Q. 103.5( 12 Votes )

# A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10^{–5} K^{–1}, steel = 1.2 × 10^{–5} K^{–1}).

Answer :

Given,

Initial temperature, T_{1} = 40°C = 40 + 273.15 K = 313.15K

Final temperature, T_{2} = 250°C = 250 + 273.15 = 523.15 K

Length of the brass rod at T_{1}, L_{1} = 50 cm = 0.5 m

Diameter of the brass rod at T_{1}, d_{1} = 3.0 mm = 3 × 10^{-3} m

Length of the steel rod at T_{2}, L_{2} = 50 cm = 0.5 m

Diameter of the steel rod at T_{2}, d_{2} = 3.0 mm = 3 × 10^{-3} m

Coefficient of linear expansion of brass, α_{1} = 2.0 × 10^{–5} K^{–1} Coefficient of linear expansion of steel, α_{2} = 1.2 × 10^{–5} K^{–1}

Change in temperature, ΔT = T_{2} – T_{1} = 523.15 K-313.15K=210 K

For the expansion in the brass rod,

Change in length (∆L_{1})/Original length (L_{1}) = α_{1}ΔT

⇒ ΔL_{1} = α_{1}L_{1}ΔT

⇒ ΔL_{1} = (2.0 × 10^{–5} K^{–1})×(0.5 m)×(210 K)

⇒ ΔL_{1} = 0.0021 m = 0.21 cm

For the expansion in the steel rod,

Change in length (∆L_{2})/Original length (L_{2}) = α_{2}ΔT

⇒ ∆L_{2} = α_{2}L_{2}ΔT

⇒ ∆L_{2} = (1.2 × 10^{–5} K^{–1})×(0.5 m)×(210 K)

⇒ ∆L_{2} = 0.00126 m = 0.126 cm

Total change in length, ΔL = ∆L_{1} + ∆L_{2}

⇒ ΔL = 0.21 cm + 0.126 cm

⇒ ∆L= 0.336 cm

No thermal stress develops at the junction since both the ends of the rod expand freely.

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