A wire has a mass 0.4±0.004 g and length 8±0.08(cm) The maximum percentage error in the measurement of its density is 4 %. The radius of the wire is r±Δ r find Δ r

Question

A wire has a mass 0.4±0.004 g and length 8±0.08(cm) The maximum percentage error in the measurement of its density is 4 %. The radius of the wire is r±Δ r find Δ r (A) 0.02 r (B) 0.01 r (C) 0.03 r (D) 0.1 r

Tardigrade Admin · Accepted Answer

ρ=(m/π r2L) (Δρ/ρ)×100=[(Δ m/m)+2(Δ r/r)+(Δ L/L)]×100 =(0.01+2(Δ r/r)+0.01)×100=4 (given) or 100(Δ r/r)=1 Δ r=0.01 r

Q. A wire has a mass $0.4 \pm 0.004 g$ and length $8 \pm 0.08 (c m)$ The maximum percentage error in the measurement of its density is $4%$ . The radius of the wire is $r \pm Δ r$ find $Δ r$

0.02 r

0.01 r

0.03 r

0.1 r

$ρ = \frac{m}{π r ^{2} L}$

$\frac{Δ ρ}{ρ} \times 100 = [\frac{Δ m}{m} + 2 \frac{Δ r}{r} + \frac{Δ L}{L}] \times 100$

$= (0.01 + 2 \frac{Δ r}{r} + 0.01) \times 100 = 4 (g i v e n)$

or $100 \frac{Δ r}{r} = 1$

$Δ r = 0.01 r$

Q. A wire has a mass 0.4±0.004g and length 8±0.08(cm) The maximum percentage error in the measurement of its density is 4%. The radius of the wire is r±Δr find Δr