Question

Two parallel forces of 4000 N are applied tangentially in opposite direction to the opposite faces of metallic cube of side length 0.25 cm. The shear modulus of the material is $80\times {10}^{9}Pa$. Then the displacement of the upper surface relative to the lower surface is

• A. $2\times {10}^{-5}m$
• B. $2\times {10}^{-6}m$
• C. $2\times {10}^{-7}m$
• D. $2\times {10}^{-8}m$

Answer 1

The correct option is C $2\times {10}^{-7}m$

$\begin{array}{c}\text{Given that}\\ \text{side of cubic material}=0.25cm\\ \text{shear modulus}=80\times {10}^{9}Pa\\ \text{two parallel forces}=4000N\\ \text{-Let lower surface is fixed}\\ \text{and displacement is}d\text{.}\end{array}$

$\begin{array}{c}\text{We know}\\ \text{shear modulus}x\text{sheared angle}=\text{shear stress}\\ \begin{array}{c}(80\times {10}^9)\times \varphi =\frac{\text{Force}}{\text{Area}}\\ 80\times {10}^9\times \frac{d}{0.25}=\frac{4000}{(0.25{)}^2}\\ d=2\times {10}^{-7}m\end{array}\end{array}$

Hence option C is correct