Question

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

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

Answer 1

The correct option is B $5\times {10}^{9}N/m^2$

Rigidity Modulus $=\frac{F/A}{Strain}$

Now, $Strain=\frac{0.1\times {10}^{-3}}{8\times {10}^{-2}}=\frac{1}{800}$

That is,

Rigidity Modulus $=\frac{4\times {10}^7}{64\times {10}^{-2}}\times 800=5\times {10}^{9}N/m^2$