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A cylindrical wire of mass (0.4±0.01) g has length (8±0.04) cm and radius (6±0.03) mm. The maximum error in its density will be

Asked in JEE Main 8th April 1st Shift 2023 · Percentage error in a derived quantity

Answer: (3) 4%

Step-by-step solution

Given: m=0.4±0.01 g, l=8±0.04 cm, r=6±0.03 mm.

Idea: ρ=m/(π r²l), so the radius enters squared.

(Δρ)/ρ=(Δ m)/m+2(Δ r)/r+(Δ l)/l

=(0.01)/(0.4)+2×(0.03)/6+(0.04)/8

=0.025+0.010+0.005=0.04, that is 4%.

Why the other options are wrong

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