Question

A cube of side length \(5\,\text{cm}\) having modulus of rigidity \(10^5\,\text{N/m}^2\). The top is pulled by a force \(10\,\text{N}\) while the bottom is fixed. Find the displacement of the upper surface of cube (in mm).

Answer 1

Correct Answer: 2

Concept: Modulus of rigidity (shear modulus) \[ \eta = \frac{\text{Shearing stress}}{\text{Shearing strain}} \] \[ \eta = \frac{F/A}{x/h} \] where \(F\) = applied force \(A\) = area of surface \(h\) = height of cube \(x\) = displacement of upper surface \includegraphics[width=0.5\linewidth]{24p ans.png} Step 1: Rearrange formula \[ x = \frac{Fh}{A\eta} \] Step 2: Substitute values Side \(= 5\,\text{cm} = 5\times10^{-2}\,\text{m}\) \[ A = (5\times10^{-2})^2 = 25\times10^{-4}\,\text{m}^2 \] \[ h = 5\times10^{-2}\,\text{m} \] \[ x = \frac{10\times 5\times10^{-2}}{25\times10^{-4}\times10^5} \] \[ x = \frac{1}{500}\,\text{m} \] \[ x = 2\times10^{-3}\,\text{m} \] \[ x = 2\,\text{mm} \]