Find the percentage error in specific resistance given by ρ = π r 2 R l where r is the radius having value ( 0.2 ± 0.02 ) cm , R is the resistance of ( 60 ± 2 ) ohm and l is the length of ( 150 ± 0.1 ) cm .

Δ ρ ρ × 100 = [ 2 ( Δ r r ) + ( Δ R R ) + ( Δ l l ) ] × 100 = [ 2 × 0.02 0.2 + 2 60 + 0.1 150 ] × 100 = [ 0.2 + 0.0333 + 0.00067 ] × 100 = 23.4 %

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Physics - Unit & Dimension Question with Solution | TestHub

PhysicsUnit & DimensionMiscellaneousMedium2 minQB

PhysicsMediumnumerical

Find the percentage error in specific resistance given by $ρ = \frac{π r^{2} R}{l}$ where $r$ is the radius having value $(0.2 \pm 0.02) cm, R$ is the resistance of $(60 \pm 2)$ ohm and $l$ is the length of $(150 \pm 0.1) cm$ .

Answer:

23.40

Solution:

$\frac{Δ ρ}{ρ} \times 100 = [2 (\frac{Δ r}{r}) + (\frac{Δ R}{R}) + (\frac{Δ l}{l})] \times 100$

$= [\frac{2 \times 0.02}{0.2} + \frac{2}{60} + \frac{0.1}{150}] \times 100$

$= [0.2 + 0.0333 + 0.00067] \times 100$

$= 23.4 %$

Subject:PhysicsTopic:Unit & DimensionSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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