If a shear force of 3000 N is applied on a cube of side 40 cm, then displacement of the top surface of the cube when bottom surface is fixed, is
`" "(eta=5xx10^(10)N//m^(2))`

Options

A. 15 mm
B. 150 nm
C. `15xx10^(-8)mu mu m`
D. `15 mu m`

Answer

B

Explanation

`eta=(Fh)/(Ax) therefore x=(Fh)/(A eta)`
`=(3000 xx 40 xx 10^(-2))/(40xx40xx10^(-4) xx 5 xx 10^(10))`
`=(3)/(2) xx 10^(-7) =1.5 xx 10^(-7)m=150xx10^(-9)m`
`=150 nm.`