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_(1)Y_(2))A])/([2(Y_(1)L_(2) +Y_(2)L_(1))])`
B. `([(Y_(1)Y_(2))A])/((L_(1)L_(2))^(2))`
C. `([(Y_(1)Y_(2))A])/((Y_(1)L_(2) +Y_(2)L_(1)))`
D. `([(Y_(1) Y_(2))^(1//2)A])/((L_(2)+L_(1))^(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)/(DeltaL) =(YA)/(L)`
where the symbols have their usual meaings.
When the two wire are connected together in series, the effective force constant is given by
`k_(eq) = (K_(1)+K_(2))/(k_(1) +k_(2))`
Substituting the corresponding lenghts, area of cross sections and the Young's moduli, we get
`k_(eq) =(((Y_(1)A)/(L_(1)))((Y_(2)A)/(L_(2))))/((Y_(1)A)/(L_(1)) +(Y_(2)A)/(L_(2)))=((Y_(1)Y_(2))A)/(Y_(1)L_(2)+Y_(2)L_(1))`