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 $10\, cm$, 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 :

• 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 $x=10 m =10 \times 10^{-2} \,m ;$
shift $d x=0.5 cm =0.5 \times 10^{-2}\, m$

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