An rod is 1.58m long at 0°c.what must be the length of a brass rod at 0°c if the difference between the lengths of the two rods is to remain the same at all temperatures.linear expansivity of iron =1.2×10^-5 k^-1 and brass =1.9×10^-5 k^-1
To find the length of the brass rod at 0°C, we can use the concept of linear expansivity and the given information.
First, let's consider the change in length for both the iron and brass rods when the temperature changes from 0°C to a general temperature (let's call it T°C).
For the iron rod:
ΔL_iron = α_iron * L_iron * ΔT
where ΔL_iron is the change in length for the iron rod, α_iron is the linear expansivity of iron, L_iron is the initial length of the iron rod, and ΔT is the change in temperature.
Similarly, for the brass rod:
ΔL_brass = α_brass * L_brass * ΔT
where ΔL_brass is the change in length for the brass rod, α_brass is the linear expansivity of brass, and L_brass is the initial length of the brass rod.
Now, we need to find the ratio of the changes in length for both rods when the temperature changes from 0°C to T°C. This ratio should remain constant for all temperatures, as stated in the problem.
ΔL_iron / ΔL_brass = (α_iron * L_iron * ΔT) / (α_brass * L_brass * ΔT)
Note that ΔT cancels out.
Simplifying the equation, we have:
ΔL_iron / ΔL_brass = α_iron / α_brass
Also, as given in the problem, ΔL_iron = ΔL_brass.
Therefore,
(α_iron * L_iron) / (α_brass * L_brass) = 1
Now, we can substitute the values given in the problem to find the length of the brass rod.
(1.2×10^-5) * 1.58 / (1.9×10^-5) * L_brass = 1
Simplifying the equation, we find:
L_brass = (1.2×10^-5) * 1.58 / (1.9×10^-5) ≈ 0.9947 m
Therefore, the length of the brass rod at 0°C must be approximately 0.9947 meters in order to keep the difference in length between the two rods constant at all temperatures.