The diameter of a brass ring at 30 ^\circC is 50.0 cm. To what temperature must this ring be heated to increase its diameter to 50.29 cm? [ linear expansivity of brass = 1.9 x 10\(^{-5}\) K\(^{-1}\)]
A
152.6 °C
B
182.6 °C
C
306.1 °C
D
335.3 °C
Verified Correct Answer: D.
Physical Principles & Numerical Solution
Area expansivity(ß) = \(\frac{ A_2 - A_1}{ A_1(\theta_2 - \theta_1})\)
Area = \(\frac{\pi d^2}{4}\)
\(d_1\) = 50.0cm, \(d_2\) = 50.29cm, \(\theta_1\) = 30ºC, \(\theta_2\) = ? and ß = 2\alpha
\(2 \times 1.9 \times 10^{-5}\) = \(\frac{50.29^2 - 50^2}{ 50^2( \theta_2 - 30)}\)
\(\theta_2 - 30\) = \(\frac{50.29^2 - 50^2}{ 50^2(2 \times 1.9 \times10^{-5})}\)
\(\theta_2\) = 30 + \(\frac{50.29^2 - 50^2}{ 50^2(2 \times 1.9 \times10^{-5})}\) = 335.3 ^\circC
note : \(\frac{\pi}{4}\) cancels out.
Source: Myschool Past Questions Archive (Record #63395)