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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 = r_{1}/ r_{2}= 1/2, Ratio of length = L_{1}/ L_{2}= 1/2, Extension equal l_{1}= l_{2}hence l_{1}/ l_{2}= 1, Material is same hence ratio of Young’s moduli Y_{1}/ Y_{2}= 1,**To Find:**Ratio of stretching force = F_{1}/ F_{2}= ?

**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 = A_{1}/A_{2}= 1, Lengths are same hence the ratio of length = L_{1}/ L_{2}= 1, Same stretching force hence F_{1}/ F_{2}= 1, Ratio of elongation l_{1}/ l_{2}= 3:1,**To Find:**Ratio of Young’s moduli = Y_{1}/ Y_{2}= ?

**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 Y_{1}/ Y_{2}= 1, Length L_{1}= 4 L_{2}hence L_{1}/ L_{2}= 4, Diameter d_{1}= 2 d_{2}hence d_{1}/d_{2}= 2, hence ratio of radii = r_{1}/r_{2}= 2, Load is same F_{1}/ F_{2}= 1**To Find:**Ratio of stresses = ? Ratio of elongations = ?

To find ratio of stresses

To find ratio of elongations

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 Y_{1}/ Y_{2}= 1, Ratio of lengths = 2: 1 i.e. L_{1}/ L_{2}= 2, Ratio of diameters = 2:1 hence d_{1}/d_{2}= 2, hence ratio of radii = r_{1}/r_{2}= 2, Load is same F_{1}/ F_{2}= 2:1**To Find:**RRatio of extensions =?

**Ans: **The extensions in two wires is equal

### Problems Based on Composite wire:

#### Example – 5:

- A brass wire (Y = 11 × 10
^{10}N/m²) and a steel wire (22 × 10^{10}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 Y_{b}= 11 × 10^{10}N/m², Young’s modulus for steel Y_{s}= 22 × 10^{10}N/m², Lengths of wire are same i.e L_{b}/ L_{s}= 1, Cross-section are same i.e. A_{b}/A_{s}= 1, Load is same F_{b}/ F_{s}= 1, Total extension in composite wire =*l*_{b}+*l*_{s}= 0.279 cm**To Find:**extension in brass wire and steel wire = ?

*l*_{b} = 2 *l*_{s}

*l*_{b} + *l*_{s} = 0.279 cm

∴ 2 *l*_{s} + *l*_{s} = 0.279 cm

∴ 3 *l*_{s} = 0.279 cm

∴ *l*_{s} = 0.093 cm

Now *l*_{b} = 2 *l*_{s} = 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 × 10
^{10}N/m²) and a uniform steel wire (Y = 20 × 10^{10}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 Y_{b}= 10 × 10^{10}N/m², Young’s modulus for steel Y_{s}= 20 × 10^{10}N/m², Lengths of wire are same each 3.14 m i.e L_{b}/ L_{s}= 1, Diameter is same each 2 × 10^{-3}m, Hence cross-section are same i.e. A_{b}/A_{s}= 1, Load is same F_{b}/ F_{s}= 1, Total extension in composite wire =*l*_{b}+*l*_{s}= 6 × 10^{-3}m**To Find:**extension in brass wire and steel wire = ?

*l*_{b} = 2 *l*_{s}

*l*_{b} + *l*_{s} = 6 × 10^{-3} m = 6 mm

∴ 2 *l*_{s} + *l*_{s} = 6 mm

∴ 3 *l*_{s} = 6 mm

∴ *l*_{s} = 2 mm

Now *l*_{b} = 2 *l*_{s} = 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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