{0.001}{1}=0.001=0.1% \] ip

Step 2: Find percentage error in period.
Period is \(2\text{ s}\), so time for \(100\) oscillations: \[ t=200\text{ s} \] Least count of stopwatch: \[ \Delta t=0.1\text{ s} \] Thus, \[ \frac{\Delta t}{t}=\frac{0.1}{200}=0.0005 \] Since \[ T=\frac{t}{100}, \] fractional error in \(T\) is same as fractional error in \(t\): \[ \frac{\Delta T}{T}=0.0005=0.05% \] ip

Step 3: Find percentage error in \(g\).
\[ \frac{\Delta g}{g} = \frac{\Delta l}{l}+2\frac{\Delta T}{T} \] \[ =0.1% + 2(0.05%) \] \[ =0.1%+0.1%=0.2% \] ip Hence, the correct answer is:
\[ \boxed{(A)\ 0.2%} \]