Question

The period of oscillation of a simple pendulum is given by $T=2\pi \sqrt{\frac{1}{g},}$ where $l$ is about $100cm$ and is known to have $1mm$ accuracy.The period is about $2s$. The time of $100$ oscillations is measured by a stopwatch of least count $0.1s$ The percentage error in g is?

• A. $0.1$%
• B. $1$%
• C. $0.2$%
• D. $0.8$%

Answer 1

Answer: (C)

$0.2$%

$\begin{array}{cc}\text{Given-}& \ast \text{Time period}=2\pm 0.18\\ & \ast \text{length}=1000\pm 1mm\end{array}$

$\begin{array}{c}\text{we have-}T=2\pi \cdot \sqrt{\frac{l}{g}}\\ \Rightarrow g=4{\pi }^2\cdot \frac{\ell }{T^2}-\left(1\right)\\ \text{for}\text{1 oscillation Time period}=2s\\ \Rightarrow \text{for}100\text{oscillations. Time}=200s\\ \Rightarrow T=200\pm 0.18\end{array}$

$\begin{array}{c}\text{Now By error analysis}\\ \text{- Taking logarithm both sides in eq}1\\ \log \left(g\right)=\log \left(e\right)-2\cdot \log \left(T\right)+\log \left(4{\pi }^2\right)\end{array}$

$\begin{array}{c}\text{Difterentiating both Sides-}\\ \frac{dg}{g}=\frac{d\ell }{l}-2\cdot \frac{dT}{T}\end{array}$

$\begin{array}{c}\text{For maximum permissible value of error}\\ \frac{dg}{g}=\frac{d\ell }{l}+2\cdot \frac{dT}{T}=\frac{1}{1000}+2\times \frac{0.1}{200}=0.002\\ \frac{dg}{g}=0.002\times 100\%=0.2\%\end{array}$