Question

The solid ball density is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is $0.5mm$ and there are $50$ divisions on the circular scale. The reading on the main scale is $2.5$ mm and that on the circular scale is $20$ divisions. If the measured mass of the ball has a relative error of $2\%$, then the relative percentage error in the density is

• A. $0.9\%$
• B. $2.4\%$
• C. $3.1\%$
• D. $4.2\%$

Answer 1

The correct option is C

$3.1\%$

Explanation for correct option

Step 1: Solve for the relative error in diameter of the ball

As we know that least count of a scale$=\frac{pitch}{no.ofdivisions}$

LC of the scale$=\frac{0.5}{50}\Rightarrow 0.01mm$

Diameter of the ball will be$=mainscalereading+(curvedsalereading\times leastcount)$

$$\begin{gathered} =2.5+(20\times 0.01) \\ =2.5+0.2 \\ =2.7 \end{gathered}$$

Step 2: Solve for relative error in density

As density$=\frac{Mass}{Volume}$

volume of a ball$=\frac{4}{3}\mathrm{\pi }{\left(\frac{\mathrm{D}}{2}\right)}^3$, where $D=$diameter of the ball

density$=\frac{Mass}{{\displaystyle \frac{4}{3}}\mathrm{\pi }{\left({\displaystyle \frac{\mathrm{D}}{2}}\right)}^3}$

relative error in density is

$\frac{∆\rho }{\rho }=\frac{∆M}{M}+\frac{3∆D}{D}$, where $\rho =density,M=mass$

Step 3 Solve for relative percentage error in density

relative percentage error in density is

$\frac{∆\rho }{\rho }\times 100=\frac{∆M}{M}\times 100+\frac{3∆D}{D}\times 100$

$$\begin{gathered} \frac{∆M}{M}\times 100=2 \\ ∆D=0.01 \\ D=2.7 \end{gathered}$$ (given)

$$\begin{gathered} \Rightarrow \frac{∆\rho }{\rho }\times 100=2+\frac{3\times 0.01}{2.7}\times 100 \\ \Rightarrow \frac{∆\rho }{\rho }\times 100=2+\frac{10}{9} \\ \Rightarrow \frac{∆\rho }{\rho }\times 100=\frac{28}{9} \\ \Rightarrow \frac{∆\rho }{\rho }\times 100=3.1\% \end{gathered}$$

Hence, option C is correct.