Two wires of the same material (Young's modulus=Y) and same length L but radii R and 2R respectively are joined end to end and a weight w is suspended from the combination as shown in the figure. The elastic potential energy in the system is

Options

A. `(3w^(2)L)/(4piR^(2)Y)`
B. `(3w^(2)L)/(8piR^(2)Y)`
C. `(5w^(2)L)/(8piR^(2)Y)`
D. `(w^(2)L)/(piR^(2)Y)`

Answer

C

Explanation

(c) We have `Deltal_(1)=(wL)/((4piR^(2))Y)`, and `Deltal_(2)=(wL)/(piR^(2)Y)`
The elastic potential energy in the system
`therefore U=(1)/(2)K_(1)(Deltal_(1))^(2)+(1)/(2)K_(2)(Deltal_(2))^(2)" "(K=(YA)/(L))`
`=(1)/(2)xx(Y(pirR^(2)))/(L)xx[(wL)/(4piR^(2)Y)]^(2)+(1)/(2)xx(Y(piR^(2)))/(L)xx[(wL)/(piR^(2)Y)]^(2)`
`(5w^(2)L)/(8piR^(2)Y)`