The diameter of a brass ring at 30 °C 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\(^{-1}\)]
- 152.6 °C
- 182.6 °C
-
306.1 °C
- 335.3 °C
The correct answer is: C
Explanation
Area expansivity(ß) = \(\frac{ A_2 - A_1}{ A_1(θ_2 - θ_1})\)
Area = \(\frac{\pi d^2}{4}\)
\(d_1\) = 50.0cm, \(d_2\) = 50.29cm, \(θ_1\) = 30ºC, \(θ_2\) = ? and ß = 2α
\(2 \times 1.9 \times 10^{-5}\) = \(\frac{50.29^2 - 50^2}{ 50^2( θ_2 - 30)}\)
\(θ_2 - 30\) = \(\frac{50.29^2 - 50^2}{ 50^2(2 \times 1.9 \times10^{-5})}\)
\(θ_2\) = 30 + \(\frac{50.29^2 - 50^2}{ 50^2(2 \times 1.9 \times10^{-5})}\) = 306.1ºC.
note : \(\frac{\pi}{4}\) cancels out.