Question

The mass and density of a solid are measured to be $(12.4\pm 0.1)$kg and $(4.6\pm 0.2)kg/m^3.$ Calculate the volume of the sphere with error limits.

• A. $\therefore V\pm \Delta V=(2.7\pm 0.14)m^3$
• B. $\therefore V\pm \Delta V=(5.4\pm 0.14)m^3$
• C. $\therefore V\pm \Delta V=(2.7\pm 0.28)m^3$
• D. None of these

Answer 1

The correct option is A $\therefore V\pm \Delta V=(2.7\pm 0.14)m^3$

Here, $m\pm \Delta m=(12.4\pm 0.1)$ kg
and, $p\pm \Delta p=(4.6\pm 0.2)kg/m^3$

Volume, $v=\frac{m}{p}=\frac{12.4}{4.6}=2.69m^3=2.7m^3$ (rounding off to one decimal place )

Now, $\frac{\Delta V}{V}=\pm (\frac{\Delta m}{m}+\frac{\Delta p}{p})$

$\Delta V=\pm (\frac{\Delta m}{m}+\frac{\Delta p}{p})\times V$
$=\pm (\frac{0.1}{12.4}+\frac{0.2}{p})\times 2.7=\pm 0.14$

$\therefore V\pm \Delta V=(2.7\pm 0.14)m^3$