Question

The density of a solid ball 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 $2.5mm$and that on the circular scale is$20$ divisions. If the measured mass of the ball has a relative error of$2\%$, 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

$3.1\%$

Step 1: Given data :

The diameter of the ball is measured with a screw gauge, whose pitch is $=0.5mm$

The circular scale divisions$=50$

The reading on the main scale $=2.5mm$

On the circular scale divisions$=20$

The ball has a relative error in mass$=2\%$

Step 2: Formula used:

Diameter of the ball, $D=MSR+\left(CSR\times LC\right)$

Density$\rho =\frac{m}{v}$

Step 3: Finding the diameter of the ball :

The least count of screw gauge = Pitch / no. of divisions on a circular scale

Then, solve and substitute the values,

$\begin{array}{l}=\frac{0.5mm}{50}\\ =0.01mm\end{array}$

Diameter of the ball, $D=MSR+\left(CSR\times LC\right)$

$\begin{array}{l}=2.5mm+20\times 0.01mm\\ =2.7mm\end{array}$

Step 4: Finding relative error in density

Density$\rho =\frac{m}{v}$

Now substitute the values,

$\rho =\frac{M}{\frac{4\pi }{3}{\left(\frac{D}{2}\right)}^3}$

Therefore relative error in the density

$\frac{\Delta \rho }{\rho }=\frac{\Delta M}{m}+\frac{3\Delta D}{D}$

Step 5: Substitute the value in the relative error formula

Therefore relative percentage error in the density

$$\begin{gathered} \frac{\Delta \rho }{\rho }\times 100=\left(\frac{\Delta M}{m}+\frac{3\Delta D}{D}\right)\times 100 \\ \left(\frac{\Delta M}{m}\times 100+\frac{3\Delta D}{D}\times 100\right)=2+3\times \frac{0.01}{2.7}\times 100 \\ \left(\frac{\Delta M}{m}\times 100+\frac{3\Delta D}{D}\times 100\right)=3.1\% \end{gathered}$$

Therefore relative percentage error in the density$3.1\%$.

Hence, the correct option is (C).