Question

A ring made from aluminum has an inner radius of 2.50000 cm and an outer radius of 3.50000 cm, giving the ring a thickness of 1.00000 cm. The thermal expansion coefficient of aluminum is 23.0 ⨯ 10-6/°C. If the temperature of the ring is increased from 20.0°C to 75.0°C, by how much does the thickness of the ring change? (answer in cm)

Answer 1

Given
aluminium ring of inner and outer radius are 2.50 cm ,3.5 cm respectively

the Initial Thickness of the ring is l1 = 1.00000 cm , final thickness l2 = ?

the temperature difference is dT = 75.0 0C - 20.0 0C = 55 0C

we know that the linear expansion of the materials upon heating is

the change in length is proportional to the difference in temperature and on its initial length, combining these

   DL = L1*alpha*dT

   l2-l1 = l1*alpha*dT

   l2 = l1(1+alpha*dT)

   l2 = 1.000*10^-2 (1+23.0*10^-6 (55)) m

   l2 = 0.01001265 m

   l2 = 1.001265 cm

so the change in Thickness of the ring is l2-l1 = 1.001265-1.000000 cm =  0.001265 cm