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Question: \(Y, K,\eta\) represent respectively the young’s modulus, bulk modulus, and rigidity modulus of a bo...

Y,K,ηY, K,\eta represent respectively the young’s modulus, bulk modulus, and rigidity modulus of a body. If the young’s modulus is twice the bulk modulus then the relation between η,K\eta, K
A.η=6K7 B.η=7K9 C.η=9K7 D.η=18K7 \begin{aligned} & A.\eta =\dfrac{6K}{7} \\\ & B.\eta =\dfrac{7K}{9} \\\ & C.\eta =\dfrac{9K}{7} \\\ & D.\eta =\dfrac{18K}{7} \\\ \end{aligned}

Explanation

Solution

Hint: The relation between the Young's modulus, bulk modulus, and rigidity modulus is used to find the relation between η,K\eta, K when young’s modulus is twice the bulk modulus

Step by step solution:
We need to consider the relationship between the three modulus and then find the relation between η,K\eta, K. We know the relationship is given by :

9Y=1K+3η\dfrac{9}{Y} = \dfrac{1}{K}+\dfrac{3}{\eta}

Now we are already given that Y=2×KY = 2 \times K. So substitute the value of Y in the above expression

92×K=1K+3η\dfrac{9}{2 \times K} = \dfrac{1}{K}+\dfrac{3}{\eta}

We see that now we have got the equation in terms of η,K\eta, K so now we can manipulate the above equation to get the value of η\eta in terms of KK we get

92×K1K=3η\Rightarrow \dfrac{9}{2 \times K} - \dfrac{1}{K} = \dfrac{3}{\eta}
72K=3η\Rightarrow \dfrac{7}{2K} =\dfrac{3}{\eta}
6K=7η\Rightarrow 6K = 7\eta
η=6K7\Rightarrow \eta = \dfrac{6K}{7}

In conclusion, we have found the relation between the η\eta and K using the relation between the rigidity modulus, bulk modulus, and the Young’s modulus. The relation between the η\eta and K value is found as η=6K7\eta = \dfrac{6K}{7}

Additional Information:
Young's modulus can be expressed in terms of Poisson's ratio and bulk modulus as

Y=3K(12μ)Y = 3K(1-2\mu)

Similarly, the young's modulus can be expressed in terms of Y=2η(1+2μ)Y = 2\eta(1+2\mu)

Now combining both the equations we get the relation between the young's modulus and bulk modulus and rigidity modulus as Y=9Kηη+3KY = \dfrac{9 K \eta}{\eta + 3K }

From the above relation, we can express the three constants as we used in the solution
9Y=1K+3η\dfrac{9}{Y} = \dfrac{1}{K}+\dfrac{3}{\eta}

Note: The possible mistake that can be done in this kind of problem is that one may start deriving the relationship between the given constants which consumes much time or one may substitute relation but will not find the way to get the relation between the required constants. We need to substitute the given values of the variables and then find the relation between the required constants by manipulating the LHS and RHS.