Question

Prove that alpha:beta:gamma=1:2:3

Answer 1

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$\boxed{\boxed{\bold{\alpha : \beta : \gamma = 1:2:3}}}$

♣  Alpha (α) is coefficient of linear expansion.
♦  Increase in length during expansion is called linear expansion.
It means object one side increases.
Hence α = 1

♣ Beta (β) is coefficient of superficial expansion.
♦ Increase in area during expansion is called superficial expansion.
We know area = Length × Breadth
No. of sides = 2
Hence, β = 2

♣ Gamma ($\gamma$) is coefficient of cubical expansion.
♦ Increase in volume during expansion is called cubical expansion.
We know, volume = Length × Breadth × Height
No. of sides = 3
Hence,$\gamma$  = 3

So, $\boxed{\boxed{\bold{\alpha : \beta : \gamma = 1:2:3}}}$

Hence Proved!

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Answer 2

We know that,

L =L• (1+αΔT)

α= coefficient of linear expansion

And,

A= A• (1+βΔT)

β= coefficient of aerial expansion

And,

V= V•(1+γΔT)

γ= coefficient of cubical expansion.

So, now

V= V• + γV•ΔT

V= V•(1+γΔT)

L³= L•³ (1+αΔT)³

L³= L•³(1+3αΔT + 3α²ΔT² +α³ΔT³)

L³= L•³(1+3αΔT)

{Neglecting 3α²ΔT² and α³ΔT³ because they are very smaller than 1}

L³= L•³(1+3αΔT)

V= L•³(1+3αΔT)

V•(1+γΔT) = V•(1+3αΔT)

1+γΔT = 1+3αΔT

γΔT = 3αΔT

γ=3α

And β=2α

A= A•(1+βΔT)

L²= L•²(1+αΔT)²

A= L•²(1+2αΔT+α²ΔT²)

A= A•(1+2αΔT)

A•(1+βΔT) = A•(1+2αΔT)

{α²ΔT² Neglecting them due to very smaller volume}

β=2α

α:β:γ=1:2:3

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