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

Options

A. `4 xx 10^(-5) m`
B. `4 xx 10^(-6) m`
C. `4 xx 10^(-7) m`
D. `4 xx 10^(-8) m`

Answer

C

Explanation

Shear modulus, `eta =(F//A)/(DeltaL//L)`
`therefore DeltaL = (FL)/(etaA)`
Here, `F =(100 kg)(10 ms^(-2)) = 1000 N`
`A = (10 cm xx 10 cm)= 100 cm^(2) = 100 xx 10^(-4) m^(2)`
`eta = 25 GPa = 25 xx 10^(9) Pa = 25 xx 10^(9) N m^(-2)`
`L = 10 cm= 10 xx 10^(-2)m`
Substituting the given values, we get
`DeltaL = ((1000N)(10xx 10^(-2)m))/((25xx10^(9)Nm^(-2))(100xx10^(-4) m^(2)))= 4xx10^(-7) m`