Question

If the length of a simple pendulum is recorded as $(90.0\pm 0.02)\text{cm}$ and period as $(1.9\pm 0.02)\text{s}$, the percentage of error in the measurement of acceleration due to gravity is

• A. $4.2\%$
• B. $2.1\%$
• C. $1.5\%$
• D. $2.8\%$

Answer 1

The correct option is B $2.1\%$

Given that
$\Delta l=0.02\text{cm},l=90\text{cm}$
$\Delta T=0.02\text{s},T=1.9\text{s}$
Fractional error in the length,
$\frac{\Delta l}{l}=\frac{0.02}{90}$
Fractional error in the time period,
$\frac{\Delta T}{T}=\frac{0.02}{1.9}$
From the formula,
$T=2\pi \sqrt{\frac{l}{g}}$
$\Rightarrow g=4{\pi }^2\frac{l}{T^2}$
Taking maximum value of fractional error in $g$
$\Rightarrow \frac{\Delta g}{g}=\frac{\Delta l}{l}+2\frac{\Delta T}{T}$
$\%$ error in $g$ measurement
$\frac{\Delta g}{g}\times 100=(\frac{\Delta l}{l}+2\frac{\Delta T}{T})\times 100$
$=(\frac{0.02}{90}+\frac{2\times 0.02}{1.9})\times 100$
$=2.12\%$
Thus, option $\left(b\right)$ is the correct answer.