Help on Uncertainty needed!

[Ash1994]

The mass of a cube of aluminium is found to be 580 g with an uncertainty in the
measurement of 10 g. Each side of the cube has a length of (6.0 ± 0.1) cm.
Calculate the density of aluminium with its uncertainty. Express your answer to an
appropriate number of significant figures.

 

[Talha]

d=m/v
d=m/l^3
d=580/6.0^3
d=2.69 g cm^-3

for uncertainity:
d=m/l^3
δ d/d = δ m/m + 3 δ l/l
= 10/580 + 3 × 0.1/6.0
= 0.0172 + 0.05
δ d/d = 0.0672
δ d = 0.0672 × 2.69
= 0.181

so, density = 2.69 +- 0.18 g cm^-3 :Yahoo!: 8)

 

[Ash1994]

d=m/v
d=m/l^3
d=580/6.0^3
d=2.69 g cm^-3

for uncertainity:
d=m/l^3
δ d/d = δ m/m + 3 δ l/l
= 10/580 + 3 × 0.1/6.0
= 0.0172 + 0.05
δ d/d = 0.0672
δ d = 0.0672 × 2.69
= 0.181

so, density = 2.69 +- 0.18 g cm^-3 :Yahoo!: 8)

buh the marking scheme says 2.69 ± 0.09 g cm-3 =S

 

[Saturation]

No, look at the ms carefully and you'll see that they say only 4 marks for 2.69 ± 0.09 g cm-3

The uncertainity is +- 0.18