#### 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%

#### Solution

(a) 1%

Let l be the length if the pendulum and T be the period.

\[\text { Also, let ∆ l be the error in the length and ∆ T be the error in the period } . \]

\[\text { We have }\]

\[\frac{∆ l}{l} \times 100 = 2\]

\[ \Rightarrow \frac{dl}{l} \times 100 = 2\]

\[\text { Now,} T = 2\pi\sqrt{\frac{l}{g}}\]

\[\text { Taking \log on both sides, we get }\]

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

\[\text { Differentiating both sides w . r . t . x, we get }\]

\[\frac{1}{T}\frac{dT}{dl} = \frac{1}{2l}\]

\[ \Rightarrow \frac{dT}{dl} = \frac{T}{2l}\]

\[ \Rightarrow \frac{dl}{l} \times 100 = 2\frac{dT}{T} \times 100\]

\[ \Rightarrow \frac{dT}{T} \times 100 = \frac{2}{2}\]

\[ \Rightarrow \frac{∆ T}{T} \times 100 = 1\]

\[\text { Hence, there is an error of 1 % in calculating the period of the pendulum } .\]