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)      

\[\Rightarrow d '_{st} = 2 . 005 + 2 . 005 \times 11 \times {10}^{- 6} \times ∆ T\]

\[ d '_{Al} = 2\left( 1 + \alpha_{Al} \times ∆ T \right)\]

\[ \Rightarrow d '_{Al} = 2 + 2 \times 23 \times {10}^{- 6} \times ∆ T\]

The steel ball will fall when both the diameters become equal.
So, d'_st = d'_Al

\[\Rightarrow 2 . 005 + 2 . 005 \times 11 \times {10}^{- 6} ∆ T = 2 + 2 \times 23 \times {10}^{- 6} ∆ T\]

\[ \Rightarrow \left( 46 - 22 . 055 \right) \times {10}^{- 6} ∆ T = 0 . 005\]

\[ \Rightarrow ∆ T = \frac{0 . 005 \times {10}^6}{23 . 945} = 208 . 81\]

Now , ΔT = T₂ -T₁ =T₂-10 ° C

\[ \Rightarrow T_2 = ∆ T + T_1 = 208 . 81 + 10\]

⇒ T₂ = 218.8 ≅ 219°C

Therefore, ​the temperature at which the ball will fall is 219 °C.