Question

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

• A. $5.40\%$
• B. $3.40\%$
• C. $4.40\%$
• D. $2.40\%$

Answer 1

The correct option is C $4.40\%$

$\text{Given, Length of simple pendulum,}l=25.0\text{cm}$

$\text{Time of 40 oscillation,}T=50\text{s}$

Time period of pendulum

$T=2\pi \sqrt{\frac{l}{g}}$

$\Rightarrow T^2=\frac{4{\pi }^{2}l}{g}\Rightarrow g=\frac{4{\pi }^{2}l}{T^2}$

$\Rightarrow \text{Fractional error in g}=\frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2\Delta T}{T}$
Here $\Delta l=0.1\text{cm},\Delta T=1\text{s}$

$\Rightarrow \frac{\Delta g}{g}=\left(\frac{0.1}{25.0}\right)+2\left(\frac{1}{50}\right)=0.044$

$\therefore \text{Percentage error in}g=\frac{\Delta g}{g}\times 100=4.4\%$