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Question: The inner diameter of a brass ring at\({\text{ }}273K{\text{ }}\)is \({\text{ }}5cm{\text{ }}\). To ...

The inner diameter of a brass ring at 273K {\text{ }}273K{\text{ }}is  5cm {\text{ }}5cm{\text{ }}. To what temperature should it be heated for it to accommodate a ball  5.01 cm {\text{ }}5.01{\text{ }}cm{\text{ }}in diameter? (α=2×10−5/∘C)\alpha = 2 \times {10^{ - 5}}{/^ \circ }C)
A) 273K273K
B) 373K373K
C) 437K437K
D) 173K173K

Explanation

Solution

The question involves the thermal properties of matter. We have to find the temperature required to expand the ring of  5 cm {\text{ }}5{\text{ }}cm{\text{ }} diameter to accommodate a ring of the radius  5.01 cm {\text{ }}5.01{\text{ }}cm{\text{ }}. When solids are heated, the length of the solid, the surface area and volume will increase. The increase in length of the solid for the rise in temperature is called linear expansion.

Formula used:
α=ΔllΔT\alpha = \dfrac{{\Delta l}}{{l\Delta T}}
Where α\alpha stands for the coefficient of linear expansion.
l l{\text{ }} stands for the original length of the material.
Δl \Delta l{\text{ }} is the change in length for the temperature rise.
ΔT \Delta T{\text{ }} is the temperature rise.

Complete step by step solution:
Let us take the inner radius of the ring as  d1=5cm{\text{ }}{d_1} = 5cm
The diameter of the ball to be accommodated inside the ring  d2=5.01cm{\text{ }}{d_2} = 5.01cm
The difference between the diameters  d2−d1=Δl=5.01−5=0.01cm{\text{ }}{d_2} - {d_1} = \Delta l = 5.01 - 5 = 0.01cm
The coefficient of linear expansion is defined as the ratio of change in length to the original length for a one-degree rise of temperature.
The coefficient of linear expansion, α=ΔllΔTK−1{\text{ }}\alpha = \dfrac{{\Delta l}}{{l\Delta T}}{K^{ - 1}}
From this equation,  ΔT=ΔlαlK{\text{ }}\Delta T = \dfrac{{\Delta l}}{{\alpha l}}K
Here, l=5×10−2m {\text{ }}l = 5 \times {10^{ - 2}}m{\text{ }}, Δl=0.01×10−2m {\text{ }}\Delta l = 0.01 \times {10^{ - 2}}m{\text{ }}, α=2×10−5/∘C{\text{ }}\alpha = 2 \times {10^{ - 5}}{/^ \circ }C
Substituting the values in the above equation,
ΔT=0.01×10−22×10−2×5×10−5K\Delta T = \dfrac{{0.01 \times {{10}^{ - 2}}}}{{2 \times {{10}^{ - 2}} \times 5 \times {{10}^{ - 5}}}}K
ΔT=100K\Delta T = 100K
The initial temperature is  T {\text{ }}T{\text{ }}= 273K273K
The temperature at which the ring is heated to accommodate the ball is given by, TF=T+ΔT{\text{ }}{T_F} = T + \Delta T
TF=273+100=373K{T_F} = 273 + 100 = 373K

The answer is Option (B):  373K{\text{ }}373K.

Note: The increase in surface area of a solid when heated is called superficial expansion. The increase in volume is called the cubical expansion. The coefficient of superficial expansion is defined as the ratio of change in surface area to the original surface area for the  1∘C {\text{ }}{1^ \circ }C{\text{ }} rise of temperature. The coefficient of cubical expansion is defined as the ratio of change in volume to the original volume for  1∘C {\text{ }}{1^ \circ }C{\text{ }} the rise of temperature.