Question

The students in the previous problem (attatched) determine the mass of the metal cube to be 21.60 pm.05 gm How should they report the density of the material?

A group of students is determining the density of a cube of metal. Using a calipers, they measure all three sides of the cube to be 2.000+ .002 cm. What is the volume they calculate, including the uncertainty? Do Expert Answer Anonymous 12 minutes later The length of the side of the cube is a = (2.000 + 0.002) cm The volume of the cube is V = a= 2.0003 = 8 cm3 And the uncertainty in the calculation of the volume can be given by u(V) 3u(a) 2 = u(V)= "u(a) = *° x 0.002 = 0.024 cm3 So, the volume of the cube with the uncertainty is V = (8.000+ 0.024) cm3

Answer 1

Hi,

Hope you are doing well.

Given that,

Volume of the cube with uncertainty,

V = (8.000 + 0.024) cm3

Mass of cube with uncertainty,

m = (21.60 = .05) 9

Hence, Density of the cube is given by,

21.60 g 8.000 cm3 V

..D= 2.700 g/cm

Uncertainty in the calculation of density is given by,

... (D) _um) u(V) m

um) u(V) 1. (D) = Dx +- ) = (2.700 g/cm) 10.05 g 0.024 cm3 (21.60 g. 8.000 cm3 m

..u (D) = 0.01435 g/cm

Rounding off to 3 significant figures,

u (D) = 0.01 g/cm

So the density of the cube with uncertainty is,

D= (2.700 + 0.01) g/cm