First slide

Thermal expansion

Question

Two thin metal strips, one of brass and the other of iron are fastened together parallel to each other. Thickness of each strip is 1 mm. If the strips are of equal length at 0°C . The inner radius of the arc formed by the bimetallic strip when heated to 80°C is (Coefficient of linear expansion of brass = 19×10^–6/⁰C & of iron = 12 × 10^–6/⁰C).

Easy

A

3.57 m
B

2.67 m
C

3.12 m
D

4.56 m

Solution

d = t₁ + t₂ = 1 + 1 = 2mm = 2 x 10^-3m
θ₁ = 0⁰C, θ₂ = 80⁰C, α₁ = 19 x 10^-6, α₂ = 12 x 10^-6
Inner radius of curvature of bimetallic strip
$\large r=\frac {d}{\left (\alpha_1-\alpha_2 \right )\Delta \theta }=\frac {2\times 10^{-3}}{(19-12)\times 10^{-6}(80-0)}=3.57m$

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