Find the percentage error in the specific resistance ρ = π r 2 R l , where r = ( 0.3 ± 0.03 ) cm , R = ( 40 ± 2 ) ohm and l = ( 120 ± 0.1 ) cm .

Δ ρ ρ × 100 = [ 2 ( Δ r r ) + ( Δ R R ) + ( Δ l l ) ] × 100 = [ 2 × 0.03 0.3 + 2 40 + 0.1 120 ] × 100 = [ 0.2 + 0.05 + 0.00083 ] × 100 = 25.1 %

Start Free Mock LoginAlready registered? Login here

Physics - Unit & Dimension Question with Solution | TestHub

PhysicsUnit & DimensionMiscellaneousMedium2 minQB

PhysicsMediumnumerical

Find the percentage error in the specific resistance $ρ = \frac{π r^{2} R}{l}$ , where $r = (0.3 \pm 0.03) cm, R = (40 \pm 2)$ ohm and $l = (120 \pm 0.1) cm$ .

Answer:

25.10

Solution:

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

$= [\frac{2 \times 0.03}{0.3} + \frac{2}{40} + \frac{0.1}{120}] \times 100$

$= [0.2 + 0.05 + 0.00083] \times 100$

$= 25.1 %$

Subject:PhysicsTopic:Unit & DimensionSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...