MHT CET · Physics · Mechanical Properties of Solids
A thick brass wire of length ' \(\mathrm{L}\) ' and density ' \(\rho\) ' is suspended from rigid support.
Due to its own weight, \(' \ell^{\prime}\) is the increase in length. Young's modulus ' \(\mathrm{Y}^{\prime}\) of brass
wire in terms of density is
\((\mathrm{g}=\) acceleration due to gravity \()\)
- A \(\mathrm{Y}=\frac{\rho \mathrm{gL}}{4 \ell}\)
- B \(\mathrm{Y}=\frac{\rho \mathrm{gL}^{2}}{4 \ell}\)
- C \(\mathrm{Y}=\frac{\rho \mathrm{gL}}{3 \ell}\)
- D \(\mathrm{Y}=\frac{\rho \mathrm{gL}^{2}}{\ell}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{Y}=\frac{\rho \mathrm{gL}^{2}}{\ell}\)
Step-by-step Solution
Detailed explanation
\(Y=(F / A) \cdot(L / \Delta L) \quad(\because \Delta L=l)\)
\(=\frac{M g L}{A \Delta L}\)
\(=\frac{\rho g L^{2}}{l}$$(\because M=\rho A l)\)
\(\because\) (option 4 incorrect)
\(=\frac{M g L}{A \Delta L}\)
\(=\frac{\rho g L^{2}}{l}$$(\because M=\rho A l)\)
\(\because\) (option 4 incorrect)
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