Question
Question: A metal wire of length \(L_1\) and area of cross section A is attached to a rigid support. Another ...
A metal wire of length L1 and area of cross section A is attached to a rigid support. Another metal wire of length L2 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 Y1 and Y2 are the Young's moduli of the wires respectively, the effective force constant of the system of two wires is
A: [2(Y1L2+Y2L1)][(Y1Y2)A]
B: (L1L2)1/2[(Y1Y2)A]
C: [(Y1L2+Y2L1)][(Y1Y2)A]
D: (L1L2)1/2[(Y1Y2)1/2A]
Solution
When two heavy wires are connected like a chain and hung up from a height with an attached mass, it experiences an elongation due to elastic force. It experiences a strain and the length is changed. This might also lead to an oscillation from the original length to the stretched length and vice versa.
Formula used:
Force constant K=ΔLF, where F is the force acting on the body and ΔLis the change in length of the body.
keq=k1+k2k1k2: The formula to find the effective force constant of two bodies connected in series.
Complete step by step answer:
We’re given the lengths of two wires as L1 and L2. Young's moduli are Y1 and Y2. They have the same area of cross section A.
We know that the force constant K=ΔLF=LYA (Since Y=AΔLFL)
keq=k1+k2k1k2 is the formula to find the effective force constant of two bodies connected in series.
Substituting the corresponding values of K, we get
keq=(L1Y1A)+(L2Y2A)(L1Y1A)(L2Y2A)=Y1L2+Y2L1(Y1Y2)A
So, the correct answer is “Option C”.
Note:
Young’s modulus is used to determine rigidity of a body .It is also used to determine the level of deformation of a body in a given load. Lower the value of Young’s modulus, more deformation is experienced by the body.