As shown in the figure, forces of $10^5 \mathrm{~N}$ each are applied in opposite directions, on the upper…
As shown in the figure, forces of $10^5 \mathrm{~N}$ each are applied in opposite directions, on the upper and lower faces of a cube of side $10 \mathrm{~cm}$, shifting the upper face parallel to itself by $0.5 \mathrm{~cm}$. If the side of another cube of the same material is, $20 \mathrm{~cm}$ then under similar conditions as above, the displacement will be:
$1.00 \mathrm{~cm}$
$0.25 \mathrm{~cm}$
$0.37 \mathrm{~cm}$
$0.75 \mathrm{~cm}$
Solution
For same material the ratio of stress to strain is same For first cube
$
\text { Stress }_1=\frac{\text { force }_1}{\text { area }_1}=\frac{10^5}{\left(0.1^2\right)}
$
For second block,
$
\begin{aligned}
&\text { stress }_2=\frac{\text { force }_2}{\text { area }_2}=\frac{10^5}{\left(0.2^2\right)} \\
&\text { strain }_2=\frac{\text { change in length }{ }_2}{\text { original length }}=\frac{x}{0.2}
\end{aligned}
$
$x$ is the displacement for second block.
For same material, $\frac{\text { stress }_1}{\text { strain }_1}=\frac{\text { stress }_2}{\text { strain }_2}$
$
\text { or, } \frac{\frac{10.5}{(0.1)^2}}{\frac{0.5 \times 10^{-2}}{0.1}}=\frac{\frac{10^5}{(0.2)^2}}{\frac{x}{0.2}}
$
$
\text { Solving we get, } x=0.25 \mathrm{~cm}
$
