Question

In an experiment, the following observations were recorded:
$L=2.820m,M=3.00kg,l=0.087cm$, diameter $D=0.041cm$. Taking $g=9.81m{s}^{-2}$ and using the formula, Y =$\frac{4Mg}{\pi D^{2}l}$, find the maximum permissible error in Y.

Answer 1

$y=\frac{4Mgh}{\Pi D^{2}l}$

Maximum possible error in $y$ is

$\frac{\Delta y}{y}\times 100=(\frac{\Delta M}{M}+\frac{\Delta g}{g}+\frac{2\Delta D}{D}+\frac{\Delta L}{L}+\frac{\Delta l}{l})\times 100$

$=(\frac{1}{3.00}+\frac{1}{9.81}+\frac{1}{41}\times 2+\frac{1}{2820}+\frac{1}{87})\times 100$

$=0.065\times 100$

$=6.5$%