An iron ring measuring 15.00 cm in diameter is a be shrunk on a pulley which is 15.05 cm in diamer. All measurements refer to the room temperature `20^0 C`. To What minimum temperature should the ring be heated to make the job possible? Calculate the strain developed in the ring when in comes to the room temperature. Coefficient of linear expansion of iron `=12 xx 10^(-6) ^0 C^(-1)`.

Explanation

The ring should be heated to increase its diameter form `15.00 cm` to `15.05cm`.
Using `l_(2) = l_(1) (1 + alpha Delta theta)`,
`= (0.05 cm)/(15.00 cm xx 12 xx 10^(-6)//"^(@)C) = 278^(@)C`
The temperature ` =20^(@)C +278^(@)C =298^(@)C`
The strain developed `= (l_(2)-l_(1))/(l_(1)) = 3.33 xx 10^(-3)`.