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Mechanical Properties of Solids — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMechanical Properties of Solids1 Mark
MCQ
A rigid bar of mass $M$ is supported symmetrically by three wires each of length $l.$ 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
    $\text{Y}_\text{copper}/\text{Y}_\text{iron}.$
  • ✓
    $\sqrt{\frac{\text{Y}_\text{iron}}{\text{Y}_\text{copper}}}.$
  • C
    $\frac{\text{Y}^2_\text{iron}}{\text{Y}^2_\text{copper}}.$
  • D
    $\frac{\text{Y}_\text{iron}}{\text{Y}_\text{copper}}.$
✓

Answer

Correct option: B.
$\sqrt{\frac{\text{Y}_\text{iron}}{\text{Y}_\text{copper}}}.$
As the bar is supported symmetrically by the three wires, therefore extension in each wire is same. Let $T$ be the tension in each wire and diameter of the wire is $D,$ then Young’s modulus is
$(\text{Y})=\frac{\text{Stress}}{\text{Strain}}=\frac{\text{F/A}}{\Delta\text{L}/\text{L}}=\frac{\text{F}}{\text{A}}\times\frac{\text{L}}{\Delta\text{L}}$
$=\frac{\text{F}}{\pi(\text{D/2})^2}\times\frac{\text{L}}{\Delta\text{L}}=\frac{4\text{FL}}{\pi\text{D}^2\Delta\text{L}}$
$\Rightarrow\ \text{D}^2=\frac{4\text{FL}}{\pi\Delta\text{LY}}\Rightarrow\ \text{D}=\sqrt{\frac{4\text{FL}}{\pi\Delta\text{LY}}}$
As $F$ and $\frac{\text{L}}{\Delta\text{L}}$ are constants.
Hence, $\text{D}\propto\sqrt{\frac{1}{\text{Y}}}$
or $\text{D}=\frac{\text{K}}{\sqrt{\text{Y}}} (K$ is the proportionality constant$)$
Now we can find ratio as $\frac{\text{D}_\text{copper}}{\text{D}_\text{iron}}=\sqrt{\frac{\text{Y}_\text{iron}}{\text{Y}_\text{copper}}}$

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