Solveeit Logo

Question

Question: Two wires with the same area of cross section and length but young's modulus E and Y respectively ar...

Two wires with the same area of cross section and length but young's modulus E and Y respectively are joined end to end. Find equivalent young's modulus.

A

(E+Y)/2

B

EY/(E+Y)

C

no change

D

E+Y

Answer

(E+Y)/2

Explanation

Solution

The problem describes two wires joined end to end. This configuration is a series combination. For wires in series, the force (tension) is the same through both wires, and the total extension is the sum of individual extensions.

Let LL be the length and AA be the cross-sectional area of each wire. Let EE and YY be their Young's moduli.

The extension of the first wire is ΔL1=FLAE\Delta L_1 = \frac{FL}{AE}.

The extension of the second wire is ΔL2=FLAY\Delta L_2 = \frac{FL}{AY}.

The total extension is ΔLtotal=ΔL1+ΔL2=FLA(1E+1Y)=FLAE+YEY\Delta L_{total} = \Delta L_1 + \Delta L_2 = \frac{FL}{A} \left(\frac{1}{E} + \frac{1}{Y}\right) = \frac{FL}{A} \frac{E+Y}{EY}.

For the equivalent wire, the total length is 2L2L and the area is AA. If EeqE_{eq} is the equivalent Young's modulus, then ΔLtotal=F(2L)AEeq\Delta L_{total} = \frac{F(2L)}{AE_{eq}}.

Equating the expressions for ΔLtotal\Delta L_{total}: F(2L)AEeq=FLAE+YEY\frac{F(2L)}{AE_{eq}} = \frac{FL}{A} \frac{E+Y}{EY}.

This simplifies to 2Eeq=E+YEY\frac{2}{E_{eq}} = \frac{E+Y}{EY}, which gives Eeq=2EYE+YE_{eq} = \frac{2EY}{E+Y}.

This result is not among the options.

However, if the wires were joined side-by-side (parallel combination), the extension for both wires would be the same, and the total force would be the sum of individual forces.

For parallel combination: F1=EAΔLLF_1 = E \frac{A \Delta L}{L} and F2=YAΔLLF_2 = Y \frac{A \Delta L}{L}.

Total force Ftotal=F1+F2=(E+Y)AΔLLF_{total} = F_1 + F_2 = (E+Y) \frac{A \Delta L}{L}.

For the equivalent wire, the total length is LL and the total area is 2A2A. Ftotal=Eeq(2A)ΔLLF_{total} = E_{eq} \frac{(2A) \Delta L}{L}.

Equating the forces: (E+Y)AΔLL=Eeq(2A)ΔLL(E+Y) \frac{A \Delta L}{L} = E_{eq} \frac{(2A) \Delta L}{L}.

This simplifies to E+Y=2EeqE+Y = 2E_{eq}, which gives Eeq=E+Y2E_{eq} = \frac{E+Y}{2}.

This result matches option (a). Given the options and common exam patterns, it is highly probable that the question intends to describe a parallel combination, despite the wording "joined end to end".