A wire has a mass 0.3 ±0.003 g, radius 0.5 ±0.005 mm and length 6 ±0.06 cm. The maximum percentage error in the measurement of its density is

Question

A wire has a mass 0.3 ±0.003 g, radius 0.5 ±0.005 mm and length 6 ±0.06 cm. The maximum percentage error in the measurement of its density is (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

Density ρ =(m/π r2 L) ∴ (Δ ρ /ρ )× 100=((Δ m/m) + 2 (Δ r/r) + (Δ L/L))× 100 After substituting the values we get the maximum percentage error in density = 4 %

Q. A wire has a mass $0.3 \pm 0.003 g,$ radius $0.5 \pm 0.005 mm$ and length $6 \pm 0.06 c m .$ The maximum percentage error in the measurement of its density is

$1$

$2$

$3$

$4$

Density
$ρ = \frac{m}{π r ^{2} L}$
$∴ \frac{Δ ρ}{ρ} \times 100 = (\frac{Δ m}{m} + 2 \frac{Δ r}{r} + \frac{Δ L}{L}) \times 100$
After substituting the values we get the maximum percentage error in density $= 4%$

Q. A wire has a mass 0.3±0.003g, radius 0.5±0.005mm and length 6±0.06cm. The maximum percentage error in the measurement of its density is

1

2

3

4

Density ρ=πr2Lm​ ∴ρΔρ​×100=(mΔm​+2rΔr​+LΔL​)×100 After substituting the values we get the maximum percentage error in density =4%

Q. A wire has a mass $0.3 \pm 0.003 g,$ radius $0.5 \pm 0.005 mm$ and length $6 \pm 0.06 c m .$ The maximum percentage error in the measurement of its density is

$1$

$2$

$3$

$4$

Density
$ρ = \frac{m}{π r ^{2} L}$
$∴ \frac{Δ ρ}{ρ} \times 100 = (\frac{Δ m}{m} + 2 \frac{Δ r}{r} + \frac{Δ L}{L}) \times 100$
After substituting the values we get the maximum percentage error in density $= 4%$