The measured length of two rods are l1=30cm±0.5cm and l2=20cm+0.1cm.The percentage error in difference of length of rods is

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The measured length of two rods are $l_{1} = 30 cm \pm 0 .5 cm$ and $l_{2} = 20 cm + 0 .1 cm$ .The percentage error in difference of length of rods is

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3 %

6 %

4 %

5 %

answer is A.

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$\frac{Δy %}{y} = \frac{{Δℓ}_{1} + {Δℓ}_{2}}{ℓ_{1} - ℓ_{2}} = \frac{0 .5 + 0 .1}{30 - 20} = \frac{0 .6}{10} = 6 %$

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