Question

If there is an error of 2% in measuring the length of a simple pendulum, then percentage error in its period is

• A. $1\%$
• B. $2\%$
• C. $3\%$
• D. $4\%$

Answer 1

The correct option is A $1\%$

Time period $T=2\pi \sqrt{\frac{l}{g}}$
$\frac{dT}{dl}=\frac{\pi }{\sqrt{gl}}$
Percentage error in measuring l = 2%
$\Rightarrow \frac{\Delta l}{l}=\frac{2}{100}$
$\Rightarrow \Delta l=\frac{2l}{100}$
Approximate error in $T=dT=\frac{dT}{dl}\Delta l$
$=\frac{T}{100}=1\%ofT$
Percentage error in $T=1\%$