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 :
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 \[x=0.25cm\]
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