Two wire of circular cross-sectionA&B are made of the same metal and have equal length but the resistance of wire A is three times of that of wire B.what is the ratio of their cross-sectional areas?how do their radii compare?
We can use the equation for the resistance of a conductor to find the ratio of their cross-sectional areas:
R = ρ * (L / A)
Where R is the resistance, ρ is the resistivity of the metal, L is the length of the wire, and A is the cross-sectional area. From the problem, we are given that wire A has three times the resistance of wire B. Therefore,
R_A = 3 * R_B
Since both wires are made of the same metal and have the same length, their resistivities and lengths are equal. So, we can rewrite the equation for each wire:
R_A = ρ * (L / A_A)
R_B = ρ * (L / A_B)
Now, we substitute R_A = 3 * R_B and ρ and L cancel out:
3 * R_B = R_B * (A_B / A_A)
Divide by R_B to solve for the ratio of their cross-sectional areas:
3 = A_B / A_A
So the ratio of their cross-sectional areas is 3:1. Now, let's find how their radii compare. The cross-sectional area (A) of a wire with circular cross-section is given by
A = π * r²
where r is the radius. So, we can write the cross-sectional areas of wires A and B as:
A_A = π * r_A²
A_B = π * r_B²
From our previous result, we have:
A_B / A_A = 3
Substitute the expressions for A_A and A_B:
(π * r_B²) / (π * r_A²) = 3
The π values cancel out:
(r_B²) / (r_A²) = 3
Taking the square root of both sides:
r_B / r_A = sqrt(3)
So the ratio of their radii is sqrt(3):1, or approximately 1.73:1.