The volumes of two bodies are measured to be $V_1=(10.2\pm 0.02)c{m}^{3}and{V}_2=(6.4\pm 0.01)c{m}^3.$ The sum and difference in volumes with error limits is

Given, $V_1=(10.2\pm 0.02)c{m}^3$

and $V_2=(6.4\pm 0.01)c{m}^3$

$\Delta V=\pm (\Delta V_1+\Delta V_2)$

$=\pm (0.02+0.01)c{m}^3=\pm 0.03c{m}^3$

$V_1+V_2=(10.2+6.4)c{m}^3=16.6c{m}^3$

and $V_1-V_2=(10.2-6.4)c{m}^3=3.8c{m}^3$

Hence, sum of volume $=(16.6\pm 0.03)c{m}^3$

and    difference of volume $=(3.8\pm 0.03)c{m}^3$

In product If Z = AB, then maximum fractional error is $\frac{\Delta Z}{Z}=\pm (\frac{\Delta A}{A}+\frac{\Delta B}{B})$

Therefore, maximum fractional error in product of two (or more) quantities is equal to sum of fractional errors in the individual quantities. 

In division If Z = A /B, then maximum fractional error is $\frac{\Delta Z}{Z}=\pm (\frac{\Delta A}{A}+\frac{\Delta B}{B})$

Therefore, maximum fractional error in product of two ( or more) quantities is equal to sum of fractional errors in the individual quantities.