Question

The measure of mass and volume of a body are 2 kg and 5 cm ${}^3$ respectively, with possible errors of 0.01 g and 0.1 $c{m}^3$. What would be the maximum permissible error in its density?
Hint: $\frac{\Delta d}{d}=\frac{\Delta m}{m}+\frac{\Delta V}{V}$=$\Delta d=dx(\frac{\Delta m}{m}+\frac{\Delta V}{V}$) where d=$\frac{m}{V}$

Answer 1

We have,

$mass=2kg,volume=5c{m}^3$

possible errors is $0.01g,0.1c{m}^3$

Density $=\frac{mass}{volume}$

$d=\frac{2}{5}g/c{m}^3$

$=0.4g/c{m}^3$

for error , $\frac{\Delta d}{d}=\frac{\Delta m}{m}+\frac{\Delta v}{v}$

$\Delta d=(\frac{0.01}{2}+\frac{0.1}{5})\times 0.4$

$=0.01$

$relativeerror=\frac{\Delta d}{d}$

$=\frac{0.01}{0.4}$

$=\frac{0.1}{4}$

$\%$ error $=100\times$ relative error

$=\frac{0.1}{4}\times 100$

$=2.5\%$