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2 wires of equal length are suspended.$Y_1=2\times 10^{11}N/m^2$ and $Y_2=3\times 10^{11}N/m^2$.Find equivalent young's modulus of combination,$Y$

$\begin{array}{1 1}(A)\;4\times 10^{11}N/m^2\\(B)\;5\times 10^{11}N/m^2\\(C)\;1\times 10^{11}N/m^2\\(D)\;8\times 10^{11}N/m^2\end{array} $

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Spring constant $K=\large\frac{F}{\Delta l}=\frac{Ya}{l}$
Equivalent K in parallel combination is written as
$K_{eq}=K_1+K_2$
$\large\frac{Y\times 2A}{l}=\frac{Y_1\times A}{l}+\frac{Y_2\times 2A}{l}$
$Y=\large\frac{1}{2}$$(2\times 10^{11}+3\times 10^{11}\times 2)$
$\;\;\;=4\times 10^{11}N/m^2$
Hence (A) is the correct answer.
answered May 8, 2014 by sreemathi.v
 

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