A square steel plate has area `1 m^(2)` and thickness 5 cm. The lower surface is fixed. A tangential force applied to top surface displace it through 0.005 cm. Find the modulus of rigidity of steel, if shearing stress of modulus of rigidity of steel. If shearing stress of `4.2 xx 10^(3) N//m^(2)` is being applied.

Options

A. `5.2 xx 10^(6) Nm^(-2)`
B. `4.2 xx 10^(6) Nm^(-2)`
C. `3.6 xx 10^(4) Nm^(-2)`
D. `4.6 xx 10^(4) Nm^(-2)`

Explanation

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)`