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

Solution:

(a) 1%
Let l be the length if the pendulum and T be the period.

Also, let $\Delta l$ be the error in the length and $\Delta T$ be the error in the period.

We have

$\frac{\Delta l}{l} \times 100=2$

$\Rightarrow \frac{d l}{l} \times 100=2$

Now, $T=2 \pi \sqrt{\frac{1}{g}}$

Taking log on both sides, we get

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

Differentiating both sides w.r.t. $x$, we get

$\frac{1}{T} \frac{d T}{d l}=\frac{1}{2 l}$

$\Rightarrow \frac{d T}{d l}=\frac{T}{2 l}$

$\Rightarrow \frac{d l}{l} \times 100=2 \frac{d T}{T} \times 100$

$\Rightarrow \frac{d T}{T} \times 100=\frac{2}{2}$

$\Rightarrow \frac{\Delta T}{T} \times 100=1$

Hence, there is an error of $1 \%$ in calculating the period of the pendulum.