Given, area `A = 1 m^(2)`
Thickness, `d = 5 cm = 5 xx 10^(-2) m`
lateral displacement = 0.005 cm
`= 5 xx 10^(-5) m`
Shearing strain `= ("Lateral displacement")/("Distance from fixed layer")`
`= (5 xx 10^(-5))/(5 xx 10^(-2)) = 10^(-3)`
Modulus of ragidity `= ("Shearing stress")/("Shearing strain")`
`= (4.2 xx 10^(2))/(10^(-3))`
Modulus of rigidity `eta = 4.2 xx 10^(6) Nm^(-2)`