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 temperature of ${30^ \circ }C$. At what temperature, the same for the ball and plate, will the ball just pass through the hole? ${\alpha _{brass}} = 19 \times {10^{ - 6}}^ \circ {C^{ - 1}}$ and ${\alpha _{iron}} = 12 \times {10^{ - 6}}^ \circ {C^{ - 1}}$

Answer 1

In this question, the ball has to pass through the plate only with the help of thermal expansion. But, when it is heated together, both the iron ball and plate expand and hence, the thermal expansion of both the ball and plate should be taken into account while calculating the change in the dimensions due to thermal expansion.

Complete step by step answer:
Whenever there is an increase in temperature, there is a tendency in the materials to change it’s shape by increasing its length, area or even volume. Here, we classify these expansions into three broad categories of expansion: i) Linear expansion: Here we consider the increase in length per unit rise in temperature. ii) Superficial expansion: Here we consider the increase in area per unit rise in temperature. iii) Volumetric expansion: Here we consider the increase in volume per unit rise in temperature.
In this problem, since the length is given, we have to consider the linear expansion of the metal.
The formula for the linear expansion is given by :
$\Delta l = l\alpha \Delta T$
where
$\Delta l$ is the change in the length $l$ is the original length at ${0^ \circ }C$ $\alpha$ is called as the coefficient of linear expansion $\Delta T$ is the temperature difference
Since the iron ball has to pass through the brass plate with the hole, let us consider the linear expansion of the combination is equal to the difference between the linear expansion of the plate and the linear expansion of the iron ball.
$\Delta L = \Delta {l_{brass}} - \Delta {l_{iron}}$
After the expansion has happened and the length increases to a point where the iron ball will pass through the hole, the change in length would be equal to the width of the original hole.
Thus,
$\implies$ $\Delta L = 0 \cdot 010mm = 0 \cdot 001cm$
$\implies$ ${\alpha _{brass}} = 19 \times {10^{ - 6}}^ \circ {C^{ - 1}}$ and ${\alpha _{iron}} = 12 \times {10^{ - 6}}^ \circ {C^{ - 1}}$
Substituting them in the equation, we get –
$\implies$ $\Delta L = \Delta {l_{brass}} - \Delta {l_{iron}}$
$\implies$ $0 \cdot 001 = l{\alpha _{brass}}\Delta T - l{\alpha _{iron}}\Delta T$
$\implies 0 \cdot 001 = l\Delta T\left( {{\alpha _{brass}} - {\alpha _{iron}}} \right)$
Here, $l = d = 6cm$ which is the diameter of the iron ball and the hole in the brass plate.
The temperature difference,
$\implies$ $\Delta T = \dfrac{{0 \cdot 001}}{{d\left( {{\alpha _{brass}} - {\alpha _{iron}}} \right)}}$
Substituting, we get –
$\implies$ $\Delta T = \dfrac{{0 \cdot 001}}{{6 \times \left( {19 - 12} \right) \times {{10}^{ - 6}}}}$
Solving,
$\implies$ $\Delta T = 23 \cdot {8^ \circ }C$
The original temperature, ${T_0} = {30^ \circ }C$
Hence, the new temperature to which the iron ball and brass plate must be raised to, is given by,
$T = {T_0} + \Delta T = {30^ \circ } + 23 \cdot {8^ \circ } = 53 \cdot {8^ \circ }C$

Note: Just like we have the coefficient of linear expansion, we have two more coefficients of expansion for area and volume, represented by $\beta$ and $\gamma$ respectively.
The relationship among these coefficients of expansion is given by:
$\dfrac{\alpha }{1} = \dfrac{\beta }{2} = \dfrac{\gamma }{3}$