Question

A cube of side 40 mm has its upper face displaced by 0.1 mm by a tangential force of 8 kN. The shearing modulus of cube is :

• A. $ 2\times {{10}^{9}}\,N{{m}^{-2}} $
• B. $ 4\times {{10}^{9}}\,N{{m}^{-2}} $
• C. $ 8\times {{10}^{9}}\,N{{m}^{-2}} $
• D. $ 16\times {{10}^{9}}\,N{{m}^{-2}} $

Answer 1

Answer: (A)

Shearing modulus of cube $ \eta =\frac{Fl}{A\Delta l}=\frac{8\times {{10}^{3}}\times 40\times {{10}^{-3}}}{(40\times {{10}^{-3}})\times (0.1\times {{10}^{-3}})} $ $ =2\times {{10}^{9}}\,N{{m}^{-2}} $