Question

D. If the length of a cylinder is 1 =
(4.00+0.001) cm, radius r = (0.0250
+0.001) cm and mass m = (6.25+0.01)
gm. Calculate the percentage error in the
determination of density.
[Ans: 8.185%​

Answer 1

Hey there,

◆ Answer -

∆ρ/ρ = 8.185 %.

● Explaination -

# Given -

h±∆h = 4 ± 0.001 cm

r±∆r = 0.025 ± 0.001 cm

m±∆m = 6.25 ± 0.01 g

# Solution -

Density of cylinder would be -

ρ = m / V

ρ = m/(πr²h)

Therefore, relative error in measurement of density will be -

∆ρ/ρ = ∆m/m + 2∆r/r + ∆h/h

∆ρ/ρ = 0.01/6.25 + 2×0.001/0.025 + 0.001/4

∆ρ/ρ = 0.08185

Percentage error in measurement of density will be ∆ρ/ρ×100% = 8.185 %.

Thanks dear...

Answer 2

Answer:

8.815%

Explanation:

length of cylinder , l = (4.00 ± 0.001) cm

radius of base of cylinder , r = (0.025 ± 0.001) cm

mass of cylinder, m = (6.25 ± 0.01) gm

we know, density , d = m/v

and volume of cylinder , v = πr²l

so, density , d = m/(πr²l)

then fractional error can be written as,

∆d/d = ∆m/m + 2 × ∆r/r + ∆l/l

given, ∆m = 0.01g , m = 6.25 g , ∆r = 0.001, r = 0.025 , ∆l = 0.001 and l = 4.00

so, ∆d/d = (0.01)/(6.25) + 2 × (0.001)/(0.025) + (0.001)/(4.00)

= 0.0016 + 0.08 + 0.00025

= 0.08185

percentage error of density = 100 × fractional error of density

= 100 × 0.08185

= 8.185 %