Question

relation between alpha and gamma in thermal expansion ​

Answer 1

Answer:alpha = beta/2 = gamma/3

Explanation:

Answer 2

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Consider a metal sphere of radius r and volume v.

Let the metal sphere be heated until its temperature changes by ΔT, volume changes by ΔV

$radius \: changes \: by \:Δ \gamma .$

$initial \: volume \: of \: the \: sphere \\ = \frac{4}{3}\pi {r}^{3}$

Final volume of the sphere

$( \gamma + Δv) = \frac{4}{3} \pi(r + Δr) \\ by \: using \: (a + b {)}^{3} formula \\ ( \gamma + Δv) = \frac{4}{3} \pi( {r}^{3} + 3 {r}^{3}(Δr) \\ + 3 {r}^{3} (Δr)(Δ {r}^{3} )) \\ = \frac{4}{3} \pi( {r}^{3} + 3 {r}^{2} (Δr) + 3rΔ {r}^{2} +Δ {r}^{3}$

$since \: Δr \: is \: very \: small \: (Δ {r}^{2} )(Δ {r}^{3} ) \\ is \: smaller \: and \: can \: be \: ignored$

$Δv = \frac{4}{3}\pi3 {r}^{2} Δr \\ multiply \: both \: sides \: by \: \frac{Δt}{v} \\Δv. \frac{Δt}{v } = \frac{4}{3} \pi3 {r}^{2}Δ {r}. \frac{Δt}{v} \\ = \frac{4}{3} \pi3 {r}^{2} . \frac{Δt}{ \frac{4}{3} } \\ =$

$\frac{Δv}{v} .Δt = \frac{3\pi.r}{r}. Δt \\ r = 3 \alpha$

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