Question

A pendulum has length l = 100 ± 0.1 cm and period T = 2 ± 0.01 s. Using g = \[\frac {4π²l}{T²}\], the max percent error in g is:

Options

• 1.1%

• 11%

• 0.11%

• 2%

Answer 1

1.1%

Explanation:

Percent error formula for multiplication/division:

\[\frac{\Delta g}{g}\times100=\left(\frac{\Delta l}{l}+2\cdot\frac{\Delta T}{T}\right)\times100\]
\[\therefore\] l = 0.001 and T = 0.005

\[\frac {Δg}{g}\] × 100 = (0.001 + 2 · 0.005) × 100 = (0.001 + 0.01) × 100 = 0.011 × 100 = 1.1%