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

Torsion of a cylinder `tau = (pi eta r^4 phi)/(2l)`
Twist produced in the cylinder `phi = (tau xx (2l))/(pi eta r^4) = ((F xx l) xx (2l))/(pi eta r^4)`
`= (4 xx 10^5 xx 2 xx (1.5)^2)/(3.14 xx 6 xx 10^(10) xx (2 xx 10^(-2))^(4)) = (18 xx 10^(5))/(3.01 xx 10^(-6) xx 10^(10))`
`phi = 59.80`