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Q.Establish relation between coefficient of linear expansion (α), coefficient of superficial expansion (β), and coefficient of volume expansion (γ). OR State and prove Bernoulli's theorem.

Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2020Subjective· 5mImportance★★★★★
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Starting from linear expansion L=L0(1+αΔT)L=L_0(1+\alpha\Delta T) and extending to area and volume, one finds β=2α\beta=2\alpha and γ=3α\gamma=3\alpha, i.e. α:β:γ=1:2:3\alpha:\beta:\gamma = 1:2:3.

Step 1 — Linear expansion.

For a rod of original length L0L_0 heated through a temperature rise ΔT\Delta T, the new length is

L=L0(1+αΔT)L = L_0(1+\alpha\Delta T)

where α\alpha is the coefficient of linear expansion.

Step 2 — Superficial (areal) expansion.

Consider a square lamina of side L0L_0, so its original area is A0=L02A_0 = L_0^2. After heating, each side becomes L=L0(1+αΔT)L=L_0(1+\alpha\Delta T), so the new area is

A=L2=L02(1+αΔT)2=L02(1+2αΔT+α2ΔT2)A = L^2 = L_0^2(1+\alpha\Delta T)^2 = L_0^2\left(1+2\alpha\Delta T+\alpha^2\Delta T^2\right)

Since α\alpha is very small, α2ΔT2\alpha^2\Delta T^2 is negligible, so

A≈L02(1+2αΔT)=A0(1+2αΔT)A \approx L_0^2(1+2\alpha\Delta T) = A_0(1+2\alpha\Delta T)

By definition, A=A0(1+βΔT)A = A_0(1+\beta\Delta T), so comparing:

β=2α\beta = 2\alpha

Step 3 — Cubical (volume) expansion.

Similarly, for a cube of original volume V0=L03V_0=L_0^3, the new volume after heating is

V=L3=L03(1+αΔT)3=L03(1+3αΔT+3α2ΔT2+α3ΔT3)V = L^3 = L_0^3(1+\alpha\Delta T)^3 = L_0^3\left(1+3\alpha\Delta T + 3\alpha^2\Delta T^2+\alpha^3\Delta T^3\right) …

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