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

The correct option is C 0.30%

Let initial diameter of coin is $d$.

Hence, area of coin $A_1=\pi (d/2{)}^2=\pi d^2/4$

Now, given that diameter increases by $0.15$%

Therefore new diameter will be:

$d^{\prime }=\frac{100.15d}{100}$

New area of coin,

$A_2=\pi (d^{\prime }/2{)}^2=\pi (100.15d/2\times 100{)}^2=\pi d^2/4(100.15/100{)}^2$

$A_2=A_1(100.15/100{)}^2$

Hence, percentage increase in area of coin,

$\Delta A^{\prime }=\frac{A_2-A_1}{A_1}\times 100$

$\Delta A^{\prime }=\frac{A_1(100.15/100{)}^2-A_1}{A_1}\times 100$

$\Delta A=0.30$%