Question

The period of oscillation of a simple pendulum is given by $T=2\pi \sqrt{\frac{l}{g}}$ where $l$ is about $100cm$ 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 $0.1s.$ The percentage error in g is

• A. $0.1\%$
• B. $1\%$
• C. $0.2\%$
• D. $0.8\%$

Answer 1

Answer: (C)

$T=2\pi \sqrt{l/g}$
$\Rightarrow T^2=4{\pi }^{2}l/g$
$\Rightarrow g=\frac{4{\pi }^{2}l}{T^2}$
Here $\%$error in $l=\frac{1mm}{100cm}\times 100$
$=\frac{0.1}{100}\times 100=0.1\%$
and $\%$ error in $T=\frac{0.1}{2\times 100}\times 100$
$=0.05\%$
$\therefore \%$ in $g=\%$ error in $l+2(\%$ error in $T)$