Question

Derive the relation between alpha and beta and alpha and gama in thermal expansion heat and thermo . Explain fully please

Answer 1

Heya friend,

Here we go:

α = coefficient of linear expansion (alpha)

β = coefficient of superficial (beta)

γ = coefficient of cubical expansion (gamma)

Now,

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For the relation between α and β :-

Consider a rectangular lamina of

length = a

breadth = b

So,

area of rectangular lamina (s) = ab

On heating the lamina through 1 K,

change in length = α a                           [α = Δ l  / l ΔT

change in breath = β b                           α l ΔT = Δ l , Here, Δt = 1 K

                                                               so, α l = Δ l ⇒ α a = Δ l ]

So,

New length = original length + change in the length

                   = a + αa ⇒  a(1+α)

New breadth = original breadth + change in the breadth

                      = b + βb ⇒ b(1+β)

New area ($s_{0}$) = ab $(1+\alpha )^{2}$

Now,

Δs = $s_{0}$ - s

   = $ab(1+\alpha )^{2}$ - ab

   = ab[1 + $\alpha ^{2} + 2\alpha$] - ab

Now,

$\alpha ^{2}$ is <<< as compared to $\alpha$ ,

So,

Δs = ab [1 + 2$\alpha$] - ab

    = 2$\alpha$ab

Now, we know,

$\beta$ = Δs / sΔT

$\beta$  = $\frac{2\alpha ba}{(ab) 1}$

β = 2α

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For the relation between α and γ

Consider a cubical lamina of

length = a

breadth = b

height = c

So,

Volume = abc

On heating through 1 K,

change in length = $\alpha a$

change in breadth = $\alpha b$

change in height = $\alpha c$

New length = a + $\alpha a$ = a(1+$\alpha$)

New breadth = b + $\alpha$b = b(1+$\alpha$)

New height = c +$\alpha$c  = c(1+$\alpha$)

New volume = abc $(1+\alpha )^{3}$

Δv = Nev volume - original volume

    =  abc$(1+\alpha )^{3}$ - abc

    = abc [1+$\alpha ^{3}+3 \alpha ^{2} +3\alpha$] - abc

   

Now,

$\alpha^{2}, \alpha ^{3}$ are too small , so they can be neglected

So,

Δv = abc (1 + 3α) - abc

    = abc + 3αabc - abc

    = abc3α

Now,

γ = Δv / vΔT

  = $\frac{abc3\alpha }{(abc)1}$                   [ΔT = 1 K]

  = 3α

γ = 3α

So,

β = 2α

γ = 3α

THANKS !

#BAL #answerwithquality