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(B) Given that Young's modulus $Y = 3\eta$,where $\eta$ is the modulus of rigidity.We know the relation $Y = 2\eta(1 + \sigma)$,where $\sigma$ is the

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Infinity

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(B) Given that Young's modulus $Y = 3\eta$,where $\eta$ is the modulus of rigidity.We know the relation $Y = 2\eta(1 + \sigma)$,where $\sigma$ is the Poisson's ratio.Substituting $Y = 3\eta$ into the relation: $3\eta = 2\eta(1 + \sigma) \implies 1.5 = 1 + \sigma \implies \sigma = 0.5$.Now,we use the relation between Young's modulus $Y$,bulk modulus $K$,and Poisson's ratio $\sigma$: $Y = 3K(1 - 2\sigma)$.Substituting $\sigma = 0.5$ into the equation: $Y = 3K(1 - 2(0.5)) = 3K(1 - 1) = 3K(0) = 0$.However,solving for $K$: $K = \frac{Y}{3(1 - 2\sigma)}$.As $\sigma 	o 0.5$,the denominator $3(1 - 2(0.5)) = 0$.Therefore,$K = \frac{Y}{0} = \infty$.

If the Young's modulus of the material is $3$ times its modulus of rigidity,then its volume elasticity (bulk modulus) will be:

Similar Questions

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