Question

The specific resistance σ of a thin wire of radius r cm, resistance R Ω and length L cm is given by σ = (πr^2R)/L. If r = 0.26 +/- 0.02 cm, R = 32 +/- 1 Ω and L = 78.00 +/- 0.01 cm, find the % error in σ.

Answer 1

Answer:

Explanation:

Answer is in attachment.

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Answer 2

Answer:

The percentage error in the specific resistance is 18.5%.

Explanation:

Given that,

The specific resistance$\sigma=\dfrac{\pi r^2R}{L}$

Radius r = 0.26±0.02 cm

Length l =78.00±0.01 cm

Resistance R =32±1 ohm

The percentage error is defined as,

The percentage error in the specific resistance is

$\dfrac{\Delta \sigma}{\sigma}\times100=2\times\dfrac{\Delta r}{r}+\dfrac{\Delta R}{R}-\dfrac{\Delta L}{L}$

$\dfrac{\Delta \sigma}{\sigma}\times100=2\times\dfrac{0.02}{0.26}+\dfrac{1}{32}-\dfrac{0.01}{78.00}$

$\dfrac{\Delta \sigma}{\sigma}\times100=0.185\times100$

$\sigma\%=18.5\%$

Hence, The percentage error in the specific resistance is 18.5%.