Question

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.

Answer 1

Given:
Diameter of the steel sphere at temperature (T₁ = 10 °C) , d_st = 2.005 cm
Diameter of the aluminium sphere, d_Al = 2.000 cm
Coefficient of linear expansion of steel, α_st = 11 × 10^$-6$ °C^$-1$
Coefficient of linear expansion of aluminium, α_Al = 23 × 10^$-$6 °C^$-$1​
Let the temperature at which the ball will fall be T_{2 ,} so that change in temperature be ΔT​.
d'_st = 2.005(1 + α_st ΔT)
$$\begin{gathered} \Rightarrow d{\text{'}}_{\mathrm{st}}=2.005+2.005\times 11\times {10}^{-6}\times ∆T \\ d{\text{'}}_{\mathrm{Al}}=2\left(1+{\alpha }_{\mathrm{Al}}\times ∆T\right) \\ \Rightarrow d{\text{'}}_{\mathrm{Al}}=2+2\times 23\times {10}^{-6}\times ∆T \end{gathered}$$
The steel ball will fall when both the diameters become equal.
So, d'_st = d'_Al
$$\begin{gathered} \Rightarrow 2.005+2.005\times 11\times {10}^{-6}∆T=2+2\times 23\times {10}^{-6}∆T \\ \Rightarrow \left(46-22.055\right)\times {10}^{-6}∆T=0.005 \\ \Rightarrow ∆T=\frac{0.005\times {10}^6}{23.945}=208.81 \\ \mathrm{Now},∆T=T_2-T_1=T_2-10°\mathrm{C} \\ \Rightarrow T_2=∆T+T_1=208.81+10 \\ \Rightarrow T_2=218.8\cong 219°\mathrm{C} \end{gathered}$$
Therefore, ​the temperature at which the ball will fall is 219 °C.