Question

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 :

• A. 0.25 cm
• B. 0.37 cm
• C. 0.75 cm
• D. 1.00 cm

Answer 1

Answer: (A)

Given: Force applied $F=10^{3}N;$
side of cube $x=10m=10\times 10^{-2}m;$
shift $dx=0.5cm=0.5\times 10^{-2}m$

On cube of side $x=10cm:$
Stress $=\frac{F}{x^2}=\frac{10^5}{{\left(10\times 10^{-2}\right)}^2}=10^{7}N/m^2$
Strain $=\frac{dx}{x}=\frac{0.5\times 10^{-2}}{10\times 10^{-2}}=0.5$
On cube of side $x^{\prime }=20cm$ :
Stress ${}^{\prime }=\frac{F}{x^{{}^{\prime }2}}=\frac{10^5}{{\left(20\times 10^{-2}\right)}^2}$
Strain' $=\frac{d{x}^{\prime }}{x^{\prime }}=\frac{d{x}^{\prime }}{20\times 10^{-2}}$
Now it is given that the material is same for both cubes, therefore,
$\frac{\text{ Stress }}{\text{ Strain }}=\frac{{\text{ Stress’ }}^{\prime }}{{\text{ Strain }}^{\prime }}$
$\Rightarrow \frac{10^7}{0.05}=\frac{10^5}{\left(20\times 10^{-2}\right)}\times \frac{20\times 10^{-2}}{d{x}^{\prime }}$
$\Rightarrow \frac{10^7\times 10^2}{5}=\frac{10^5}{20\times 10^{-2}}\frac{1}{d{x}^{\prime }}$
$\Rightarrow d{x}^{\prime }=\frac{10^5}{20\times 10^{-2}}\times \frac{5}{10^7\times 10^2}=\frac{5\times 10^{-2}}{20}=0.25\times 10^{-2}m$
$\Rightarrow d{x}^{\prime }=0.25cm$