A solid steel ball with radius 3.00 cm and an aluminum ring with inner radius 2.98 cm are initially at 20 °C. If the ball and ring are heated together, at what common minimum temperature will the ball fit through the ring? A. 226 °C B. 403 °C C. 461 °C D. 552 °C E. 583 °C

635 degree.

Explanation:

L = L₁ + L₁ X α X ΔT

L is length after expansion, L₁ is length before expansion,α is coefficient o linear expansion and ΔT is rise in temperature.

For steel, α is 13 x 10⁻⁶ and for aluminium it is 24 x 10⁻⁶. Let common temperature required be T.

Using the formula for steel

L (steel) = 3 + 3 x 13 x 10⁻⁶ (T - 20)

Using the formula for aluminium

L (Aluminium) = 2.98 + 2.98 x 24 x 10⁻⁶ x (T-20)

L (Steel) = L (Aluminium)

3 + 3 x 13 x 10⁻⁶ (T - 20) = 2.98 + 2.98 x 24 x 10⁻⁶ x (T-20)

T = 635 degree.