Question

A square lead slab of side 50 cm and thinness 10 cm is subject to ashearing force (on its narrow face) of $4\times {10}^{4}N$ as shown. Lower edge is $\eta$ is ${10}^{1}0N/m^2$, then upper edge is displaced by

• A. 0.1 mm
• B. 0.02 mm
• C. 0.04 mm
• D. 0.004 mm

Answer 1

The correct option is C 0.04 mm

$\begin{array}{c}\text{Shear force}=\eta \text{( shear strain)}\\ \begin{array}{cc}\frac{F_{11}}{A}=& \eta \left(\frac{\Delta l}{l}\right)\\ \frac{4\times {10}^4}{500\times {10}^{-4}}& ={10}^{10}\left(\frac{x}{50\times {10}^{-2}}\right)\\ x& =4\times {10}^{-5}m\\ \therefore \text{Elongation on the ether side}0.04mm& \end{array}\end{array}$