Question

In an experiment, on the measurement of g using a simple pendulum, the time period was measured with an accuracy of 0.2% while the length was meausred with an accuracy of 0.5%. The percentage accuracy in the value of g thus obtained is

• A. 0.0025
• B. 0.001
• C. 0.007
• D. 0.009

Answer 1

Answer: (D)

0.009

Answer 2

Given $\frac{\Delta T}{T} = 0.2\% , \frac{\Delta L}{L} = 0.5\%$
Since $T = 2 \pi \sqrt{\frac{L}{g}}$
$\Rightarrow \:\:\:\:\: g = \frac{4 \pi^2 L}{T^2}$
$\therefore \:\:\:\: \frac{\Delta g}{g} = \frac{\Delta L}{L} + \frac{2 \Delta T}{T}$
$= 0.5\% + 2(0.2 \%)$
$= 0.9\%$