An iron ball of diameter `6cm` and is `0.01 mm` too large to pass through a hole in a brass plate when the ball and plate are at a temperature of `20^(0)C`. The temperature at which (both for ball and plate) the ball just passes through the hole is
`(alpha_(iron) = 12 xx 10^(-6 //0)C, alpha_(brass) = 18 xx 10^(-6 //0)C)`

Options

A. `68^(0)C`
B. `48^(0)C`
C. `28^(0)C`
D. `40^(0)C`

Answer

B

Explanation

`d'_(iron) = d_(brass)` (after heating)
`d_(iron)(1 + alpha_(iron)Deltat) = d_(brass)(1 + alpha_(brassDeltat)`