Question

If the length of a simple pendulum is recorded as $(90.00\pm 0.02)$ cm and period as $(1.90\pm 0.02)$ s, the percentage of error in the measurement of acceleration due to gravity is :

• A. $4.2$
• B. $2.1$
• C. $1.5$
• D. $2.8$

Answer 1

The correct option is B $2.1$

Time period of a simple pendulum is given as: $T=2\pi \sqrt{\frac{l}{g}}$

So, $g=4{\pi }^2\frac{l}{T^2}$

$\therefore$ % error in the measurement of $g$ is given by:

$\frac{\Delta g}{g}\times 100=(\frac{\Delta l}{l}+\frac{2\Delta T}{T})\times 100$

$=(\frac{0.02}{90}+\frac{2\times 0.02}{1.90})\times 100$

$=2.1$ %