An aluminium plate fixed in a horizontal position has a hole of diameter 2.000 cm. A steel sphere of diameter 2.005 cm rests on this hole. All the lengths refer to a temperature of 10°C. The temperature of the entire system is slowly increased. At what temperature will the ball fall down? Coefficient of linear expansion of aluminium is 23 × 10–6 °C–1 and that of steel is 11 × 10–6 °C–1.
Given:
Diameter of the hole in the aluminium plate : dAl= 2.000 cm
Diameter of the Steel sphere resting on the hole: dst= 2.005 cm
Initial temperature : T1= 10 ° C
Coefficient of linear expansion of aluminium : α Al =23 × 10–6 °C–1
Coefficient of linear expansion of Steel : αst = 11 × 10–6 °C–1
Now, when the temperature of the entire system is increased thermal expansion will take place.
Formula used:
The formula for linear expansion is:
where L’ is the changed length.
Similarly, change in diameter of aluminium plate is:
change in diameter of Steel sphere is:
Where, 
and 
is the changed diameter of aluminum plate and steel sphere respectively.
Now, for the steel ball to fall through the hole the changed diameter of the steel ball and the aluminum plate should be equal.
Now, ΔT = T1 -T2
Hence the temperature at which the ball will fall down is T2 = 218.811 ° C