As shown in the figure, forces of 10^5 ~N each are applied in opposite directions, on the upper and lower faces of a cube of side 10 ~cm , shifting the upper face parallel to itself by 0.5 ~cm . If the side of another cube of the same material is, 20 ~cm then under similar conditions as above, the displacement will be:

Step-by-step solution

For same material the ratio of stress to strain is same For first cube Stress ₁= force ₁ area ₁ = 10^5 (0.1^2 ) For second block, aligned & stress ₂= force ₂ area ₂ = 10^5 (0.2^2 ) & strain ₂= change in length ₂ original length = x 0.2 aligned x is the displacement for second block. For same material, stress ₁ strain ₁ = stress ₂ strain ₂ or, 10.5 (0.1)^2 0.5 10⁻² 0.1 = 10^5 (0.2)^2 x 0.2 Solving we get, x=0.25 ~cm