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Question

Physics Question on elastic moduli

Two rods AA and BB of the same material and length have radii r1andr2r_{1} and \,r_{2} respectively. When they are rigidly fixed at one end and twisted by the same torque applied at the other end, the ratio Two rods AA and BB of the same material and length have radii r1andr2r_{1} and\, r_{2} respectively. When they are rigidly fixed at one end and twisted by the same torque applied at the other end, the ratio (the angle of twist at the end of Athe angle of twist at the end of B)\left(\frac{\text{the angle of twist at the end of A}}{\text{the angle of twist at the end of B}}\right)

A

r12r12\frac{r_{1}^{2}}{r_{1}^{2}}

B

r13r23\frac{r_{1}^{3}}{r_{2}^{3}}

C

r24r14\frac{r_{2}^{4}}{r_{1}^{4}}

D

r14r24\frac{r_{1}^{4}}{r_{2}^{4}}

Answer

r24r14\frac{r_{2}^{4}}{r_{1}^{4}}

Explanation

Solution

A rod of length LL and radius rr fixed at one end and twisted by applying a torque τ\tau at the other end, then the angle of twist θ\theta is given by θ\theta =2τLπr4η=\frac{2\tau L}{\pi r^{4}\eta} where η\eta is shear modulus of the material of the wire.As LL, τ\tau and η\eta are same for both the wires \therefore \quad θ1r4\theta \propto\frac{1}{r^{4}} or θ1θ2\frac{\theta_{1}}{\theta_{2}} =(r2r1)4=\left(\frac{r_{2}}{r_{1}}\right)^{4}