Question

A wire has a mass $$(0.3\pm 0.003)$$ g, radius$$(0.5\pm 0.005)$$mm and length$$(0.6\pm 0.006)$$cm. The maximum percentage error in the measurement of its density is:
     A. 1
     B. 2
     C. 3
     D. 4

Answer 1

Hint: We will use the formula of density given as the ratio of mass to the volume of the wire. The density of a wire is given by the formula, $\rho ={\displaystyle \frac{m}{\pi r^{2}l}}$ . Then we will find the error propagation through the parameters present in the formula for finding density. The propagation of error in measurement of a certain physical quantity is given as the sum of relative errors of the parameters on which that quantity depends. For example, the relative error in mass is given by ${\displaystyle \frac{\mathrm{\Delta }m}{m}}$ , where $\mathrm{\Delta }m$ is the magnitude of error.in certain cases, it will be equal to the least count of the instrument used for measurement.
Formula Used:
$\begin{array}{rl}& \rho ={\displaystyle \frac{m}{\pi r^{2}l}}\\ & {\displaystyle \frac{\mathrm{\Delta }\rho }{\rho }}\text{\%}={\displaystyle \frac{\mathrm{\Delta }m}{m}}\times 100+2({\displaystyle \frac{\mathrm{\Delta }r}{r}})\times 100+{\displaystyle \frac{\mathrm{\Delta }l}{l}}\times 100\end{array}$

Complete answer:
Let’s start by discussing the propagation of error (or) uncertainty.
The error propagation formulas are based upon the consideration that the errors are random or not correlated to each other.
Now, let’s find out the density of the wire as asked in the problem. Density $(\rho )$ is given by mass (m) divided by volume (V). The wire is in the shape of a cylinder. Hence, the volume of a wire of length (l) and radius (r) will be given by, $V=\pi r^{2}l$ .
Hence, the density of the wire is given by, $\rho ={\displaystyle \frac{m}{\pi r^{2}l}}$ .
Now, using formula of error propagation in density which is given as,
${\displaystyle \frac{\mathrm{\Delta }\rho }{\rho }}={\displaystyle \frac{\mathrm{\Delta }m}{m}}+2({\displaystyle \frac{\mathrm{\Delta }r}{r}})+{\displaystyle \frac{\mathrm{\Delta }l}{l}}$
The percentage error will be the same as above, with multiplying 100 to each of the terms. That is, ${\displaystyle \frac{\mathrm{\Delta }\rho }{\rho }}\text{ \%}={\displaystyle \frac{\mathrm{\Delta }m}{m}}\times 100+2({\displaystyle \frac{\mathrm{\Delta }r}{r}})\times 100+{\displaystyle \frac{\mathrm{\Delta }l}{l}}\times 100$
Hence the percentage error in density is,
$\begin{array}{rl}& {\displaystyle \frac{\mathrm{\Delta }\rho }{\rho }}\text{ \%}={\displaystyle \frac{0.003}{3}}\times 100+2({\displaystyle \frac{0.005}{5}})\times 100+{\displaystyle \frac{0.006}{6}}\times 100\\ & \Rightarrow {\displaystyle \frac{\mathrm{\Delta }\rho }{\rho }}\text{\%}=[1+2(1)+1]=4\text{\%}\end{array}$
Hence, the percentage error in the measurement in density of wire is 4%.

Note:
Each of the multiple formulas shown in the error propagation can be clubbed into each other as per the requirement of the variable. The percentage values of each term has to be calculated for each term individually.