Question

The apparent coefficient of expansion of liquid when heated in a copper vessel is $C$ and when heated in a silver vessel is $S$ . If $A$ is the coefficient of linear expansion of copper, what is the coefficient of linear expansion of silver?
(A) $\dfrac{{C + S - 3A}}{3}$
(B) $\dfrac{{C + 3A - S}}{3}$
(C) $\dfrac{{S + 3A - C}}{3}$
(D) $\dfrac{{C + S + 3A}}{3}$

Answer 1

Hint : It is to be remembered that both liquid and the vessels expand on heating. We need to use the relation between linear expansion coefficient and volume expansion coefficient in order to find the solution here.

Formula Used: The following formulae are used to solve this question,
$\Rightarrow {\alpha _l} = \dfrac{1}{3}{\alpha _V}$ where ${\alpha _l}$ is the linear expansion coefficient and ${\alpha _V}$ is the volume expansion coefficient.

Complete step by step answer
Expansion is the change in volume and shape and when this change is due to change in temperature, that is, transferring of heat energy, it is known as thermal expansion. Thermal expansion results in an increase in the length of dimensions of a body. Three types of expansion are known in solids, namely linear expansion, surface expansion and volume expansion.
Linear expansion occurs when there is a change in length of a body. Linear expansion is represented by ${\alpha _l}$ .
$\Rightarrow {\alpha _l}\vartriangle T = \dfrac{{\vartriangle l}}{{{l_0}}}$ where ${\alpha _l}$ is the linear expansion coefficient, $\vartriangle T$ is change in temperature, $\vartriangle l$ is the change in length and ${l_0}$ is the original length of the body.
Volume expansion is the increase in volume, represented by ${\alpha _V}$ .
$\Rightarrow {\alpha _V}\vartriangle T = \dfrac{{\vartriangle V}}{{{V_0}}}$ where ${\alpha _V}$ is the volume expansion coefficient, $\vartriangle V$ is the change in length, ${V_0}$ is the original area of the body.
When liquid is heated in a copper or silver vessel, both the liquid and vessel expand. The apparent coefficient of volume expansion is actually the difference between the volumetric expansion of the liquid and the volume expansion of the vessel.
Given that, the apparent coefficient of expansion of liquid when heated in a copper vessel is $C$ .
$\therefore {\alpha _{{V_L}}} - {\alpha _{{V_C}}} = C$ where ${\alpha _{{V_L}}}$ and ${\alpha _{{V_C}}}$ are the real coefficient of volumetric expansion of liquid and copper respectively.
Given that, the apparent coefficient of expansion of liquid when heated in a silver vessel is $S$ .
$\therefore {\alpha _{{V_L}}} - {\alpha _{{V_S}}} = S$ where ${\alpha _{{V_L}}}$ and ${\alpha _{{V_S}}}$ are the real coefficient of volumetric expansion of liquid and silver respectively.
It is known to us that linear expansion is equal to one-third of its volumetric expansion.
Thus, ${\alpha _l} = \dfrac{1}{3}{\alpha _V}$ .
It can be written that, ${\alpha _{{V_C}}} = 3A$ .
To find the difference between the apparent coefficient of expansion of liquid when heated in a copper and silver vessel respectively,
$\therefore \left( {{\alpha _{{V_L}}} - {\alpha _{{V_C}}}} \right) - \left( {{\alpha _{{V_L}}} - {\alpha _{{V_S}}}} \right) = C - S$
$\Rightarrow {\alpha _{{V_S}}} = C - S + {\alpha _{{V_C}}}$
Substituting, ${\alpha _{{V_C}}} = 3A$ ,
$\Rightarrow {\alpha _{{V_S}}} = C - S + 3A$
We know that the volumetric expansion of silver is three times its linear expansion.
Let us assume that the linear expansion of silver is ${\alpha _l}$ .
Thus , $3{\alpha _l} = {\alpha _{{V_S}}}$
$\Rightarrow {\alpha _{{V_S}}} = 3{\alpha _l} = C - S + 3A$
$\Rightarrow {\alpha _l} = \dfrac{{C - S + 3A}}{3} = \dfrac{{C + 3A - S}}{3}$
$\therefore$ The correct answer is Option B.

Note
The temperature of one vessel is more than that of the other, depending on its material, in practical cases. But this fact is ignored here in the solution as it makes the process more complicated and the calculations get more difficult.