Question

The length of a simple pendulum is about $100cm$ known to have an accuracy of $1mm$. Its period of oscillation is $2s$ determined by measuring the time for $100$ oscillations using a clock of $0.1s$ resolution. What is the accuracy in the determined value of $g$ ?

• A. 0.2%
• B. 0.5%
• C. 0.1%
• D. 2%

Answer 1

Answer: (A)

Time period of a simple pendulum
$T=2\pi \sqrt{\frac{L}{g}}$ or $g=2{\pi }^2\frac{L}{T^2}$ ...(i)
Taking log on both sides, we have
$\log g=4{\pi }^2[\log L-2\log T]$
Differentiating, we get $\frac{\Delta g}{g}=\frac{\Delta L}{L}-\frac{2\Delta T}{T}$
${\left|\frac{\Delta g}{g}\right|}_{\max }=\frac{\Delta L}{L}+\frac{2\Delta T}{T}$ .. (ii)
Given, $L=100cm,T=2s,\Delta T=\frac{0.1}{100}=0.001s$
$\Delta L=1mm=0.1cm$
Substituting the given values in Eq. (ii), we get
$\frac{\Delta g}{g}=\frac{0.1}{100}+2\times \frac{0.001}{2}$
Thus, maximum percentage error
${\left|\frac{\Delta g}{g}\right|}_{\max }\times 100=\left(\frac{0.1}{100}\times 100\right)+\left(2\times \frac{0.001}{2}\times 100\right)$
$=0.1$%$+0.1\%$
$=0.2$%