To calculate the percentage error in the resistivity of the material of the wire, we need to understand the formula for resistivity. The resistivity $$ \rho $$ of a wire is given by:

$\rho = \frac{RA}{l}$

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:

$\rho = \frac{R \pi r^2}{l}$

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 ($$ \rho = \frac{R \pi r^2}{l} $$), 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:

$(\delta r)_{\%} = \left(\frac{0.05}{0.35}\right) \times 100$

• The percentage error in resistance $$ (\delta R)_{\%} $$ is:

$(\delta R)_{\%} = \left(\frac{10}{100}\right) \times 100$

• The percentage error in length $$ (\delta l)_{\%} $$ is:

$(\delta l)_{\%} = \left(\frac{0.2}{15}\right) \times 100$

Now let's calculate each:

$(\delta r)_{\%} = \left(\frac{0.05}{0.35}\right) \times 100 \approx 14.29\%$

$(\delta R)_{\%} = \left(\frac{10}{100}\right) \times 100 = 10\%$

$(\delta l)_{\%} = \left(\frac{0.2}{15}\right) \times 100 \approx 1.33\%$

However, since the area $$ A $$ is proportional to $$ r^2 $$, the percentage error in $$ A $$ will be twice the percentage error in $$ r $$. Thus:

$(\delta A)_{\%} = 2 \times (\delta r)_{\%} = 2 \times 14.29\% \approx 28.58\%$

Finally, we add the percentage errors to find the percentage error in resistivity:

$(\delta \rho)_{\%} = (\delta R)_{\%} + (\delta A)_{\%} + (\delta l)_{\%}$

$(\delta \rho)_{\%} = 10\% + 28.58\% + 1.33\% \approx 39.91\%$

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 $39.9\%$