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

-0.005

We have,

$l+\Delta l=0.992 l$ and $g+\Delta g=1.002 g$

$\Rightarrow \Delta l=-0.008 l$ and $\Delta g=0.002 g$

$\Rightarrow \frac{\Delta l}{l}=-0.008$ and $\frac{\Delta g}{g}=0.002$

Now, $T=2 \pi \sqrt{\frac{l}{g}}$

$\Rightarrow \log T=\log 2 \pi+\frac{1}{2}(\log l-\log g)$

$\Rightarrow d(\log T)=d(\log 2 \pi)+\frac{1}{2} d(\log l-\log g)$

$\Rightarrow \frac{d T}{T}=\frac{1}{2}\left(\frac{d l}{l}-\frac{d g}{g}\right)$

$\Rightarrow \frac{\Delta T}{T}=\frac{1}{2}\left(\frac{\Delta l}{l}-\frac{\Delta g}{g}\right)$             $[∵ d T \cong \Delta t, d l \cong \Delta l$ and $d g \cong \Delta g]$

$\Rightarrow \frac{\Delta T}{T}=\frac{1}{2}(-0.008-0.002)=-0.005$