There are two wires of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wire by applying same load will be
Load refers to the external force or combination of forces applied to an object or structure. It can cause deformation or motion depending on the type and magnitude of the load. Loads are classified as static dynamic or impact and are crucial in designing structures and machines for safety and efficiency.
Class 11 Physics Chapter 8 focuses on mechanical properties of solids covering stress strain relationships Young’s modulus bulk modulus and shear modulus. It explores elasticity Hooke’s law factors affecting elasticity and their applications. These concepts are vital for understanding material behavior and designing structures in engineering and construction effectively.
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In formula format:
ΔL = (F × L) / (A × Y)
Where:
– F= applied load which is the same for both the wires
– L= length of wire which is the same for both
– A= cross sectional area of the wire A = πd² / 4
– Y = Young’s modulus (same material, so constant for both wires)
The extension is **inversely proportional to the cross-sectional area (A)
Step 1: Calculate the area ratio
For the first wire (diameter = d):
A₁ = (π × d²) / 4
For the second wire (diameter = 2d):
A₂ = (π × (2d)²) / 4 = (π × 4d²) / 4 = πd²
Area ratio is:
A₁ : A₂ = (πd² / 4) : πd² = 1 : 4
Step 2: Ratio of extensions
Since extension (ΔL) is inversely proportional to the area:
ΔL₁ : ΔL₂ = A₂ : A₁ = 4 : 1
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