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 20⁰C. The temperature at which (both for ball and plate) the ball just pass through the hole is
$\left({\alpha }_{\text{iron }}=12\times {10}^{-6}/1^0\mathrm{C};{\alpha }_{\text{brass }}=18\times {10}^{-6}/1^0\mathrm{C}\right)$