Question

The time period of a pendulum is given by $T=2\pi \sqrt{\frac{L}{g}}$. The length of pendulum is $20$cm and is measured up to $1$mm accuracy. The time period is about $0.6$s. The times of $100$ oscillations is measured with a watch of $1/10s$ resolution. The accuracy in the determination of g is given by the expression $(x-0.2)$. Find the value of $x$ ?

Answer 1

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

$\frac{\Delta g}{g}=\frac{\Delta l}{l}+2\frac{\Delta T}{T}=\frac{0.1}{20}+2\left(\frac{1}{600}\right)=\frac{1}{120}$

$\frac{\Delta g}{g}\times 100=0.833$

The expression is given by $(x-0.2)$

$=1-0.2=0.8$

So the value of $x$ is $1$

$=0.8\%$