Question

If the measured value of resistance R = 1.05 Ω wire diameter $\text{d = 0.60 mm,}$ mm, and length$\text{l = 75.3 cm}$ then find the maximum permissible error in resistivity, ρ=R(πd2/4)l\rho = \frac{R(\pi d^2 /4)}{l}ρ=lR(πd2/4)​

Answer 1

Given Data:

Resistance, R = 1.05 Ω

Diameter, $\text{d = 0.60 mm,}$

Length of wire, $\text{l = 75.3 cm}$

Step 1: Error in resistivity:

$\frac{d\text{ρ}}{\text{ρ}}=\frac{\Delta R}{R}+\frac{2\Delta d}{d}+$ $+\frac{\Delta l}{l}$

For resistance least count will be, $\Delta R=0.01$ Ω

For Diameter least count will be, $\Delta d=0.01mm$

For length least count will be, $\Delta l=0.1cm$

$\frac{d\text{ρ}}{\text{ρ}}=\frac{\Delta R}{R}+\frac{2\Delta d}{d}+$ $+\frac{\Delta l}{l}⟹$

$\frac{d\text{ρ}}{\text{ρ}}=\frac{0.01}{1.05}+\frac{2\left(0.01mm\right)}{0.06mm}+\frac{0.1cm}{75.3cm}$

$\frac{d\text{ρ}}{\text{ρ}}=(\frac{0.01}{1.05}+\frac{2\left(0.01mm\right)}{0.06mm}+\frac{0.1cm}{75.3cm})\times 100=4\%$