A rubber cube of side $5 \mathrm{~cm}$ has one face fixed, while a tangential force $1800 \mathrm{~N}$ is…

A rubber cube of side $5 \mathrm{~cm}$ has one face fixed, while a tangential force $1800 \mathrm{~N}$ is applied on its opposite face. If modulus of rigidity of rubber is $2.4 \times 10^6 \mathrm{Nm}^{-2}$, then the lateral displacement of the strained face is ________
  1. 3 mm
  2. 5 mm
  3. 15 mm
  4. 1.5 mm

Solution

Shear stress $ =\frac{\text { Tangential force }}{\text { Area }}=\frac{1800 \mathrm{~N}}{(0.05)^2} $ Shear strain $=\tan \theta$ $ \begin{aligned} & =\frac{x}{h}=\frac{\text { Lateral displacement }}{\text { Height of cube }}=\frac{x}{5} \mathrm{~cm} \\ & =\frac{x \mathrm{~m}}{0.05 \mathrm{~m}} \end{aligned} $ Modulus of rigidity, $ \eta=\frac{\text { Shear stress }}{\text { Shear strain }} $ $ 2.4 \times 10^6 \frac{\mathrm{N}}{\mathrm{m}^2}=\frac{1800}{25 \times 10^{-4}} \times \frac{0.05}{x} \mathrm{~N} / \mathrm{m}^2 $ or $ \begin{aligned} x & =\frac{1800 \times 0.05}{25 \times 10^{-4} \times 2.4 \times 10^6} \\ & =15 \times 10^{-3} \mathrm{~m} \\ & =15 \mathrm{~mm} \end{aligned} $

Asked in: AP EAMCET 2018 (23 Apr Shift 1)

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