A metal wire of length `L_1` and area of cross section A is ttached to a rigid support. Another metal wire of length `L_2` and of the same cross sectional area is attached to the free end of the first wire. A body of mass M is then suspended from the free end of the second wire, if `Y_1` and `Y_2` are the Young's moduli of the wires respectively the effective force constant of the system of two wires is

Options

A. `([(Y_1Y_2)A])/([2(Y_1L_2+Y_2L_1)])`
B. `([Y_1Y_2)A)]/((L_1L_2)^((1)/(2)))`
C. `([(Y_1Y_2)A])/((Y_1L_2+Y_2L_1))`
D. `([(Y_1Y_2)^((1)/(2))A])/((L_1L_2)^((1)/(2)))`

Answer

C

Explanation

Using the usual expression for the Young's modulus the force constant for the wire can be written as
`k=(F)/(triangleL)=(YA)/(L)`
Where the symbols have their usul meanings, When the two are connected together in series, the effective force constant is given by
`k_(eq)=(k_1k_2)/(k_1+k_2)`
Substituting the corresponding lengths, area of cross section and the Young's moduli, we get
`k_(eq)=(((Y_1A)/(L_1))((Y_2A)/(L_2)))/((Y_1A)/(L_1)+(Y_2)/(L_2))=((Y_1Y_2)A)/(Y_1L_2+Y_2L_1)`