The specific resistance ρ of a circular wire of radius, resistance R and length l is given by ρ=πr2Rl. Given r = 0.24 ± 0.02 cm, R = 30 ± 1Ω, and l = 4.80±0.01 cm The percentage error in ρ is nearly:

∆ρρ=2∆rr+∆RR+∆ll = 2×.020.24+130+.014.80 =.16+.033+ .002 =0.195 Percentage error in specific resistance= 20%

The specific resistance ρ of a circular wire of radius, resistance R and length l is given by ρ=πr2Rl. Given r = 0.24 ± 0.02 cm, R = 30 ± 1Ω, and l = 4.80±0.01 cm The percentage error in ρ is nearly:

∆ρρ=2∆rr+∆RR+∆ll = 2×.020.24+130+.014.80 =.16+.033+ .002 =0.195 Percentage error in specific resistance= 20%

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The specific resistance ρ of a circular wire of radius, resistance R and length l is given by $ρ = \frac{{πr}^{2} R}{l}$ . Given r = 0.24 ± 0.02 cm, R = 30 ± 1Ω, and l = 4.80 $\pm$ 0.01 cm The percentage error in ρ is nearly:

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$\frac{∆ ρ}{ρ} = 2 \frac{∆ r}{r} + \frac{∆ R}{R} + \frac{∆ l}{l} = 2 \times \frac{. 02}{0.24} + \frac{1}{30} + \frac{. 01}{4.80} = . 16 + . 033 + . 002 = 0.195$

Percentage error in specific resistance= 20%

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The specific resistance ρ of a circular wire of radius, resistance R and length l is given by ρ=πr2Rl. Given r = 0.24 ± 0.02 cm, R = 30 ± 1Ω, and l = 4.80±0.01 cm The percentage error in ρ is nearly:

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The specific resistance ρ of a circular wire of radius, resistance R and length l is given by $ρ = \frac{{πr}^{2} R}{l}$ . Given r = 0.24 ± 0.02 cm, R = 30 ± 1Ω, and l = 4.80 $\pm$ 0.01 cm The percentage error in ρ is nearly: