The edges of an aluminum cube are `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. Shear modulus of aluminum is `25 xx 10^(9) Pa`, the vertical deflection in the face to which mass is attached is

Explanation

`L = 10cm = 10^(-1)m`,
`A = L^(2) = 10^(-2) m^(2)`
`F = 100kg wt = 100xx9.8N = 9.8N = 9.8xx10^(2)N`
`eta = 25GPa = 25xx10^(9) a = 25xx10^(9) N//m^(2)`
As `eta = ("shear stress")/("Shear strain") = (F//A)/(Delta x//L) = (FL)/(A Delta x)`
`Delta x = (FL)/(A eta) = ((9.8xx10^(2))(10^(-1)))/((10^(-2))(25xx10^(9)))`
`= 03.4xx10^(-6)m = 4xx10^(-7)m`.