Question

Calculate the percentage error in the specific resistance, $\rho = \pi {r^2}R/l$, where $r =$radius of the wire $= 0.26 \pm 0.02\,{\text{cm}}$, $l =$length of wire $= \,\,156.0 \pm 0.1\,{\text{cm}}$, and $R =$Resistance of the wire$= 64 \pm 2\Omega$.

Answer 1

Use the formula
$\dfrac{{\Delta \rho }}{\rho } = 2 \times \dfrac{{\Delta r}}{r} + \dfrac{{\Delta R}}{R} + \dfrac{{\Delta l}}{l}$, then convert to percentage.

Complete step by step solution:
In some data the approximation error is the difference between an exact value and some approximation to it. An approximation error can occur because: Data measurement due to the instruments is not accurate or uses approximations instead of the actual details. Relative error (RE), when used as a measure of accuracy, is the ratio of a measurement's absolute error to the measurement being taken. This kind of error, in other words, is proportional to the size of the object being measured. Relative error is expressed in percentage terms and does not have units.

In this problem the given quantities are:
Specific resistance,
$\rho = \pi {r^2}R/l$
Radius of the wire,
$r = 0.26 \pm 0.02\,{\text{cm}}$
Length of the wire,
$l = \,\,156.0 \pm 0.1\,{\text{cm}}$
Resistance of the wire,
$R = 64 \pm 2\Omega$
The formula which relates the specific resistance and the resistance is:

$$R = \dfrac{{\rho L}}{A} \\\ R = \dfrac{{\rho L}}{{\pi {r^2}}} \\\$$

Rearranging the above equation:
$\rho = \dfrac{{\pi {r^2} \times R}}{L}$ …… (1)
We differentiate both the sides in terms of relative error:
Differentiation of ${r^2}$gives a multiplier of $2$.
To find the percentage error in the specific resistance, apply the formula:
$\dfrac{{\Delta \rho }}{\rho } \times 100\%$ …… (2)
Now, do the following operation:
First add all the errors of the quantities mentioned in the equation (1):

$$\dfrac{{\Delta \rho }}{\rho } = 2 \times \dfrac{{\Delta r}}{r} + \dfrac{{\Delta R}}{R} + \dfrac{{\Delta l}}{l} \\\ = 2 \times \dfrac{{0.02}}{{0.26}} + \dfrac{2}{{64}} + \dfrac{{0.1}}{{156}} \\\ = 2 \times 0.076 + 0.03 + 0.0006 \\\ = 0.1826 \\\$$

The error in specific resistance is found to be $0.1826$.
To convert it into percentage, substitute the values in the equation (2):

$$0.1826 \times 100\% \\\ = 18.26\% \\\$$

Hence, the percentage error in the specific resistance is $18.26\%$.

Note: In this given problem, you are asked to find the percentage error in specific resistance. For this, you need to first differentiate the quantities in terms of relative error. It is important that a multiplier $2$ comes, as the radius is seen as square. Not using the multiplier will definitely affect the result.