Question

A brass disc at 0°C has a diameter of 50 cm and a hole of diameter 10 cm. Find the diameter of the hole when the disc is heated to 100°C. (For brass, $\alpha = 18 \times {10^{ - 6}}/{\text{C}}$ ).
A) 10.018 cm
B) 10.18 cm
C) 10.8 cm
D) 10.81 cm

Answer 1

Hint: As the disc is heated, the dimensions of the disc change causing the disc to expand. This change in the length of a material when it is heated is referred to as linear expansion. The change in the diameter of the hole will be proportional to the change of temperature while heating.

Formula Used:
Linear expansion of a material is given by, $\delta L = {L_0}\alpha \left( {{T_1} - {T_0}} \right)$ where ${L_0}$ is the original length of the sample, $\alpha $ is the coefficient of linear expansion, ${T_1}$ is the final temperature and ${T_0}$ is the initial temperature.

Complete step by step answer:
Step 1: List the parameters known from the question.
At ${T_0} = 0^\circ {\text{C}}$, the diameter of the hole is ${d_0} = 10{\text{cm}} = 0.1{\text{m}}$ .
The diameter of the disc is 50 cm.
The disc is heated to ${T_1} = 100^\circ {\text{C}}$ .
The coefficient of linear expansion for brass is $\alpha = 18 \times {10^{ - 6}}/{\text{C}}$ .
Step 2: Express the relation for linear expansion of the brass disc.
Linear expansion of a material is given by, $\delta L = {L_0}\alpha \left( {{T_1} - {T_0}} \right)$ where ${L_0}$ is the original length of the sample, $\alpha $ is the coefficient of linear expansion of the material, ${T_1}$ is the final temperature and ${T_0}$ is the initial temperature.
Based on the general relation, the change in the diameter of the hole can be expressed as $\delta d = {d_0}\alpha \left( {{T_1} - {T_0}} \right)$ --------- (1)
Step 3: Find the new diameter of the hole using equation (1).
Equation (1) gives the change in the hole diameter on heating as $\delta d = {d_0}\alpha \left( {{T_1} - {T_0}} \right)$ .
Substituting values for ${d_0} = 0.1{\text{m}}$, ${T_0} = 0^\circ {\text{C}}$, ${T_1} = 100^\circ {\text{C}}$ and $\alpha = 18 \times {10^{ - 6}}/{\text{C}}$ in the above equation we get, $\delta d = 0.1 \times 18 \times {10^{ - 6}} \times \left( {100 - 0} \right) = 0.018{\text{cm}}$
Thus the change in the diameter of the hole is positive and the diameter of the hole has an increase of 0.018 cm.
So, the new diameter of the hole will be ${d_1} = {d_0} + \delta d$
Substituting values for ${d_0} = 10{\text{cm}}$ and $\delta d = 0.018{\text{cm}}$ in the above relation we get, ${d_1} = 10 + 0.018 = 10.018{\text{cm}}$
Thus the diameter of the hole at ${T_1} = 100^\circ {\text{C}}$ is ${d_1} = 10.018{\text{cm}}$
So, the correct option is A.

Note: The S.I. unit of temperature is Kelvin. Here the temperature is expressed in the units of degree Celsius. Conversion of the unit is not necessary since the temperature difference $\left( {{T_1} - {T_0}} \right)$ is considered. The temperature difference will be the same irrespective of the unit in which it is expressed. We can check if it is true, ${T_0} = 0^\circ {\text{C}} = 273{\text{K}}$ and ${T_1} = 100^\circ {\text{C}} = 373{\text{K}}$ then $\left( {{T_1} - {T_0}} \right) = 373 - 273 = 100{\text{K}}$. The value remains the same. If the change in diameter $\delta d$ is negative it implies that the diameter has decreased.