Question

Define alpha, beta and gamma. Derive relation between alpha and gamma.

Answer 1

Explanation:

This is regarding linear , areal, volume expansion of substances.

ΔL = α L ΔT

Area of a square sheet of dimensions L * L:

 = A+ΔA = (L+ α L ΔT)² = L² + 2 α L² ΔT   ignoring α² terms (too small).

ΔA = 2 α A ΔT

β = ΔA/A  = 2 α

Volume = V = L³

Changed volume = V+ ΔV = (L + α L ΔT)³

         = L³ + 3 α L³ ΔT  +  terms with α² or α³  ignored (too small)

ΔV = 3 α V ΔT

=>  gamma = ΔV/V = 3 α

Answer 2

According to Linear , Superficial or Area and Cubical or volume expansion.

Suppose a cube of each length (l) and it's initial volume is (v)

When we heat this cube by δT it's each length will be increased by δl.

$\: l' = l + \delta \: l .... \sf(i)$

We know that,

$\alpha = \frac{ \delta \: l}{l \times \delta \: T}$

$\delta \: l = \alpha l \delta T ... \sf(ii)$

put the volume of the cube is V'

$V' = {(l' )}^{3} .... \sf(iii)$

From equation (ii) we will find volume of (l')³

${(l')}^{3} = {l}^{3} (1 + \alpha \delta \: {T)}^{3} \\ \\ {(l')}^{3} = {l}^{3} ( {1}^{3} + \alpha ' \delta \: {T)}^{3} + 3 \times \alpha \delta \: T(1 + \alpha \delta \: T) \\ \\ {(l')}^{3} = {l}^{3} ( {l}^{3} + { \alpha }^{3} \delta \: {T}^{3} )+ 3 \times \alpha\delta \: T + 3 \times 1 \times { \alpha }^{2} \delta {T}^{2}$

${(l')}^{3} = V(1 + 3 \alpha \delta \: T).... \sf(iv)$

Now neglecting higher power....

put the value of (l')³ in equation (iii)

$V' = \: {(l')}^{3}$

$V' =V(1 + 3 \alpha \delta \: T)$

According to the coefficient of cubical expansion :-

$\gamma = \frac{ \delta \: V}{V \delta \: T}$

$\gamma = \frac{V' - V}{V' \delta T}$

$\gamma = \frac{V(1 + 3 \alpha \delta T) - V}{V \delta T}$

$\gamma = \frac{ \cancel{V}(1 + 3 \alpha \delta T - 1)}{ \cancel{V}\delta T}$

$\gamma = \frac{3 \alpha \cancel{\delta T}}{ \cancel{\delta T}}$

$\gamma = 3 \alpha$

$\frac{ \alpha }{1} = \frac{ \gamma }{3}$

Similarly,

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