Two wires have their lengths in the ratio 1:3 diameters in the ratio 2:3 and their materials have specific resistances in the ratio 2:1. Then the ratio of their resistances is :A1:3B3:1C2:3D3:2
Question
Two wires have their lengths in the ratio 1:3 diameters in the ratio 2:3 and their materials have specific resistances in the ratio 2:1. Then the ratio of their resistances is :A1:3B3:1C2:3D3:2
Solution
The resistance (R) of a wire can be calculated using the formula:
R = ρL/A
where:
- R is the resistance,
- ρ is the resistivity (or specific resistance) of the material,
- L is the length of the wire, and
- A is the cross-sectional area of the wire.
The cross-sectional area (A) of a wire can be calculated using the formula:
A = π(d/2)^2
where:
- A is the area, and
- d is the diameter of the wire.
Given that the lengths of the wires are in the ratio 1:3, the diameters are in the ratio 2:3, and the specific resistances are in the ratio 2:1, we can substitute these ratios into the resistance formula to find the ratio of their resistances.
Let's denote the resistances of the two wires as R1 and R2, the lengths as L1 and L2, the diameters as d1 and d2, and the specific resistances as ρ1 and ρ2.
Then we have:
R1/R2 = (ρ1L1/A1) / (ρ2L2/A2)
Substituting the given ratios, we get:
R1/R2 = (21/π(2/2)^2) / (13/π(3/2)^2)
Solving this equation gives:
R1/R2 = 4/9
So, the ratio of their resistances is 4:9. However, this option is not given in your question. Please check the question again.
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