Question

The length of a simple pendulum is about $100cm$, known to have an accuracy of $1mm$. Its period of oscillation is $2s$, determined by measuring the time for $100$ oscillations using a clock of $0.1s$ resolution. What is the accuracy in the determined value of $g$?

• A. $0.2\%$
• B. $0.5\%$
• C. $0.1\%$
• D. $2\%$

Answer 1

The correct option is A $0.2\%$

$T=\frac{t}{n}$ and $\Delta T=\frac{\Delta t}{n}$

$\therefore \frac{\Delta T}{t}\times 100=\frac{\Delta t}{t}\times 100$

$=\frac{0.1}{2\times 100}\times 100=0.05$%

$\frac{\Delta l}{l}\times 100=\frac{0.1cm}{100cm}\times 100=0.1$%

Now, $T=2\pi \sqrt{\frac{l}{g}}$
$\Rightarrow g=\frac{4{\pi }^{2}l}{T^2}$
$\therefore$ % error in $g=($% error in $l)+2($% error in $T)$
$=0.1$% $+2(0.05$%$)=0.2$%