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If L₁ = (3.03±0.02)m and L₂ = (2.01±0.02)m then L₁ + 2L₂ is (in m)

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7.05 ± 0.06

6.05 ± 0.06

6.05 ± 0.02

7.05 ± 0.02

answer is A.

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Detailed Solution

$x = L_{1} + 2 L_{2} = 7 .05; Δx = {ΔL}_{1} + 2 {ΔL}_{2}$

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