Experimental Physics Question 6
Question 6 - 2024 (31 Jan Shift 2)
The measured value of the length of a simple pendulum is $20 cm$ with $2 mm$ accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is $N \%$. The value of $N$ is:
(1) 4
(2) 8
(3) 6
(4) 5
Show Answer
Answer: (3)
Solution:
$$ \begin{aligned} T & =2 \pi \sqrt{\frac{\ell}{g}} \\ g & =\frac{4 \pi^{2} \ell}{T^{2}} \\ \frac{\Delta g}{g} & =\frac{\Delta \ell}{\ell}+\frac{2 \Delta T}{T} \\ & =\frac{0.2}{20}+2\left(\frac{1}{40}\right) \\ = & \frac{0.3}{20} \end{aligned} $$
Percentage change $=\frac{0.3}{20} \times 100=6 \%$
