Question

A simple pendulum is being used to determine the value of gravitational acceleration at a certain place. The length of the pendulum is $25.0cm$ measured with accuracy of $0.1cm$ and a stopwatch with $1s$ resolution measures the time taken for $40$ oscillations to be $50s$. The accuracy in $g$

• A. $5.40\%$
• B. $3.40\%$
• C. $2.2\%$
• D. $4.40\%$

Answer 1

The correct option is D

$4.40\%$

Step 1. Given:

$∆T=1s$ ,

$T=50s$

$l=25cm$

Step 2.Finding the accuracy in $g$ :

$$\begin{gathered} T=2\pi \sqrt{\frac{l}{g}} \\ g=\frac{4{\pi }^{2}l}{T^2} \end{gathered}$$

Where $g$ is the acceleration due to gravity, $T$ is the time period,$l$is the length of string

Taking log both side and differentiating, we get the following equation

$\frac{∆g}{g}=2\frac{∆T}{T}+\frac{∆l}{l}$

$\frac{∆g}{g}=2\frac{1}{50}+\frac{0.1}{25}=\frac{1.1}{25}$

Accuracy in $g$ is $=\frac{∆g}{g}\times 100=\frac{1.1\times 100}{25}=4.4\%$

Hence, the correct option is D.