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 $4\times {10}^{9}N$ is applied at the top with bottom surface fixed. The rigidity modulus of the material of the cube is

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

Answer 1

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

Given,

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

$L=8cm=8\times {10}^{-2}m$

$\Delta L=0.1mm={10}^{-4}m$

$A=L^2=(8\times {10}^{-2}{)}^2{m}^2$

Rigid modulus of the material,

$Y=\frac{F}{A}\times \frac{L}{\Delta L}$

$Y=\frac{4\times {10}^9}{(8\times {10}^{-2}{)}^2}\times \frac{8\times {10}^{-2}}{{10}^{-4}}$

$Y=0.5\times {10}^{15}N/m^2$

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

The correct option is B.