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(C) For a unit cube (side length $L = 1$),the tensile stress on each face is $\sigma_{stress} = \frac{F}{A} = \frac{F}{1^2} = F$.The longitudinal stra

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$\frac{F(1 - 2\sigma)}{Y}$

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(C) For a unit cube (side length $L = 1$),the tensile stress on each face is $\sigma_{stress} = \frac{F}{A} = \frac{F}{1^2} = F$.The longitudinal strain along one axis due to the force on that axis is $\epsilon_{long} = \frac{F}{Y}$.Due to the tensile forces on the other two perpendicular faces,there will be lateral strains. The lateral strain due to one perpendicular force is $-\sigma \epsilon_{long} = -\sigma \frac{F}{Y}$.Since there are two such perpendicular faces,the total lateral strain is $2 	imes (-\sigma \frac{F}{Y}) = -\frac{2\sigma F}{Y}$.The net strain (change in length $\Delta L$ for $L=1$) is the sum of the longitudinal strain and the lateral strains:$\Delta L = \frac{F}{Y} - \frac{2\sigma F}{Y} = \frac{F(1 - 2\sigma)}{Y}$.

On all the six surfaces of a unit cube,an equal tensile force of $F$ is applied. The increase in length of each side will be ($Y =$ Young's modulus,$\sigma =$ Poisson's ratio).

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