Question

Which statement is true for a metal?
A. $Y < \eta $
B. $Y = \eta $
C. $Y > \eta $
D. $Y < \dfrac{1}{\eta }$

Answer 1

Hint:This problem is based on the Mechanical properties of solids, we know that young’s modulus and modulus of rigidity play a significant role in establishing a relationship between different parameters hence, in order to explain the solution, carefully consider each alternative and evaluate which one seems to be most relevant.



Formula used:
$Young's\,Modulus\,(Y) = \dfrac{{Stress(\sigma )}}{{Strain( \in )}}$
$Modulus{\text{ }}of{\text{ }}Rigidity(\eta ) = \dfrac{{ShearStress({\sigma _s})}}{{ShearStrain(\theta )}}$
$Y = 2\eta \left( {1 + \sigma } \right)$ where, $\sigma $is Poisson’s ratio.




Complete answer:
We know that the formula of young’s modulus;
$Young's\,Modulus\,(Y) = \dfrac{{Stress(\sigma )}}{{Strain( \in )}}$
 Here, we have:
 $Stress(\sigma ) = \dfrac{{Force}}{{Area}} \\$
 $= \dfrac{Weight}{Area} $
 $= \dfrac{mg}{πr^2} $
And, $Strain( \in ) = \dfrac{{\Delta l}}{l}$
Similarly, we consider the formula of modulus of rigidity; $Modulus{\text{ }}of{\text{ }}Rigidity(\eta ) = \dfrac{{ShearStress({\sigma _s})}}{{ShearStrain(\theta )}}$
Where, $ShearStress({\sigma _s}) = \dfrac{{Force}}{{Area}}$
And, $ShearStrain = \theta = \dfrac{{\Delta x}}{l}$
We also know that there is a relationship between Young’s Modulus $Y$and modulus of rigidity $\eta $, and it can be given as: -
For Metals,
$Y = 2\eta \left( {1 + \sigma } \right)$
Here, $\sigma $is the Poisson’s ratio
It can be clearly observed from the above expression that $\eta $ (modulus of rigidity) is always less than $Y$(Young’s modulus) for metals.
Thus, $Y > \eta $ is true for metals.
Hence, the correct option is (C) $Y > \eta $.



Note: Young's modulus of a material is essential to know to determine how a material would react when subjected to a force. When the application of a shear force causes lateral deformation, the elastic coefficient is the modulus of rigidity. It allows us to estimate how rigid a body is.