Question

A thin Brass sheet at ${10^ \circ }C$and thin steel sheet at ${20^ \circ }C$ have the same surface area. Find the common temperature at which both would have the same area:
(linear expansion coefficient for brass and steel are $19 \times {10^{ - 6}}{/^ \circ }C$ and $11 \times {10^{ - 6}}{/^ \circ }C$ respectively.)

(a) $ - {3.75^ \circ }C$
(b) $ - {2.75^ \circ }C$
(c) ${2.75^ \circ }C$
(d) ${3.75^ \circ }C$

Answer 1

Hint: Derive the areal expansion coefficient from linear expansion coefficient. Then, use the thermal expansion relation to get expansion of the sheet for both the bodies.

Formula Used:
1.expansion produced due to given expansion coefficients is given by
$A(T) = {A_0}(1 + \beta (T - {T_0}))$ …… (1)
Where,
A(T) =area at any temperature T
T= Temperature (It is variable)
${T_0}$= initial temperature of the respective sheet
$\beta $= Coefficient of areal expansion for the sheet.

2. Relation between Coefficient of area expansion and coefficient of linear expansion is given by: \[\beta = 2\alpha \] …… (2)
Where,
$\beta $= Coefficient of areal expansion for the sheet.
$\alpha $= Coefficient of linear expansion for the sheet along each dimension.


Complete step by step answer:
Given,
Initial area of brass sheet: ${A_0}$
Coefficient of linear thermal expansion for the brass sheet: ${\alpha _b} = 19 \times {10^{ - 6}}{/^ \circ }C$ .
initial temperature of the brass sheet: ${T_0}$=${10^ \circ }C$

Initial area of steel sheet: ${A_0}$ (initially they have same area)
Coefficient of linear thermal expansion for the steel sheet: ${\alpha _s} = 11 \times {10^{ - 6}}{/^ \circ }C$
initial temperature of the steel sheet: ${T_0}$=${20^ \circ }C$
Let the final new temperature of the two bodies be ${T_f}$ where both bodies have the same area.

Step 1:
Using equation (2) we can find a real expansion coefficient from a given linear expansion coefficient.
For brass sheet:
Coefficient of areal thermal expansion for the brass sheet:
${\beta _b} = 2 \times {\alpha _b} = 2 \times 19 \times {10^{ - 6}}{/^ \circ }C$. …… (3)
Similarly, for steel sheet:
Coefficient of areal thermal expansion for the steel sheet:
 ${\beta _s} = 2 \times {\alpha _s} = 2 \times 11 \times {10^{ - 6}}{/^ \circ }C$. …… (4)

Step 2:
Using equation (1) and (2) we get areal expansion for both metal sheets-
For brass sheet:
Putting value given above for the brass sheet in equation (1) we get-
$A(T) = {A_0}(1 + \beta (T - {T_0}))$
$ \Rightarrow A{(T)_b} = {A_0}(1 + 2 \times 19 \times {10^{ - 6}}(T - 10))$ ……. (5)
For steel sheet:
Putting value given above for the brass sheet in equation (1) we get-
$A(T) = {A_0}(1 + \beta (T - {T_0}))$
$ \Rightarrow A{(T)_s} = {A_0}(1 + 2 \times 11 \times {10^{ - 6}}(T - 20))$ ……. (6)

Step 3:
Dividing equation (5) by (6) we get,

$ \Rightarrow \dfrac{{A{{(T)}_b}}}{{A{{(T)}_s}}} = \dfrac{{{A_0}(1 + 2 \times 19 \times {{10}^{ - 6}}(T - 10))}}{{{A_0}(1 + 2 \times 11 \times {{10}^{ - 6}}(T - 20))}}$
\[ \Rightarrow \dfrac{{A{{(T)}_b}}}{{A{{(T)}_s}}} = \dfrac{{(1 + 2 \times 19 \times {{10}^{ - 6}}(T - 10))}}{{(1 + 2 \times 11 \times {{10}^{ - 6}}(T - 20))}}\] ……. (7)

Step 4:
We know finally both have same temperature ${T_f}$ and same area
So\[\dfrac{{A{{(T)}_b}}}{{A{{(T)}_s}}}\]=1 ……(8)
Putting final temperature and values from equation (8) in (7) we get-
\[ \Rightarrow 1 = \dfrac{{(1 + 2 \times 19 \times {{10}^{ - 6}}({T_f} - 10))}}{{(1 + 2 \times 11 \times {{10}^{ - 6}}({T_f} - 20))}}\]
\[ \Rightarrow (1 + 2 \times 11 \times {10^{ - 6}}({T_f} - 20)) = (1 + 2 \times 19 \times {10^{ - 6}}({T_f} - 10))\]
Simplifying both sides, we get-
\[
   \Rightarrow 11({T_f} - 20) = 19({T_f} - 10) \\
   \Rightarrow - 220 + 190 = (19 - 11){T_f} \\
   \Rightarrow - 30 = 8{T_f} \\
   \Rightarrow {T_f} = \dfrac{{ - 30}}{8} = - {3.75^ \circ }C \\
 \]

Final Answer
Hence, option (a) $ - {3.75^ \circ }C$ is correct.

Note: It is important to convert coefficient of linear expansion into coefficient of areal expansion, as there are three different kinds of expansion coefficient for a material i.e Linear, Areal and Volumetric.