Question

In an experimental set up, the density of a small sphere is to be determined. The diameter of the small sphere is measured with the help of a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the sphere has a relative error of 2%, the relative percentage error in the density is

• A. $0.03$%
• B. $3.11$%
• C. $0.08$%
• D. $8.2$%

Answer 1

The correct option is B $3.11$%

$\begin{array}{c}\text{Pitch of the Screw gauge}=0.5mm\\ \text{So circular Scale Divisions (CSD)}\\ \text{Reading}=2.5mm\text{and}20CSD\end{array}$

$\begin{array}{cc}\text{Diameter}& =2.5mm+\frac{\text{pitch}}{50}\times CSD\\ & =2.5mm+\frac{0.5}{50}\times 20mm=2.7mm\end{array}$

$\text{Least count}=\frac{\text{Pitch}}{\text{Total CSD}}=\frac{0.5mm}{50}=0.01mm$

$\text{Diameter}\Rightarrow D=2.7mm\pm 0.01mm$

$\text{we know}\rho =\frac{M}{V}=\frac{6M}{\pi D^3}$

$\begin{array}{c}\text{* By error analysis -}\\ \text{we get -}\\ \begin{array}{c}\frac{\Delta \rho }{\rho }=\frac{\Delta M}{M}+3\cdot \frac{\Delta D}{D}\\ \frac{\Delta \rho }{\rho }=2\%+3\times \frac{0.01}{2.7}\times 100\%=3.11\%\end{array}\end{array}$