Question

An iron ball has a diameter of 6 cm and is 0.010 mm too large to pass through a hole in a brass when the ball and plate are at a temperature of ${30}^{\circ }C$. At what temperature, the same for ball and plate, will the ball just pass through the hole ?

Answer 1

Since the linear expansion is given by the expression $\Delta l=l\alpha \Delta T$

Hence, in the given situation, since both iron ball and brass plate will expand. Hence,

$d({\alpha }_{Brass}-{\alpha }_{Iron})\Delta T=0.01mm$

$60({\alpha }_{Brass}-{\alpha }_{Iron})\Delta T=0.01$

$60(19-12){10}^{-6}(T-30)=0.01$

$T={50.8}^{0}C$

Answer is ${50.8}^{0}C$