Question

A metal cube of side length $8.0cm$ has its upper surface displaced with respect to the bottom by $0.10mm$ when a tangential force of 4 x 10${}^9$ N is applied at the top with bottom surface fixed. The rigidity modulus of the material of the cube is :

• A. $4\times {10}^{9}N/m^2$
• B. $5\times {10}^{14}N/m^2$
• C. $8\times {10}^{9}N/m^2$
• D. $1\times {10}^{9}N/m^2$

Answer 1

The correct option is B $5\times {10}^{14}N/m^2$

Given,

Side length, $L=8cm$

$\Delta L=0.1mm$

Force, $F=4\times {10}^{9}N$

Shearing strain $=\frac{0.01}{8}$

Stress $=\frac{4\times {10}^9}{8\times 8\times {10}^{-4}}N/m^2$

Rigidity Modulus, $y=\frac{\frac{F}{A}}{\frac{\Delta X}{X}}=\frac{\frac{4\times {10}^9}{9\times 8\times {10}^{-4}}}{\frac{0.01}{8}}$

$y=\frac{1}{2}\times {10}^{14}N/m^2$

$=5\times {10}^{14}N/m^2$