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

The ball fall down at  the temperature of $\bold{219 ^{\circ} C}$

Explanation:

Metal Sphere Diameter at Temperature ($T_1$ = 10 °C) , $d_{st}$ = 2.005 cm

Aluminium sphere diameter,$d_{Al}$ = 2.000 cm

Linear expansion ratio for steel, αst  =$11 \times 10^{-6}$ $^{\circ}C^{-1}$

Linear expansion coefficient for aluminium, $\alpha _{Al}$  $= 23 \times 10^{-6} ^{\circ}C^{-1}$

Let the temperature at which the ball drops be $T_2$ so that the temperature changes are equal to  ΔT

$\mathrm{d}_{\mathrm{st}}^{\mathrm{ \prime}}=2.005\left(1+\alpha_{\mathrm{st}} \Delta \mathrm{T}\right)$

$\mathrm{d}_{\mathrm{st}}^{\prime}=2.005+2.005 \times 11 \times 10^{-6} \times \Delta T$

$d_{A l}^{\prime}=2\left(1+\alpha_{A l} \times \Delta T\right)$

$\mathrm{d}_{\mathrm{Al}}^{\prime}=2+2 \times 23 \times 10^{-6} \times \Delta \mathrm{T}$

When both the diameters are equal, the steel ball will fall.

So,    $d^{\prime} \text { st }=d^{\prime} A$

$2.005+2.005 \times 11 \times 10^{-6} \times \Delta T &=2+2 \times 23 \times 10^{-6} \times \Delta T$

$2.005+2.2055 \times 10^{-5} & \times \Delta T=2+4.6 \times 10^{-5} \times \Delta T$

$2.005-2=4.6 \times 10^{-5} \times \Delta T-2.2055 \times 10^{-5} \times \Delta T$

$2.3945 \times 10^{-5} \times \Delta T=5 \times 10^{-3}$

$\Delta T=\frac{5 \times 10^{-3}}{2.3945 \times 10^{-5}}$

$\Delta T=\frac{5 \times 10^{2}}{2.3945}$

$\Delta T=208.81$

$&\Delta \mathrm{T}=\mathrm{T}_{2}-\mathrm{T}_{1}=\mathrm{T}_{2}-10^{\circ} \mathrm{C}$

$&\mathrm{T}_{2}=\Delta \mathrm{T}+\mathrm{T}_{1}=208.81+10$

$&\mathrm{T}_{2}=218.8 \cong 219^{\circ} \mathrm{C}$