Question

The time period of oscillation of a simple pendulum is given by $T=2\pi \sqrt{\frac{l}{g}}$ . The length of the pendulum is measured as $l=10\pm0.01 \, cm \, $ and the time period as $T=0.5\pm0.02 \, s$ . The percentage error in the value of $g$ is

• A. $5\%$
• B. $8\%$
• C. $7\%$
• D. None of these

Answer 1

Answer: (B)

$\frac{\Delta g}{g}\times 100=\frac{\Delta l}{l}\times 100+2\left(\frac{\Delta T}{T} \times 100\right)$
$\Delta l=0.01 \, cm$
$\Delta T=0.02s$
$l=10 \, cm$
$T=0.5 \, s$
$\frac{\Delta l}{l} \times 100 = \frac{0.01}{10} \times 100 = 0.1$
$\frac{\Delta T}{T} \times 100 = \frac{0.02}{0.5} \times 100 = 4$
$\frac{Δg}{g}\times 100=0.1+2\times 4\approx8\%$