Question

The period of oscillation $T$ of a pendulum of length $l$ at a place of acceleration due to gravity $g$ is given by $T=2\pi \sqrt{\frac{l}{g}}$. If the calculated length is $0.992$ times the actual length and if the value assumed for $g$ is $1.002$ times its actual value, the relative error in the computed value of $T$ is

• A. $0.005$
• B. $-0.005$
• C. $0.003$
• D. $-0.003$

Answer 1

The correct option is B $-0.005$

Relative error will calculate by
$\frac{\Delta T}{T}=\frac{1}{2}[\frac{\Delta L}{L}-\frac{\Delta g}{g}]$
$\frac{\Delta T}{T}=\frac{1}{2}[\frac{0.992L}{L}-\frac{1.002g}{g}]$
$\Delta T=-0.005T$