Question : The specific resistivity ρ of a circular wire of radius r, resistance R and length l is give by equation ρ=πr2R/l. Given r=(0.25±0.02) cm, R=(20±1)Ω and l=(4.80±0.12) cm. The percentage error in ρ is nearly
a) 7%
b) 18%
c) 23.5%
d) None of these.
Doubt by Ananya
Solution :
ρ=πr2R/l (Given)
Where
ρ = Specific Resistivity
r = Radius of the wire
r = Radius of the wire
R = Resistance of Wire
l=Length of the wire
l=Length of the wire
r=(0.25±0.02) cm,
R=(20±1)Ω and
l=(4.80±0.12) cm
Relative Error
Δρ/ρ = 2Δr/r+ΔR/R+Δl/l
Relative Error
Δρ/ρ = 2Δr/r+ΔR/R+Δl/l
Percentage Error
(Δρ/ρ)×100%
= 2(Δr/r)×100%+(ΔR/R)×100% +(Δl/l)×100%
(Δρ/ρ)×100%
= 2(Δr/r)×100%+(ΔR/R)×100% +(Δl/l)×100%
= 2(0.02/0.25)×100% + (1/20)×100% +(0.12/4.80)×100%
= 16% + 5%+2.5%
= 23.5%
= 23.5%
Hence, c) would be the correct answer.