Question

The length of a simple pendulum is about $100 \,cm$ known to have an accuracy of $1 \,mm$. Its period of oscillation is $2 \,s$ determined by measuring the time for $100$ oscillations using a clock of $0.1\, s$ 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=100\text{ }cm,\text{ }T=2s,\text{ }\Delta T=\frac{0.1}{100}=0.001s $
$ \Delta L=1\text{ }mm=0.1\text{ }cm $
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) $
$=\text{ }0.1$%$\text{ }+\text{ }0.1\% $
$=\text{ }0.2$%