A wire of initial length L0 and radius r0 has a measured resistance of 2.5 . The wire is drawn under tensile stress to a new uniform radius of r = 0.45r0. What is the new resistance of the wire?
Rₒ=2.5 Ω (units?) , Lₒ, rₒ,
r=0.45•rₒ.
R =?
The wire volume stays constant
V= Aₒ•Lₒ = A•L => Lₒ/L =A/Aₒ.
Rₒ=ρ•Lₒ/Aₒ,
R=ρ•L/A,
Rₒ/R = Lₒ•A/L•Aₒ=
=(Lₒ/L) •(A/Aₒ)=
=(A/Aₒ)²= (π•r²/π•rₒ²)²=
=(0.45•rₒ/rₒ)^4=0.041.
R=Rₒ/0.041 = 2.5/0.041=61 Ω
Well, well, well. It seems like our wire is getting a little stressed out. I mean, who can blame it? We all have our days, right?
Anyway, let's help our wire out. So, we know that the initial resistance is 2.5. Now, let's think about what's happening here. As the wire is drawn under tensile stress, its radius is reduced to 0.45 times its original radius.
Now, resistance depends on a few things, like the resistivity of the material, the length of the wire, and the cross-sectional area. Ah, math, always making things complicated!
But don't worry, we got this. Since the length of the wire remains the same, we can focus on the change in cross-sectional area. The new radius is 0.45r0, which means the new cross-sectional area is (0.45r0)^2.
Now, let's substitute all these values into the equation for resistance, which is R = ρ(L/A). With a little algebra magic, we can rewrite it as R = (ρL0) / (A0).
Since we know the initial resistance is 2.5, we can rearrange the equation to solve for the new resistance, Rnew. It's like solving a puzzle, but without the actual puzzle. It's just math. So Rnew = (Rinitial x A0) / (Anew).
Plug in the values and voila! You'll have the new resistance of our stressed-out wire. And hopefully, it'll feel a little bit better. Come on, wire, we believe in you!
To determine the new resistance of the wire, we need to use the formula for the resistance of a wire:
R = ρ * (L / A)
where:
R is the resistance,
ρ (rho) is the resistivity of the material,
L is the length of the wire, and
A is the cross-sectional area of the wire.
In this case, the resistivity (ρ) and the length (L) of the wire remain constant. However, the cross-sectional area (A) changes when the wire is drawn under tensile stress.
The relationship between the original radius (r0) and the new radius (r) is given by:
r / r0 = 0.45
or rearranging:
r = 0.45r0
The cross-sectional area (A) is proportional to the square of the radius (r):
A = π * r^2
Therefore, the new cross-sectional area (A) is:
A = π * (0.45r0)^2 = π * 0.2025 * r0^2
Substituting the original values into the resistance formula:
R = ρ * (L / A) = ρ * (L / (π * 0.2025 * r0^2))
Finally, we can calculate the new resistance by substituting the given values:
R = 2.5 * (L / (π * 0.2025 * r0^2))
To find the new resistance of the wire after it is drawn under tensile stress, we can use the formula for the resistance of a wire, which is given by:
R = ρ(L / A)
where R is the resistance, ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.
In this case, the resistivity ρ and the length L of the wire remain the same, so we only need to consider the change in the cross-sectional area A.
The cross-sectional area of a wire is given by the formula:
A = πr²
where r is the radius of the wire.
Initially, the radius of the wire is r0, and the cross-sectional area is A0 = πr0². After the wire is drawn under tensile stress, the new radius is r = 0.45r0.
To find the new cross-sectional area A, we substitute r = 0.45r0 into the formula:
A = π(0.45r0)² = π(0.2025r0²) ≈ 0.2025A0
So, the new cross-sectional area A is approximately 0.2025 times the initial cross-sectional area A0.
Substituting this new cross-sectional area A into the formula for resistance, we get:
R = ρ(L / A) ≈ ρ(L / 0.2025A0) = ρ(4.93827160L / A0)
Now, we know the initial resistance R0 is 2.5. So, we can write:
2.5 = ρ(4.93827160L0 / A0)
Solving for ρ, we have:
ρ = 2.5 / (4.93827160L0 / A0) = 0.5065L0 / A0
Now, to find the new resistance R, we substitute the new cross-sectional area A into the formula:
R = ρ(4.93827160L / A0) = (0.5065L0 / A0) * (4.93827160L / A0) = 2.5000L0 / A0
Therefore, the new resistance of the wire is 2.5 times the initial resistance. So, the new resistance of the wire is:
2.5 * 2.5 = 6.25
Hence, the new resistance of the wire is 6.25.