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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 L1L_1​ and area of cross section A is attached to a rigid support. Another metal wire of length L2L_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 Y1Y_1​ and Y2Y_2​ are the Young's moduli of the wires respectively, the effective force constant of the system of two wires is
A: [(Y1Y2)A][2(Y1L2+Y2L1)]\dfrac{[({{Y}_{1}}{{Y}_{2}})A]}{[2({{Y}_{1}}{{L}_{2}}+{{Y}_{2}}{{L}_{1}})]}
B: [(Y1Y2)A](L1L2)1/2\dfrac{[({{Y}_{1}}{{Y}_{2}})A]}{{{({{L}_{1}}{{L}_{2}})}^{1/2}}}
C: [(Y1Y2)A][(Y1L2+Y2L1)]\dfrac{[({{Y}_{1}}{{Y}_{2}})A]}{[({{Y}_{1}}{{L}_{2}}+{{Y}_{2}}{{L}_{1}})]}
D: [(Y1Y2)1/2A](L1L2)1/2\dfrac{[{{({{Y}_{1}}{{Y}_{2}})}^{1/2}}A]}{{{({{L}_{1}}{{L}_{2}})}^{1/2}}}

Explanation

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=FΔLK=\dfrac{F}{\Delta L}, where F is the force acting on the body and ΔL\Delta Lis the change in length of the body.
keq=k1k2k1+k2{{k}_{eq}}=\dfrac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}: 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 L1L_1 and L2L_2. Young's moduli are Y1Y_1 and Y2Y_2. They have the same area of cross section A.
We know that the force constant K=FΔL=YALK=\dfrac{F}{\Delta L}=\dfrac{YA}{L} (Since Y=FLAΔLY=\dfrac{FL}{A\Delta L})
keq=k1k2k1+k2{{k}_{eq}}=\dfrac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}} is the formula to find the effective force constant of two bodies connected in series.

Substituting the corresponding values of K, we get
keq=(Y1AL1)(Y2AL2)(Y1AL1)+(Y2AL2)=(Y1Y2)AY1L2+Y2L1{{k}_{eq}}=\dfrac{(\dfrac{{{Y}_{1}}A}{{{L}_{1}}})(\dfrac{{{Y}_{2}}A}{{{L}_{2}}})}{(\dfrac{{{Y}_{1}}A}{{{L}_{1}}})+(\dfrac{{{Y}_{2}}A}{{{L}_{2}}})}=\dfrac{({{Y}_{1}}{{Y}_{2}})A}{{{Y}_{1}}{{L}_{2}}+{{Y}_{2}}{{L}_{1}}}

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.