Two wires A and B of same length and of same | vedclass

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(B) The twisting couple (torque) $	au$ applied to a wire is given by the formula: $	au = \frac{\pi \eta r^4 \phi}{2L}$,where $\eta$ is the modulus o

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$r_2^4 / r_1^4$

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(B) The twisting couple (torque) $	au$ applied to a wire is given by the formula: $	au = \frac{\pi \eta r^4 \phi}{2L}$,where $\eta$ is the modulus of rigidity,$r$ is the radius,$\phi$ is the angle of twist,and $L$ is the length of the wire.Since the wires have the same length $L$,same material (same $\eta$),and are subjected to the same twisting couple $	au$,we have: $	au = \frac{\pi \eta r_1^4 \phi_A}{2L} = \frac{\pi \eta r_2^4 \phi_B}{2L}$.Equating the two expressions: $r_1^4 \phi_A = r_2^4 \phi_B$.Therefore,the ratio of the angle of twist at the end of $A$ to the angle of twist at the end of $B$ is: $\frac{\phi_A}{\phi_B} = \frac{r_2^4}{r_1^4}$.

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