A wire whose cross-sectional area is 2 mm² is stretched by 0.1 mm by a certain load, and if a similar wire of triple the area of cross-section is stretched by the same load, then the elongation of the second wire would be
Cross-sectional refers to the shape or area of an object when it is cut perpendicular to its length. In physics, it is used to describe the area through which forces, fluids, or electrical currents flow. It is important in calculating stress, strain, and other material properties in solid mechanics.
Chapter 8 of Class 11 Physics discusses mechanical properties of solids including stress strain and Hooke’s law. It covers Young’s modulus bulk modulus shear modulus and elastic potential energy. The chapter explores Poisson’s ratio and its significance in material behavior. These concepts are essential for understanding deformation and the mechanical response of solids to external forces.
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The formula for the elongation of a wire is as follows:
ΔL = (F L) / (A Y)
Where:
ΔL is the elongation,
F is the force,
L is the original length of the wire,
A is the cross-sectional area,
Y is Young’s Modulus.
For the first wire, the elongation is given by:
ΔL1 = (F L) / (A1 Y)
For the second wire, the elongation is given by:
ΔL2 = (F L) / (A2 Y)
Since the area of the second wire is three times that of the first wire, A2 = 3 A1, we can compare the elongations:
ΔL2 / ΔL1 = A1 / A2 = 1 / 3
Thus, the elongation of the second wire is one-third that of the first wire:
ΔL2 = ΔL1 / 3 = 0.1 mm / 3 = 0.033 mm
The correct answer is:
0.033 mm
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