QJEEM24Phebe0: JEE Main 2024 Physics - Class 11 JEEM24Phebe0 Solution: The radius , length and resistance of a metal wire was measured …

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 $39.9\%$