The time period of a simple pendulum is given by T = 2√(l/g). The measured value of the length of the pendulum is 10 cm known to a 1 mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 seconds using a clock of 1 s resolution. The percentage accuracy in the determination of ‘g’ using this pendulum is ‘x’. The value of ‘x’ to the nearest integer is.
a. 5%
b. 4%
c. 3%
d. 2%
Answer:(c)
\(\begin{array}{l}T = 2\pi \sqrt{\frac{l}{g}}\Rightarrow T^{2}= 2\pi (l/g)\end{array} \)
\(\begin{array}{l}g = 2\pi \frac{l}{T^{2}}\end{array} \)
\(\begin{array}{l}\frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2\Delta T}{T}\end{array} \)
\(\begin{array}{l}\frac{\Delta g}{g}=\frac{1\times 10^{-3}}{10\times 10^{-2}}+\frac{2\times 1}{100}\end{array} \)
\(\begin{array}{l}\frac{\Delta g}{g}= 0.01 + 0.02 = 0.03\end{array} \)
\(\begin{array}{l}100\times \frac{\Delta g}{g}= 0.03\times 100 = 3%\end{array} \)
