Figure shows elongation versus temperature for two different materials. If these metals be employed to form a bimetallic strip of thickness 6cm and heated it bends into an arc, the radius of curvature changes with temperature as shown. The linear expansivity of the metals are

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$3 \times 10^{- 6} /^{0} C$

$8 \times 10^{- 6} /^{0} C$

$2 \times 10^{- 6} /^{0} C$

$6 \times 10^{- 6} /^{0} C$

answer is A.

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Detailed Solution

$\begin{array}{l} R = \frac{d}{(α_{1} - α_{2}) Δt} \Rightarrow 75 = \frac{3}{(α_{1} - α_{2}) 50} \\ \Rightarrow α_{1} - α_{2} = 8 \times 10^{- 6} /^{\circ} C \\ \tan θ_{1} = (\frac{Δ ℓ}{Δt}) = ℓ_{1} α_{1} = \frac{4}{4} = 1 \\ \tan θ_{2} = {(\frac{Δ ℓ}{Δt})}_{2} = ℓ_{2} α_{2} = \frac{2}{4} = \frac{1}{2} ℓ_{1} = ℓ_{2} \\ \frac{α_{1}}{α_{2}} = 2 Solving \\ α_{1} = 16 \times 10^{- 6} /^{\circ} C, α_{2} = 8 \times 10^{- 6} /^{\circ} C \end{array}$