If the lengths of two bars of different solids are inversely proportional to their respective coefficients of linear expansion at the same initial temperature, the difference in length between them will be constant at all temperatures. That means one way to construct a device with two points whose separation remains the same in spite of temperature changes is to bolt together one end of two bars having different coefficients of linear expansion as shown below. What should be the lengths of a steel (α = 11 x 10-6 / C°) and a brass (α = 19 x 10-6 / C°) bar at 0°C so that at all temperatures their difference in length is 0.30 m
If the lengths of two bars of different solids are inversely proportional to their respective coefficients of linear expansion at the same initial temperature, the difference in length between them will be constant at all temperatures. That means one way to construct a device with two points whose separation remains the same in spite of temperature changes is to bolt together one end of two bars having different coefficients of linear expansion as shown below. What should be the lengths of a steel (α = 11 x 10-6 / C°) and a brass (α = 19 x 10-6 / C°) bar at 0°C so that at all temperatures their difference in length is 0.30 m
Related questions
Question
. If the lengths of two bars of different solids are inversely proportional to their respective coefficients of linear expansion at the same initial temperature, the difference in length between them will be constant at all temperatures. That means one way to construct a device with two points whose separation remains the same in spite of temperature changes is to bolt together one end of two bars having different coefficients of linear expansion as shown below. What should be the lengths of a steel (α = 11 x 10-6 / C°) and a brass (α = 19 x 10-6 / C°) bar at 0°C so that at all temperatures their difference in length is 0.30 m?
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 2 images