Q 11-01-033JEE MainMedium
In a simple pendulum experiment, the length is measured as $1.000\ \text{m}$ with an error of $1\ \text{mm}$, and the time period as $2.00\ \text{s}$ with an error of $0.02\ \text{s}$. Find the maximum percentage error in the value of $g$.
Numerical value type. Enter your answer.
Answer: 2.1
$g = \dfrac{4\pi^2 L}{T^2}$, so
$$\frac{\Delta g}{g}\times 100 = \frac{\Delta L}{L}\times 100 + 2\,\frac{\Delta T}{T}\times 100$$
$$= \frac{0.001}{1.000}\times 100 + 2\times\frac{0.02}{2.00}\times 100 = 0.1\% + 2\% = 2.1\%$$
Solution by Sreeraj P, M.Sc Physics