A rigid bar of mass \[M\] is supported symmetrically by three wires each of length one. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to.
A. $\sqrt {\dfrac{{{Y_{copper}}}}{{{Y_{iron}}}}} $
B. $\sqrt {\dfrac{{{Y_{iron}}}}{{{Y_{copper}}}}} $
C. $\dfrac{{{Y^2}_{iron}}}{{{Y^2}_{copper}}}$
D. $\dfrac{{{Y_{iron}}}}{{{Y_{copper}}}}$
Answer
526.8k+ views
Hint: The ratio of applied force F to a cross-section area is known as the stress which is described as "force per unit area". Tensile stress is defined as the stress that contributes to stretch or lengthen the material this acts normal to the stressed area
Compressive stress is defined as the stress that contributes to compress or shorten the material that acts normal to the stressed area.
Complete step by step solution:
The force acting on the unit area of a material is called stress. The stress effect on the body is known as the strain. Deformation of the body can be done by stress. The experienced force by a material is measured using the stress units. Stress is classified into three categories based on the direction of the deforming forces acting on the body.
Let's consider T is the tension of each wire, hence the bar is supported symmetrically by the three wires here the extension in the wire is the same,
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}}$
When \[D\] is the diameter of the wire then,
$Y = \dfrac{{\dfrac{F}{{\pi {{\left( {\dfrac{D}{2}} \right)}^2}}}}}{{\dfrac{{\Delta L}}{L}}}$
After simplification,
$Y = \dfrac{{4FL}}{{\pi {D^2}\Delta L}}$
From the conditions of the problem tension, length, extension is the same for each wire,
$Y \propto \dfrac{1}{{{D^2}}}$
$D \propto \sqrt {\dfrac{1}{Y}} $
From this,
$\dfrac{{{D_{copper}}}}{{{D_{iron}}}} = \sqrt {\dfrac{{{Y_{iron}}}}{{{Y_{copper}}}}} $
So, the correct answer is, option (B).
Note:
"Deformation of a solid because of stress" is called a strain.
A contraction of a line segment is known as the normal strain.
A change in angle between two line segments originally perpendicular is known as the shear strain.
Compressive stress is defined as the stress that contributes to compress or shorten the material that acts normal to the stressed area.
Complete step by step solution:
The force acting on the unit area of a material is called stress. The stress effect on the body is known as the strain. Deformation of the body can be done by stress. The experienced force by a material is measured using the stress units. Stress is classified into three categories based on the direction of the deforming forces acting on the body.
Let's consider T is the tension of each wire, hence the bar is supported symmetrically by the three wires here the extension in the wire is the same,
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}}$
When \[D\] is the diameter of the wire then,
$Y = \dfrac{{\dfrac{F}{{\pi {{\left( {\dfrac{D}{2}} \right)}^2}}}}}{{\dfrac{{\Delta L}}{L}}}$
After simplification,
$Y = \dfrac{{4FL}}{{\pi {D^2}\Delta L}}$
From the conditions of the problem tension, length, extension is the same for each wire,
$Y \propto \dfrac{1}{{{D^2}}}$
$D \propto \sqrt {\dfrac{1}{Y}} $
From this,
$\dfrac{{{D_{copper}}}}{{{D_{iron}}}} = \sqrt {\dfrac{{{Y_{iron}}}}{{{Y_{copper}}}}} $
So, the correct answer is, option (B).
Note:
"Deformation of a solid because of stress" is called a strain.
A contraction of a line segment is known as the normal strain.
A change in angle between two line segments originally perpendicular is known as the shear strain.
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