Question

An iron ball has a diameter of 12cm and is 0.02 cm too large to pass through a hole in a brass plate when the ball and plate are at the same temperature of 25°C.At what temperature the same for both ball and plate would the ball just pass through the hole.

(linear expansivties for iron and brass are 1.32×10^-5 K ^-1 and 2.0×10^-5 K^-1​

Answer 1

270°C

Step-by-step explanation:

Since αB > αI

We want LB - LI = 0.02 cm at a given temperature t

But

ΔLB= LBαBΔt

ΔLI = LIαIΔt

Where;

LI = length of the iron

LB= length of he brass

αB = Linear expansivity of brass

αI = Linear expansivity of iron

Δt = Change in temperature

ΔLB - ΔLI = (LBαB - LIαI)Δt = 0.02

LB = LI = 12 cm

Δt = 0.02/12( 2.0×10^-5 - 1.32×10^-5)

Δt = 0.02/12(0.68 ×10^-5)

Δt = 245°C

25°C + 245°C = 270°C