Question

The temperature of a metal coin is increased by ${100}^{\circ }C$ and its diameter increases by 0.15%. Its area increases by nearly:

• A. 0.15%
• B. 0.60%
• C. 0.30%
• D. 0.0225%

Answer 1

Answer: (C)

$\delta \left(L\right)$= 0.15%= ${\alpha }_0\delta \left(T\right)$
A'= $L^{\prime 2}=L_0^2(1+{\alpha }_0\delta (T){)}^2$
Ignoring higher powers; $A^{\prime }=A_0(1+2{\alpha }_0(\delta \left(T\right)\left)\right)$
$\delta \left(A\right)=2{\alpha }_0\delta \left(T\right)$
Hence, Area increases by 0.30%