Question

The time period of a simple pendulum is given by the relation $T=2\pi \sqrt{\frac{L}{g}}$. The measured values of length $L$ and period of $50$ oscillations are $(20\pm 0.1)$ $cm$ and $(50\pm 1)$ $s$ respectively. The percentage error in the determination of $g$ is

• A. $2.5\%$
• B. $4.5\%$
• C. $6\%$
• D. $7.5\%$

Answer 1

The correct option is B $4.5\%$

$T=2\pi \sqrt{\frac{L}{g}}$

$\Rightarrow T^2=4{\pi }^2\frac{L}{g}$

$\Rightarrow g=\frac{4{\pi }^{2}L}{T^2}$

$\frac{\Delta g}{g}=\frac{\Delta L}{L}+\frac{2\Delta T}{T}$

$\Rightarrow \frac{\Delta g}{g}=\frac{0.1}{20}+\frac{2}{50}$

$\Rightarrow \frac{\Delta g}{g}=4.5\%$