Question

The mass and density of a solid sphere are measured to be $(12.4\pm 0.1)\text{kg}$ and $(4.60\pm 0.2){\text{kg/m}}^3$. The volume of the sphere expressed with error limits is:

• A. $(15\pm 0.1){\text{m}}^3$
• B. $(2.7\pm 0.3){\text{m}}^3$
• C. $(14.6\pm 0.2){\text{m}}^3$
• D. $(2.69\pm 0.1){\text{m}}^3$

Answer 1

The correct option is D $(2.69\pm 0.1){\text{m}}^3$

Given that
$m\pm \Delta m=(12.4\pm 0.1)\text{kg}$
So, the fractional error in mass will be.
$\frac{\Delta m}{m}=\frac{0.1}{12.4}=0.008$
Similarly for density $\rho \pm \Delta \rho =(4.60\pm 0.2){\text{kg/m}}^3$
So, fraction error in density will be
$\frac{\Delta \rho }{\rho }=\frac{0.2}{4.60}=0.04$
Volume of sphere $V=\frac{m}{\rho }=\frac{12.4}{4.60}=2.69{\text{m}}^3$
From this, Fractional error in volume can be written as,
$\frac{\Delta V}{V}=\pm (\frac{\Delta m}{m}+\frac{\Delta \rho }{\rho })$
$\Rightarrow \Delta V=\pm V(\frac{\Delta m}{m}+\frac{\Delta \rho }{\rho })$
$\Rightarrow \Delta V=\pm 2.69(0.008+0.04)=\pm 2.7\times 0.05$
$\Rightarrow \Delta V=\pm 0.1$
Volume of sphere with error limits is,
$V\pm \Delta V=(2.69\pm 0.1){\text{m}}^3$
Hence, option (d) is the correct answer.