Question

In an experiment for determining the value of acceleration due to gravity (g) using a simple pendulum, the following observations were recorded:
Length of the string (l) = $98.0$cm
Diameter of the bob (d) = $2.56$cm
Time for $10$ oscillations (T) = $20.0s$
Calculate the value of g with maximum permissible absolute error and the percentage relative error.

• A. $(9.80\pm 0.11)m{s}^{-2}$
• B. $(9.80\pm 0.12)m{s}^{-2}$
• C. $(10.80\pm 0.11)m{s}^{-2}$
• D. $(9.9\pm 0.11)m{s}^{-2}$

Answer 1

Answer: (A)

$(9.80\pm 0.11)m{s}^{-2}$

$\begin{array}{c}L=98.0cm\\ dL=0.01cm\\ t=20.0s\\ dt=0.01s\\ n=10\end{array}$

$\begin{array}{c}T=2\pi \sqrt{\frac{l}{g}}\\ g=\frac{4{\pi }^{2}L}{T^2}⟶\left(i\right)\\ \text{Taking Log}\end{array}$

$\begin{array}{c}\log g=\log 4+2\log \pi +\log l-2\log T\\ \\ \text{differentiating}\\ \frac{\delta g}{g}=\frac{\delta l}{l}+2\frac{\delta T}{T}⟶\left(ii\right)\text{(error is always added)}\\ T=\frac{t}{n}\Rightarrow T=2\text{sec.}\end{array}$

$\begin{array}{c}\text{Differentiating}\to \frac{\delta T}{T}=\frac{\delta t}{n}=\frac{0.01}{10}⟶\text{(iv)}\\ \text{from(i)}\\ g=9.80m/s^2\\ \text{From (ii)}\\ \frac{dg}{g}=\frac{0.11}{9.8}\Rightarrow dg=0.11m/s^2\end{array}$