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Physics Question on thermal properties of matter

A solid rectangular sheet has two different coefficients of linear expansion α1\alpha_{1} and α2\alpha_{2} along its length and breadth respectively. The coefficient of surface expansion is (for α1t<<1,α2t<<1\alpha_{1} t < < 1, \alpha_{2} t < < 1)

A

α1+α22\frac{\alpha_{1}+\alpha_{2}}{2}

B

2(α1+α2)2\left(\alpha_{1}+\alpha_{2}\right)

C

4α1α2α1+α2\frac{4\alpha_{1}\alpha_{2}}{\alpha _{1}+\alpha _{2}}

D

α1+α2\alpha_{1}+\alpha_{2}

Answer

α1+α2\alpha_{1}+\alpha_{2}

Explanation

Solution

The coefficient of linear expansion along its length =α1= \alpha_1

The coefficient of linear expansion along its breadth =α2=\alpha_{2}
Increase in length,
Lt=l0(1+α1Δt)L_{t}=l_{0}\left(1+\alpha_{1} \Delta t\right)
Increase in breadth,
Bt=b0(1+α2Δt2)B_{ t }=b_{0}\left(1+\alpha_{2} \Delta t_{2}\right)
Let coefficient of surface expansion is β\beta
Area = length × \times breadth
=l0(1+α1Δt)×b0(1+α2Δt)=l_{0}\left(1+\alpha_{1} \Delta t\right) \times b_{0}\left(1+\alpha_{2} \Delta t\right)
=l0b0(1+α1Δt)(1+α2Δt)=l_{0} b_{0}\left(1+\alpha_{1} \Delta t\right)\left(1+\alpha_{2} \Delta t\right)
=S0(1+α1Δt+α2Δt+…)=S_{0}\left(1+\alpha_{1} \Delta t+\alpha_{2} \Delta t+\ldots\right)
where, S0=l0⋅b0S_{0}=l_{0} \cdot b_{0}
== Initial area of surface
In state of expansion,
St=Lt×BtS_{t} =L_{t} \times B_{t}
=l0b0(1+α1Δt)(1+α2Δt)=l_{0} b_{0}\left(1+\alpha_{1} \Delta t\right)\left(1+\alpha_{2} \Delta t\right)
=S0(1+α1Δt+α2Δt+…)=S_{0}\left(1+\alpha_{1} \Delta t+\alpha_{2} \Delta t+\ldots\right)
St=S0(1+βΔt)S_{t} =S_{0}(1+\beta \Delta t)
∴S0(1+βΔt)=S0(1+α1Δt+α2Δt+…)\therefore S_{0}(1+\beta \Delta t) =S_{0}\left(1+\alpha_{1} \Delta t+\alpha_{2} \Delta t+\ldots\right)
β⋅Δt=α1Δt+α2Δt\beta \cdot \Delta t =\alpha_{1} \Delta t+\alpha_{2} \Delta t
β=α1+α2\beta =\alpha_{1}+\alpha_{2}