A copper collar is to fit tightly about a steel shaft whose diameter is 6.0000 cm at 19°C. The inside diameter of the copper collar at that temperature is 5.9800 cm. To what temperature must the copper collar be raised so that it will just slip on the steel shaft, assuming the temperature of both the steel shaft and copper collar are raised simultaneously?

T' = 865.15 °C

Explanation:

In order for the copper collar to just slip on the steel shaft the, assuming are heated simultaneously, we must find the final parameters of both and equate them. Because the final diameters of both must be same for the slipping to occur.

FOR COPPER COLLAR:

dc' = dc(1 + ∝c*ΔT)

where,

dc' = final diameter of copper ring

dc = initial diameter of copper ring = 5.98 cm

∝c = coefficient of linear expansion for copper = 16 x 10⁻⁶ °C⁻¹

ΔT = Change in Temperature

Therefore,

dc' = (5.98 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT]   equation (1)

FOR STEEL SHAFT:

ds' = ds(1 + ∝s*ΔT)

where,

ds' = final diameter of steel shaft

ds = initial diameter of steel shaft = 6 cm

∝s = coefficient of linear expansion for steel = 12 x 10⁻⁶ °C⁻¹

ΔT = Change in Temperature

Therefore,

dc' = (6 cm)[1 + (12 x 10⁻⁶ °C⁻¹)ΔT]   equation (2)

Comparing equation (1) and equation (2):

(5.98 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] = (6 cm)[1 + (12 x 10⁻⁶ °C⁻¹)ΔT]

(5.98 cm/6 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] = [1 + (12 x 10⁻⁶ °C⁻¹)ΔT]

0.9967 + (1.59 x 10⁻⁵ °C⁻¹)ΔT = 1 + (12 x 10⁻⁶ °C⁻¹)ΔT

1 - 0.9967 = [(15.9 -12) x 10⁻⁶ °C⁻¹]ΔT

0.0033/3.9 x 10⁻⁶ °C⁻¹ = ΔT

ΔT = 846.15 °C

but,

ΔT = T' - T = T' - 19°C = 846.15°C

T' = 846.15 °C + 19 °C

T' = 865.15 °C

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

see earlier