Question

In an experiment of simple pendulum, the errors in the measurement of length of the pendulum (L) and time period (T) are 3% and 2% respectively. The maximum percentage error in the value of $\frac{L}{T^2}$ is _____ %

Answer 1

Answer: 7

Time period of simple pendulum is
$T=2\pi \sqrt{\frac{L}{g}}$
Squaring both sides, we get
$\therefore 8=4{\pi }^2\frac{L}{g}$
or $8=4{\pi }^2\frac{L}{T^2}$
The maximum percentage error in g is
$\frac{\Delta g}{g}\times 100=\frac{\Delta L}{L}\times 100+2\left(\frac{\Delta T}{T}\right)\times 100=3\%+2\times 2\%=7\%$
From $\left(i\right)$, we get $\frac{L}{T^2}=\frac{g}{4{\pi }^2}$
The maximum percentage error in $\frac{L}{T^2}$ is
$\frac{\Delta \left(\frac{L}{T^2}\right)}{\frac{L}{T^2}}\times 100$ $=\frac{\Delta g}{g}\times 100=7\%$