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  • If Y, K and  are the values of Young's modulus, bulk modulus and molculus of rigidity of any material respectively. Choose the correct relation for these paramet

If Y, K and \eta are the values of Young's modulus, bulk modulus and molculus of rigidity of any material respectively. Choose the correct relation for these parameters.
Option: 1 \eta = \frac{3YK}{9K +Y}N/m^2
Option: 2 Y = \frac{9K\eta }{2 \eta + 3K}N/m^2
Option: 3 K = \frac{Y \eta }{9 \eta -3Y}N/m^2
Option: 4 Y = \frac{9K\eta }{3K- \eta}N/m^2

Answers (1)

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\\ \text{Let}\ \alpha =\text{longitudinal strain per unit stress}=\frac{\Delta l}{l}\\ And \ \beta =\text{lateral strain per unit stress.}=\frac{\Delta d}{d}\\ \\ \text{Then Poisson's ratio,}\ \sigma=\frac{\text { lateral strain }}{\text { longitudinal strain }}=\frac{\beta}{\alpha}\\\\ \text{ Young's modulus,}\ Y=\frac{\text { longitudinal stress }}{\text { longitudinal strain }}=\frac{1}{\frac{\text{longitudinal stress}}{\text{longitudinal strain}}}=\frac{1}{\alpha} \\ \\ \text{And Bulk modulus,}\ K=\frac{\text { hydrostatic strain }}{\text { volumetric strain }}=\frac{1}{3(\alpha-2 \beta)}

\\ K=\frac{1}{3(\alpha-2 \beta)}=\frac{1}{3 \alpha\left(1-2 \frac{\beta}{\alpha}\right)}\\ \\ \text{Since}\ \sigma=\frac{\beta}{\alpha} \ and\ Y=\frac{1}{\alpha},\ \text{so by substituting these values in above equation, we get} \\ \\ K=\frac{Y}{3(1-2 \sigma)} \cdots \cdots(1) \\ \text{Now the options are given in terms of Y, K and modulus of rigidity}\ (\eta). \\ \text{For this we can use this formula}\\ Y=2 \eta(1+\sigma) \cdots \cdots(2)

\\ \text{From equation( 1)}\\ 1-2 \sigma=\frac{Y}{3 K}\\ \Rightarrow 2 \sigma=1-\frac{Y}{3 K} \\ \Rightarrow \sigma=\frac{1}{2}\left(1-\frac{Y}{3 K}\right) \cdots \cdots(3) \\ \text{Now from equation (2)} \\ \sigma=\frac{Y}{2 \eta}-1 \cdots \cdots(4)

\\ \text{By comparing equations (3) and (4), we have}\\ \\ \frac{1}{2}\left(1-\frac{Y}{3 K}\right)=\frac{Y}{2 \eta}-1 \\ \\ \Rightarrow \frac{3}{2}-\frac{Y}{6 K}=\frac{Y}{2 \eta}\\ \\ \Rightarrow 3-\frac{Y}{3 K}=\frac{Y}{\eta}\\ \\ \text{On further solving, we get}\\ 3=\frac{Y}{\eta}+\frac{Y}{3 K} \\ \\ \Rightarrow \frac{Y}{3K}=3- \frac{Y}{\eta}\\ \\ \Rightarrow \frac{Y}{3K}=\frac{3 \eta-Y}{\eta}\\ \\ \Rightarrow K=\frac{Y \eta}{9 \eta-3Y}

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avinash.dongre

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