Question

The radius (r), the length (l) and resistance (x) of a thin wire are\[\left( {0.2 \pm 0.02} \right)cm\], \[\left( {80 \pm 0.1} \right)cm\] and \[\left( {30 \pm 1} \right)\Omega \] respectively. The percentages error in the specific resistance is-
1. \[23.4\% \]
2. \[25.4\% \]
3. \[26\% \]
4. \[27.5\% \]

Answer 1

Hint: In this question radius (r), the length (l) and resistance (x) of a thin wire are given so by using the resistance formula and by differentiating it we find the percentages error in the specific resistance.

Complete step by step answer:
The radius of the thin wire \[r = \left( {0.2 \pm 0.02} \right)cm\]
Length of the thin wire \[l = \left( {80 \pm 0.1} \right)cm\]
Resistance of the thin wire \[x = \left( {30 \pm 1} \right)\Omega \]
We know the resistance of a wire is given by the formula \[R = \dfrac{{\rho l}}{A} - - (i)\], where \[\rho \]is the resistivity or the specific resistance of a wire
So the equation (i) in the form of specific resistance can be written as
\[\rho = \dfrac{{AR}}{l} - - (ii)\]
Where resistance of the wire is x
Where A is the area of cross section of the thin wire which is\[A = \pi {r^2}\], hence we can further write equation (ii) as,
\[\rho = \dfrac{{\left( {\pi {r^2}} \right)x}}{l} - - (iii)\]
Now we take logarithm on both side of the equation, so we get
\[\ln \left( \rho \right) = \ln \left( \pi \right) + 2\ln \left( r \right) + \ln \left( x \right) - \ln \left( l \right) - - (iv)\][By using logarithmic properties]
Now we differentiate the equation (iv), hence we get
\[\dfrac{{\Delta \rho }}{\rho } = 2\dfrac{{\Delta r}}{r} + \dfrac{{\Delta x}}{x} + \dfrac{{\Delta l}}{l} - - (v)\]
Now we substitute the values of radius, length and the resistance of the wire in equation (v), so we get
\[
  \dfrac{{\Delta \rho }}{\rho } = 2\dfrac{{\Delta r}}{r} + \dfrac{{\Delta x}}{x} + \dfrac{{\Delta l}}{l} \\
\Rightarrow \dfrac{{\Delta \rho }}{\rho } = 2\dfrac{{0.02}}{{0.2}} + \dfrac{1}{{30}} + \Rightarrow \dfrac{{0.1}}{{80}} \\
\Rightarrow \dfrac{{\Delta \rho }}{\rho } = 0.2 + 0.033 + 0.00125 \\
\Rightarrow \dfrac{{\Delta \rho }}{\rho } = 0.234 \\
\]
Hence the error in the specific resistance \[ = 0.234\]
Also the percentages error will be
\[
  \dfrac{{\Delta \rho }}{\rho } \times 100\% = 0.234 \times 100 \\
  \Rightarrow \% \; error= 23.4\% \\
 \]
Hence the percentages error in the specific resistance will be \[ = 23.4\% \]
Option 1 is correct.

Note: The term with the raised positive power will exert more resultant error in comparison to the terms which are being added or subtracted. It is important to note that the error for the product of the terms will be added together while for the ratio of the terms are subtracted.