Question

To find the density of a given material, a solid cuboid of the material was taken and its dimensions as well as its mass were measured. The result of the measurement were

l = (5.10 ± 0.05) cm 

b = (3.25 ± 0.05) cm 

h = (1.40 ± 0.01) cm 

m = (185.450 ± 0.002) g

The volume (V), of the cuboid, and the density p , of the material equal, respectively.

(1) (32.2 ± 0.74) cm³ and (7.99 ± 0.24) g/cm³

(2) (23.21 ± 0.74) cm³ and (8.0 ± 0.2) g/cm³

(3) (23.21 ± 0.74) cm³ and (7.99 ± 0.24) g/cm³

(4) (23.2 ± 0.7) cm³ and (8.0 ± 0.2) g/cm³

Answer 1

(4) (23.2 ± 0.7) cm³ and (8.0 ± 0.2) g/cm³

We have, V = l b h

= (5.10 x 3.25 x 1.40) cm³ = 23.205 cm³

Also,

\(\frac{ΔV}{V}=\frac{Δl}{l}+\frac{Δb}{b}+\frac{Δh}{h}\)

\(=\frac{0.05}{5.10}+\frac{0.05}{3.25}+\frac{0.01}{1.40}\)

= 0.0098 + 0.0154 +0.0071 = 0.032

∴ ΔV = 0.302 x 23.205cm³ = 0.74 cm³

We, therefore, write

V = (23.2 ± 0.7) cm³ 

Now density,

d = mass/volume = 185.45/23.2 g/cm³

= 7.99 g/cm³

Also,

\(\frac{Δ{d}}{d}=\frac{Δm}{m}+\frac{ΔV}{V}\)

= \(\frac{0.002}{185.450}+\frac{0.7}{23.2}\)

= 0.00001 + 0.03 = 0.03

∴ Δd = 0.03 x 7.99 g/cm³ = 0.24 g/cm³

We, therefore, write

d = (8.0 ± 0.2) g/cm³