A wire of length and radius and mass Maximum percentage error in density is [2004]

Question

A wire of length $ℓ = 6 \pm 0.06 cm$ and radius $r = 0.5 \pm 0.005 cm$ and mass $m = 0.3 \pm 0.003 gm.$ Maximum percentage error in density is [2004]

Smart Achievers · Accepted Answer

(1)

$Density, ρ = \frac{m}{v} = \frac{m}{A ℓ} = \frac{m}{ℓ π r^{2}}$

Taking log and differentiating for errors,

$\frac{Δ ρ}{ρ} = \frac{Δ m}{m} + 2 \frac{Δ r}{r} + \frac{Δ ℓ}{ℓ} \dots (i)$

From question, putting the values of

$Δ ℓ = 0.06 cm, ℓ = 6 cm; Δ r = 0.005 cm, r = 0.5 cm,$

$m = 0.3 g; Δ m = 0.003 g$ in eqn. (i) we get $\frac{Δ ρ}{ρ} = \frac{4}{100}$

$∴$ $Percentage error in density,$

$\frac{Δ ρ}{ρ} \times 100 = \frac{4}{100} \times 100 = 4 % .$

Solution

Q.
A wire of length $ℓ = 6 \pm 0.06 cm$ and radius $r = 0.5 \pm 0.005 cm$ and mass $m = 0.3 \pm 0.003 gm.$ Maximum percentage error in density is [2004]

1 4

2 2

3 1

4 6.8

Want Us to Email you About Special Offers & Updates?

Solution

Q. A wire of length ℓ=6±0.06 cm and radius r=0.5±0.005 cm and mass m=0.3±0.003 gm. Maximum percentage error in density is [2004]

1 4

2 2

3 1

4 6.8

Want Us to Email you About Special Offers & Updates?

Q.
A wire of length $ℓ = 6 \pm 0.06 cm$ and radius $r = 0.5 \pm 0.005 cm$ and mass $m = 0.3 \pm 0.003 gm.$ Maximum percentage error in density is [2004]