Question

A piece of metal floats on mercury. The coefficient of expansion of the metal and
mercury are γ1 and γ2 respectively. If the temperature of both mercury and metal are increased by ∆T, by what factor does the fraction of the volume of the metal submerged in mercury change?​

Answer 1

AnSwer :-

Let the total volume of metal in air and mercury be V and V$\sf _s$ respectively, ρ and σ be the densities of metal and mercury respectively.

When the metal floats in equilibrium,

Fraction of the volume submerged,

$\sf f_s = \dfrac{V_s}{V} = \dfrac{\rho}{\sigma} \: \: \: \: \: \: \: \: ......(1)$

When the temperature changes, the fraction of volume submerged changes as densities change.

$\sf \dfrac{ \Delta f _{s} }{f_{s} } = \dfrac{f'_s}{f_s} - 1 = \dfrac{ \rho^{1} }{ \sigma {}^{1} } \times \dfrac{ \sigma}{ \rho} - 1 \: \: \: ....(2)$

Densities of liquid and metal decrease as temperature increases

$\sf\rho' = \dfrac{ \rho}{1 + \gamma _{ 1} \Delta T} \: \sigma' = \dfrac{ \sigma}{1 + \gamma_{2} \Delta T}$

which on substitution in eqn. (2) yields

$\sf \dfrac{\Delta f_s}{f_s} = \dfrac{1 + \gamma_2 \Delta T}{1 + \gamma_1 \Delta T} - 1 = \dfrac{(1 + \gamma_2 \Delta T)(1 + \gamma_1 \Delta T) }{1 + \gamma_1 \Delta T}$

$\sf = \dfrac{ (\gamma_2 - \gamma _{1} )\Delta T}{1 + \gamma_1 \Delta T} = (\gamma_2 - \gamma _{1} )\Delta T (1 - \gamma _{1} \Delta T)$

$\sf = (\gamma_2 - \gamma _{1} )\Delta T$

$\sf \bigg[As \dfrac{1}{1 + \gamma_1 \Delta T} =( 1 + \gamma_1 \Delta T) {}^{ - 1} = (1 - \gamma_1 \Delta T) \\ \sf and \gamma_1 \gamma_2 \: small \: number \: it \: has \: been \: neglected \bigg] \: \: \:$

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