Question

To find the value of 'g' using simple pendulum T=$2.00\pm 0.01$ sec; $50.0\pm 0.1$ cm was measured. The maximum permissible error in 'g' is:

• A. $1.4\%$
• B. $1.1\%$
• C. $1.5\%$
• D. $1.2\%$

Answer 1

The correct option is D $1.2\%$

$\begin{array}{c}T=2\pi \sqrt{\frac{l}{g}}⟶\left(ii\right)\\ \text{taking log;}\\ \text{log}T=\log 2\pi +\frac{1}{2}\log l-\frac{1}{2}\log g\end{array}$

$\begin{array}{c}\text{differentiating;}\\ \frac{dT}{T}=\frac{1}{2}\frac{dl}{l}-\frac{1}{2}\frac{dg}{g}\\ \text{error is always added:}\\ \frac{dT}{T}=\frac{1}{2}\frac{dl}{l}+\frac{1}{2}\frac{dg}{g}⟶\left(ii\right)\end{array}$

$\begin{array}{c}dT=0.01s\\ T=2s\\ dl=0.1cm\\ l=50cm\end{array}$

$\begin{array}{c}\text{pulting in (ii)}\\ \frac{dg}{g}=\frac{2\times 0.01}{2}+\frac{0.1}{50}={0.012}^{}=1.2\%\end{array}$