question_answer

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\[10cm\], shifting the upper face parallel to itself by\[0.5cm\]. If the side of another cube of the same material is \[20cm\], then under similar conditions as above, the displacement will be:

[JEE Online 15-04-2018 (II)]

A)  \[1.00\,\,cm\]    

B)       \[0.25\,\,cm\]

C)  \[0.37\,\,cm\]     D)       \[0.75\,\,cm\] Correct Answer: B Solution : [b] For same material the ration of stress to strain is same

For first cube

\[Stres{{s}_{1}}=\frac{{{10}^{5}}}{({{0.1}^{2}})}\]

\[strai{{n}_{1}}=\frac{0.5\times {{10}^{-2}}}{0.1}\]

For second block,

\[stres{{s}_{2}}=\frac{{{10}^{5}}}{{{(0.2)}^{2}}}\]

\[strai{{n}_{2}}=\frac{x}{0.2}\]

where\[x\] is the displacement for second block.

For same material,

\[\frac{stres{{s}_{1}}}{strai{{n}_{1}}}=\frac{stres{{s}_{2}}}{strai{{n}_{2}}}\]

From this \[x=0.25cm\]