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))^(1//2))`
C. `([(Y_(1)Y_(2))A])/((Y_(1)L_(2)+Y_(2)L_(1)))`
D. `((Y_(1)Y_(2))^(1//2)A)/((L_(1)L_(2))^(1//2))`

Answer

C

Explanation

Using the 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 meanings.
When the two wires are connected together in series, the effective force constant is `k_(eff)=(k_(1)k_(2))/(k_(1)+k_(2))`
Substituting the corresponding lengths, area of cross sections and the Young.s moduli, we get
`k_(eff)=(((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))`