Q. The period of oscillation of a simple pendulum is given by $T = 2\pi\sqrt{\frac{l}{g}}$ where $l$ is about $100\ cm$ and is known to have $1mm$ accuracy. The period is about $2s$. The time of $100$ oscillations is measured by a stop watch of least count $0.1\ s$ . The percentage error in $g\$is
A) $0.1\%$
B) $1\%$
C) $0.2\%$
D) $0.8\%$
Explanation
$T = 2\pi\sqrt{l/g}\ \Rightarrow T^{2} = 4\pi^{2}l/g \Rightarrow g = \frac{4\pi^{2}l}{T^{2}}$ Here, $\%$ error in $l = \frac{1mm}{100cm} \times 100 = \frac{0.1}{100} \times 100 = 0.1\%$ And $\%$ in error in $T = \frac{0.1}{2 \times 100} \times 100 = 0.05\%$ $\therefore\%$ error in $g = \%$ error in $l + 2(\%\ error\ in\ T)$ $= 0.1 + 2 \times 0.05 = 0.2\ \%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.