The specific resistance σ of a thin wire of radius r cm, resistance R Ω and length L cm is given by σ=(πr²R)/L. If r=0.26±0.02cm, R=32±1Ω and L=78 ± 0.01 cm,find the percentage error in σ.

The correct option is 18%.

Step-by-step Derivation:

The specific resistance $\sigma$ of a thin wire is given by the formula:

$\sigma =\frac{\pi r^{2}R}{L}$

where:
- $r$ is the radius of the wire,
- $R$ is the resistance of the wire, and
- $L$ is the length of the wire.

To find the relative error, we take the natural logarithm on both sides and differentiate:

$\frac{\Delta \sigma }{\sigma }=\pm \left(2\frac{\Delta r}{r}+\frac{\Delta R}{R}+\frac{\Delta L}{L}\right)$

The maximum percentage error in $\sigma$ is therefore given by:

$\frac{\Delta \sigma }{\sigma }\times 100\%=\left(2\frac{\Delta r}{r}+\frac{\Delta R}{R}+\frac{\Delta L}{L}\right)\times 100\%$

Given the measured values and their absolute errors:
- Radius, $r=0.26\pm 0.02\text{cm}$ (so $\Delta r=0.02$)
- Resistance, $R=32\pm 1\Omega$ (so $\Delta R=1$)
- Length, $L=78\pm 0.01\text{cm}$ (so $\Delta L=0.01$)

Now, we calculate the individual percentage errors for each quantity:

- Percentage error due to radius $r$:

$2\times \frac{\Delta r}{r}\times 100\%=2\times \frac{0.02}{0.26}\times 100\%\approx 15.38\%$

- Percentage error due to resistance $R$:

$\frac{\Delta R}{R}\times 100\%=\frac{1}{32}\times 100\%\approx 3.13\%$

- Percentage error due to length $L$:

$\frac{\Delta L}{L}\times 100\%=\frac{0.01}{78}\times 100\%\approx 0.01\%$

Adding the percentage errors together gives the total percentage error in specific resistance $\sigma$:

$\text{Percentage error}\approx 15.38\%+3.13\%+0.01\%=18.52\%$

Rounding to the nearest integer representing the options provided, the percentage error is approximately 18%.