Question

The period of oscillation of a simple pendulum is $T=2\pi \sqrt{L/g}$. The measured value of $L$ is $20.0cm$ known to $1mm$ accuracy and time for $100$ oscillations of the pendulum is found to be $90s$ using a wristwatch of $1s$ resolution. If the accuracy in the determination of $g$ is $n\%$, then find $20n$.

Answer 1

Given :
$T=2\pi \sqrt{L/g}\dots \left(i\right)$

Obtaining $g$ from $\left(i\right)$ equation we get,

$g=\frac{4{\pi }^{2}L}{T^2}$

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

Therefore, $\frac{\Delta T}{T}=\frac{\Delta t}{t}$,

The errors in both $L$ and $t$ are the least count errors, Therefore,
$\left(\frac{\Delta g}{g}\right)=\left(\frac{\Delta L}{L}\right)+2\left(\frac{\Delta T}{T}\right)$

$\left(\frac{\Delta g}{g}\right)=\frac{0.1}{20.0}+2\left(\frac{1}{90}\right)=0.027$

Thus, the percentage error in $g=2.7\%$

Hence, $n\%=2.7\%$

Therefore,
$20n=20\times 2.7=54$

Final answer: $54$