Question

A metal washer has a hole of diameter ${d_1}$​ and an external diameter ${d_2}$​, where ${d_2} = 3{d_1}$​. On heating, ​ ${d_2}$ increases by $0.3\,\% $. Then ${d_1}$​ will increase or decrease by what percentage
(A) decrease by $0.1\,\% $.
(B) decrease by $0.3\,\% $.
(C) increase by $0.1\,\% $.
(D) increase by $0.3\,\% $.

Answer 1

Hint: The given problem can be solved by using the formula that incorporates the thermal expansion, in which the increase in the length of the diameter and the percentage in the length of the diameter with respect to change in the temperature of each of the diameter of the metal washer.

Formulae Used:
Increase in the length of the diameter of a metal washer is given by;
$\Delta d = d \times \alpha \times \Delta T$
Where, $d$ denotes the diameter of the hole in metal washer, $\Delta T$ denotes the temperature in the metal washer and $\alpha $denotes the linear thermal coefficient of expansion.

Complete step-by-step solution:
The data given in the problem is;
Internal diameter of the hole in metal washer is, ${d_1}$,
External diameter of the hole in metal washer is, ${d_2}$,
${d_2} = 3{d_1}$.
Increase in the length of the internal diameter of a metal washer is;
$\Delta {d_1} = {d_1} \times \alpha \times \Delta T$
Percentage of Increase in the length of the internal diameter, ${d_1}$;
$\dfrac{{\Delta {d_1}}}{{{d_1}}} \times 100 = \alpha \times \Delta T \times 100\,..........\left( 1 \right)$
Increase in the length of the external diameter of a metal washer is;
$\Delta {d_2} = {d_2} \times \alpha \times \Delta T$
Percentage of Increase in the length of the internal diameter, ${d_1}$;
$\dfrac{{\Delta {d_2}}}{{{d_2}}} \times 100 = \alpha \times \Delta T \times 100\,\,..........\left( 2 \right)$
Since we already know that the,
${d_2} = 3{d_1}$
Substitute the value of ${d_2}$ in the equation (2);
$3 \times \dfrac{{\Delta {d_1}}}{{{d_1}}} \times 100 = \alpha \times \Delta T \times 100$
Substitute the value of equation (1) in the equation (2);
That is, we get;
$3 \times \left( {\alpha \times \Delta T \times 100} \right) \times 100 = \alpha \times \Delta T \times 100$
In the equation we get $0.3\,\% $.
Therefore, the percentage increase in inner and outer diameter is the same and is equal to $0.3\,\% $.
Hence, the option (D) increase by $0.3\,\% $ is the correct answer.

Note:- Thermal expansion is the rise or fall of the size of length or area or volume of the body due to a difference in temperature. Thermal expansion is greater for gases and rather small for liquids and solids. Linear thermal growth is $\Delta l = l \times \alpha \times \Delta T$, where $\Delta L$ is the change in length $L$, $\Delta T$ is the change in temperature, and $\alpha $ is the coefficient of linear expansion, which differs greatly with temperature.