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

Edge of the aluminium cube, L = 10 cm = 0.1 m
The mass attached to the cube, m = 100 kg
Shear modulus `(eta)` of aluminium = 25 GPa = 25`xx` `10^(9)`
Shear modulus, `eta=("Shear stress")/("Shear strain")=((F)/(A))/((L)/(DeltaL))`
Where, F = Applied force = mg = 100 `xx`9.8 = 980 N
A = Area of one of the faces of the cube = 0.1`xx`0.1= 0.01 `"m"^(2)`
`DeltaL` = Vertical deflection of the cube
`thereforeDeltaL=(FL)/(Aeta)`
=`(980xx0.1)/(10^(-2)xx(25xx10^(9)))`
= `3.92xx10^(-7)` m
The vertical deflection of this face of the cube is 3.92`xx``10^(-7)`m.