Question 87: Units , Dimensions and Error Analysis

Question

The radius $(\mathrm{r})$, length $(l)$ and resistance $(\mathrm{R})$ of a metal wire was measured in the laboratory as
$$
\begin{aligned}
& \mathrm{r}=(0.35 \pm 0.05) ~\mathrm{cm} \\
& \mathrm{R}=(100 \pm 10) ~\mathrm{ohm} \\
& l=(15 \pm 0.2)~ \mathrm{cm}
\end{aligned}
$$

The percentage error in resistivity of the material of the wire is :

Accepted Answer

To calculate the percentage error in the resistivity of the material of the wire, we need to understand the formula for resistivity. The resistivity $$ ho $$ of a wire is given by:
$ho = \frac{RA}{l}$
where:
$$ R $$ is the resistance
$$ A $$ is the cross-sectional area of the wire
$$ l $$ is the length of the wire
The cross-sectional area $$ A $$ of the wire with radius $$ r $$ is:
$A = \pi r^2$
We can plug this into the equation for resistivity to get:
$ho = \frac{R \pi r^2}{l}$
Now, to find the percentage error in resistivity, we need to find the percentage errors in $$ R $$, $$ r $$, and $$ l $$ and then use the following rule for combining errors:
For a given function, $$ f = f(x,y,z,...) $$, where $$ x, y, z,... $$ are the measured quantities with possible errors, the percentage error in $$ f $$, denoted as $$ (\delta f)_{\%} $$, can be approximated by adding the relative percentage errors of the input quantities. If $$ f $$ has the form of a product and quotient of the measured quantities as in our case ($$ ho = \frac{R \pi r^2}{l} $$), the percentage error in $$ f $$ is given by:
$(\delta f)_{\%} = (\delta x)_{\%} + (\delta y)_{\%} + (\delta z)_{\%} + ...$
Where $$ (\delta x)_{\%} $$, $$ (\delta y)_{\%} $$, and $$ (\delta z)_{\%} $$ are the percentage errors in each measured quantity $$ x, y, z, ... $$ respectively.
For our case:
The percentage error in radius $$ (\delta r)_{\%} $$ is given by the error in $$ r $$ divided by the average radius and then multiplied by 100:
$(\delta r)_{\%} = \left(\frac{0.05}{0.35}ight) 	imes 100$
The percentage error in resistance $$ (\delta R)_{\%} $$ is:
$(\delta R)_{\%} = \left(\frac{10}{100}ight) 	imes 100$
The percentage error in length $$ (\delta l)_{\%} $$ is:
$(\delta l)_{\%} = \left(\frac{0.2}{15}ight) 	imes 100$
Now let's calculate each:
$(\delta r)_{\%} = \left(\frac{0.05}{0.35}ight) 	imes 100 \approx 14.29\%$
$(\delta R)_{\%} = \left(\frac{10}{100}ight) 	imes 100 = 10\%$
$(\delta l)_{\%} = \left(\frac{0.2}{15}ight) 	imes 100 \approx 1.33\%$
However, since the area $$ A $$ is proportional to $$ r^2 $$, the percentage error in $$ A $$ will be twice the percentage error in $$ r $$. Thus:
$(\delta A)_{\%} = 2 	imes (\delta r)_{\%} = 2 	imes 14.29\% \approx 28.58\%$
Finally, we add the percentage errors to find the percentage error in resistivity:
$(\delta ho)_{\%} = (\delta R)_{\%} + (\delta A)_{\%} + (\delta l)_{\%}$
$(\delta ho)_{\%} = 10\% + 28.58\% + 1.33\% \approx 39.91\%$
This calculation gives us a value close to 39.91%, which means the correct option is closest to this value. Thus, the best answer is:
Option D $39.9\%$

Answer

$37.3 \%$

Answer

$25.6 \%$

Answer

$35.6 \%$

Answer

$39.9 \%$

Units , Dimensions and Error Analysis

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