Question

A thin square aluminium sheet of side 1 cm is heated to change its temperature from 20℃ to 120℃. Find the final area of the sheet. (${\alpha }_{aluminium}=24\times {10}^{-6}{K}^{-1}$)

• A.
  $1.0048c{m}^2$
• B.
  $1.048c{m}^2$
• C.
  $1.0024c{m}^2$
• D.
  $1.024c{m}^2$

Answer 1

The correct option is A

$1.0048c{m}^2$
Given,
Side length of square sheet of aluminium = 1 cm
Area of aluminium sheet $A_0=1c{m}^2$
${\beta }_{aluminium}=2{\alpha }_{aluminium}=48\times {10}^{-6}{K}^{-1}$
$\Delta T=120°C-20°C=100°Cor100K$
Change in the area of aluminium sheet $\Delta A=A_0\beta \Delta T$= $\left(1c{m}^2\right)\times \left(48\times {10}^{-6}{K}^{-1}\right)\times \left(100K\right)$= $0.0048c{m}^2$
Final area = $initialarea\left(A_0\right)+increaseinarea(\Delta A)=1c{m}^2+0.0048c{m}^2=1.0048c{m}^2$
Thus the final area of aluminium sheet is $1.0048c{m}^2$