Question

The edge of an aluminium cube is $10 \,cm$ long. One face of the cube is firmly fixed to a vertical wall. $A$ mass of $100 \,kg$ is then attached to the opposite face of the cube. The vertical deflection of this face is (Shear modulus of aluminium = $25 GPa$, $g = 10\, m$ $s^{-2}$)

• A. $4\times10^{-5}\, m$
• B. $4\times10^{-6}\, m$
• C. $4\times10^{-7}\, m$
• D. $4\times10^{-8}\, m$

Answer 1

Answer: (C)

$4\times10^{-7}\, m$

Answer 2

Shear modulus, $\eta$ $=\frac{F /A}{\Delta L/ L}$ $\therefore\quad$ $\Delta L$ $=\frac{FL}{\eta A}$ Here, $F=\left(100\,kg\right)$ $\left(10 m s^{-2}\right)$ $=1000 \, N$ $A=\left(10 cm\times10cm\right)$ $=100 cm^{2}$ $=100\times10^{-4} m^{2}$ $\eta=25 Gpa$ $=25\times10^{9}pa=25\times10^{9}N m^{-2}$ $L=10 \,cm$ $=10\times10^{-2}\,m$ Substituting the given values, we get $\Delta L$ $=\frac{\left(1000\,N\right)\left(10\times10^{-2} m\right)}{\left(25\times10^{9} N m^{-2}\right)\left(100\times10^{-4} m^{2}\right)}$ $=4\times10^{-7} \,m$