Question

Prove that α /1 = B/2 = Y/3 where α ,B,Y are coefficient of linear,
superficial and cubical expression

Answer 1

Given :

Coefficient of linear, superficial and cubical expression

To Find :

To prove $\alpha$ = $\frac{\beta }{2}$ = $\frac{Y}{3}$

Solution :

We know,

L = L₀ (1 + $\alpha$.t)

Where, $\alpha$ is the coefficient of linear expansion; L₀ = Initial Length; L = Final Length; t = Change in Temperature

A = A₀ (1 + β.t)

Where, β is the coefficient of surface expansion; A₀ = Initial Area; A = Final Area; t = Change in Temperature

V = V (1 + Y.t)

Where, Y is the coefficient of cubical expansion; V₀ = Initial Volume; V = Final Volume; t = Change in Temperature

Now,

L = L₀ (1 + $\alpha$.t)

Cubing both the sides,

L³ = L₀³ (1 + $\alpha$.t)³

As Volume = (Length)³

V  = V₀ (1 + 3$\alpha$t + 3$\alpha$²t² + $\alpha$³t³)

Neglecting 3$\alpha$²t² and $\alpha$³t³ as  $\alpha$<<1

V  =  V₀ (1 + 3$\alpha$t) ------------ (i)

But from the definition of the coefficient of volume expansion Y,

V  =  V₀ (1 + Yt)  --------------(ii)

Comparing (i) and (ii)

Y = 3$\alpha$

Now,

L = L₀ (1 + $\alpha$.t)

Squaring both the sides,

L² = L₀² (1 + $\alpha$.t)²

A = A₀ (1 + 2$\alpha$t + $\alpha$²t²)

Neglecting $\alpha$²t² as $\alpha$<<1

A = A (1 + 2$\alpha$t)  ---------- (iii)

But from the definition of the coefficient of superficial expansion β,

A = A₀ (1 + βt)  ---------(iv)

Comparing (iii) and (iv)

β = 2$\alpha$

Hence, $\alpha$ = $\frac{\beta }{2}$ = $\frac{Y}{3}$  and $\alpha$ : β : Y = 1 : 2 : 3