A cylinder of length 1.5 m and diameter 4 cm is fixed at one end. A tangential force of `4 xx 10^(5) N` is applied at the other end. If the rigidity modulus of the cylinder is `6 xx 10^(10) Nm^(-2)` then, calculate the twist produce in the cylinder.

Explanation

Length of a cylinder = 1.5 m
Diameter = 4 cm, Tangential force `F = 4 xx 10^(5) N`
Rigidity modulus `eta = 6 xx 10^(10) Nm^(-2)`
Twist produced `theta = ?`
Rigidity Modulus `eta_(2) = (("Tangential force")/("Area"))/("Shear Angle (Twist Angle)")`
Angle Twist `theta = ((F)/(A))/(eta_(R )) = ((4 xx 10^(5))/(pi xx r^(2)))/(6 xx 10^(10))`
`= (4 xx 10^(5))/(3.14 xx (2 xx 10^(-2))^(2) xx 6 xx 10^(10)) = (4 xx 10^(5) xx 10^(4) xx 10^(-10))/(24 xx 3.14)`
`theta = 0.053 xx 10^(-1) = 53 xx 10^(-4)`