Correct option is B
Given:
We have three different configurations of steel wires used for bearing a load:
A: Two wires of 1 mm diameter each, together.
B: One wire of 2 mm diameter.
C: Four wires of 1 mm diameter each, together.
We need to compare their load-bearing capability.
Concept Used:
The load-bearing capacity of a wire is proportional to its cross-sectional area.
The cross-sectional area A of a wire with diameter d is given by:
A = π(d/2)²
This means the load-bearing capability depends on the sum of the cross-sectional areas of the wires in each configuration.
Solution:
Configuration A: Two 1 mm Wires
Each wire has a diameter of 1 mm, so its cross-sectional area is:
A₁ = π(1/2)² = π(1/4) = π/4
For two wires:
Total Area = 2 × (π/4) = π/2
Configuration B: One 2 mm Wire
A single wire with 2 mm diameter has a cross-sectional area:
A₂ = π(2/2)² = π(1) = π
Configuration C: Four 1 mm Wires
Each 1 mm wire has an area of π/4.
For four wires:
Total Area = 4 × (π/4) = π
Comparison of Load-Bearing Capacity:
A (2 wires of 1 mm each) → π/2
B (1 wire of 2 mm) → π
C (4 wires of 1 mm each) → π
Thus, B = C > A.
Final Answer:
(B) A < B = C
