Problems on Composite Wires

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Problem – 1:

  • For two wires of the same material, both the radii and lengths are in the ratio 1:2. What should be the ratio of stretching forces on the wires if equal extensions are to be produced in the two?
  • Solution:
  • Given: Ratio of radii = r1 / r2 = 1/2, Ratio of length = L1 / L2 = 1/2, Extension equal l1 = l2 hence l1 / l2 = 1, Material is same hence ratio of Young’s  moduli Y1 / Y2 = 1,
  • To Find: Ratio of stretching force = F1 / F2 = ?

Composite Wires 02

Ans: The ratio of stretching forces is 1:2.

Problem – 2:

  • Two wires of different material, but of the same cross-section and length are stretched by the same force. Find the ratio of Young’s moduli are of their material if the elongations produced in them are in the ratio 3:1.
  • Solution:
  • Given: Same Coss-section, hence the ratio of area = A1 /A2 = 1, Lengths are same hence the ratio of length = L1 / L2 = 1, Same stretching force hence F1 / F2 = 1, Ratio of elongation l1 / l2 = 3:1,
  • To Find: Ratio of Young’s moduli = Y1 / Y2 = ?

Composite Wires 03

Ans: The ratio of Young’s moduli is 1:3.



Problem – 3:

  • Two wires are of the same material. Wire 1 if four times longer than the wire 2 but wire 1 has a diameter double that of wire 2. Compare stresses and elongations produced in the wires when under the same load.
  • Solution:
  • Given: Material is same hence ratio of Young’s  moduli Y1 / Y2 = 1, Length L1 = 4 L2 hence L1 / L2 = 4, Diameter d1 = 2 d2 hence d1/d2 = 2, hence ratio of radii = r1/r2 = 2, Load is same F1 / F2 = 1
  • To Find: Ratio of stresses = ? Ratio of elongations = ?

To find ratio of stresses

Composite Wires 04

To find ratio of elongations

Composite Wires 05



Hence elongations are equal



Example – 4:

  • Two wires of the same material have lengths in the ratio 2:1 and diameters are in the ratio 2:1 Find the ratio of extensions produced in the wires when the stretching forces acting on them are in the ratio 2:1
  • Solution:
  • Given: Material is same hence ratio of Young’s  moduli Y1 / Y2 = 1, Ratio of lengths = 2: 1 i.e.  L1 / L2 = 2, Ratio of diameters = 2:1 hence d1/d2 = 2, hence ratio of radii = r1/r2 = 2, Load is same F1 / F2 = 2:1
  • To Find: RRatio of extensions =?

Composite wires 06

Ans: The extensions in two wires is equal

Problems Based on Composite wire:

Example – 5:

  • A brass wire (Y = 11 × 1010 N/m²) and a steel wire (22 × 1010 N/m²) of the same length and cross-section are joined end to end. The composite wire is hung from a rigid support and a weight is suspended from the free end. Find the extension in each wire if the increase in the length of the composite wire is 0.279 cm.
  • Solution:
  • Given: Young’s modulus for brass Yb = 11 × 1010 N/m², Young’s modulus for steel Ys = 22 × 1010 N/m², Lengths of wire are same i.e  Lb / Ls = 1, Cross-section are same i.e. Ab/As = 1, Load is same Fb / Fs = 1, Total extension in composite wire =  lb + ls = 0.279 cm
  • To Find: extension in brass wire and steel wire = ?

lb =  2 ls



lb + ls = 0.279 cm

∴ 2 ls + ls = 0.279 cm

∴ 3 ls  = 0.279 cm

∴  ls  = 0.093 cm

Now lb =  2 ls  = 2 × 0.093 = 0.186 cm



Ans: Extension in brass wire = 0.186 cm and extension in steel wire = 0.o93 cm

Example – 6:

  • A uniform brass wire (Y = 10 × 1010 N/m²) and a uniform steel wire (Y = 20 × 1010 N/m²) each of length 3.14 m and diameter 2 × 10-3 m are joined end to end to form a composite wire is hung from a rigid support and a load suspended from the free end. If the increase in the length of the composite wire is 6 × 10-3 m, find the increase in the length of each wire.
  • Solution:
  • Given: Young’s modulus for brass Yb = 10 × 1010 N/m², Young’s modulus for steel Ys = 20 × 1010 N/m², Lengths of wire are same each 3.14 m i.e  Lb / Ls = 1, Diameter is same each 2 × 10-3 m, Hence cross-section are same i.e. Ab/As = 1, Load is same Fb / Fs = 1, Total extension in composite wire =  lb + ls = 6 × 10-3 m
  • To Find: extension in brass wire and steel wire = ?

Composite wire 08

lb =  2 ls

lb + ls = 6 × 10-3 m = 6 mm

∴ 2 ls + ls = 6 mm



∴ 3 ls  = 6 mm

∴  ls  = 2 mm

Now lb =  2 ls  = 2 × 2 = 4 mm

Ans: Extension in brass wire = 4 mm and extension in steel wire = 2 mm



Example – 7:

  • A light rod 1 m long is suspended horizontally by two wires of the same length and of the same cross-section but of different materials. The Young’s moudulus of material of one wire is 30 x 1010N/m2 and that of the other is 20 x 1010N/m2. At what point should a weight W be hung from the rod so that it still remains horizontal? Ans : (0.4 m from first wire)

Example – 8:

  • A weightless rod 105 cm long is suspended horizontally by two wires P and Q of equal length. The cross- section of P is 1 mm2 and that of Q is 2 mm2. From what point on the rod should a weight be suspended in order to produce equal strains in P and Q? Yp= 2 x 1011N/m2; YQ = 1011 N/m2.          Ans : (Mid –point)

Example – 9:

  • A brass wire of length 5 m and of sectional area 1 mm2 is hung from a rigid support with a brass weight of volume 1000 cc hanging from the other end. Find the decrease in length of the wire when the brass weight is completely immersed in water. Y = 1011N/m2. Take g = 10 m/s2. Density of brass = 8400 kg/m3 and of water = 1000 kg/m3. Ans : (0.5 mm)
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