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Question: A steel sphere is to be passed through a circular brass ring at \[20^\circ {\rm{C}}\] the outer diam...

A steel sphere is to be passed through a circular brass ring at 20∘C20^\circ {\rm{C}} the outer diameter of sphere is 25cm25{\rm{ cm}} and the inner diameter of the ring is 24.9cm24.9{\rm{ cm}}. If both are heated together, then find at which temperature the sphere will just pass through the ring. Take coefficient of linear expansion of steel =12×10−5/∘C = 12 \times {10^{ - 5}}{\rm{ /}}^\circ {\rm{C}} and that of brass =20×10−6/∘C = 20 \times {10^{ - 6}}{\rm{ /}}^\circ {\rm{C}}.
A. 475∘C475^\circ {\rm{C}}
B. 525∘C525^\circ {\rm{C}}
C. 575∘C575^\circ {\rm{C}}
D. 625∘C625^\circ {\rm{C}}

Explanation

Solution

We will use the concept of an increase in length when a body is subjected to the rise in temperature. The increase in diameter of sphere and ring can be expressed as the product of initial diameter, change in temperature and the corresponding value of the coefficient of linear expansion.

Complete step by step answer:
Given:
The initial temperature of the sphere and the circular ring is T1=20∘C{T_1} = 20^\circ {\rm{C}}.
The initial outer diameter of the sphere is D1=25cm{D_1} = 25{\rm{ cm}}.
The initial inner diameter of the ring is d1=24.9cm{d_1} = 24.9{\rm{ cm}}.
The coefficient of linear expansion of steel is αs=12×10−5/∘C{\alpha _s} = 12 \times {10^{ - 5}}{\rm{ /}}^\circ {\rm{C}}.
The coefficient of linear expansion of brass is αb=20×10−6/∘C{\alpha _b} = 20 \times {10^{ - 6}}{\rm{ /}}^\circ {\rm{C}}.
We have to find the value of temperature at which the sphere will pass through the ring.
Let us write the expression for the increased outer diameter of the sphere when it is subjected to a higher temperatureT2{T_2}.
D2=D1[1+αs(T2−T1)]{D_2} = {D_1}\left[ {1 + {\alpha _s}\left( {{T_2} - {T_1}} \right)} \right]
On substituting 25cm25{\rm{ cm}} for D1{D_1}, 12×10−5/∘C12 \times {10^{ - 5}}{\rm{ /}}^\circ {\rm{C}} for αs{\alpha _s} and 20∘C20^\circ {\rm{C}} for T2{T_2} in the above expression, we get:
D2=25cm[1+12×10−5/∘C(T2−20∘C)]{D_2} = 25{\rm{ cm}}\left[ {1 + 12 \times {{10}^{ - 5}}{\rm{ /}}^\circ {\rm{C}}\left( {{T_2} - 20^\circ {\rm{C}}} \right)} \right]….…(1)
The expression for the increased inner diameter of the ring when is subjected to temperature T′T' is given as:
d2=d1[1+αb(T′−T1)]{d_2} = {d_1}\left[ {1 + {\alpha _b}\left( {T' - {T_1}} \right)} \right]
On substituting 24.9cm24.9{\rm{ cm}} for D1{D_1}, 20×10−6/∘C20 \times {10^{ - 6}}{\rm{ /}}^\circ {\rm{C}} for αs{\alpha _s} and 20∘C20^\circ {\rm{C}} for T2{T_2} in the above expression, we get:
d2=24.9cm[1+20×10−6/∘C(T′−20∘C)]{d_2} = 24.9{\rm{ cm}}\left[ {1 + 20 \times {{10}^{ - 6}}{\rm{ /}}^\circ {\rm{C}}\left( {T' - 20^\circ {\rm{C}}} \right)} \right]
The sphere can pass through the ring only when they are subjected to the same temperature that means T2{T_2} must be equal to T’. Therefore, we will substitute T2{T_2} for T’ in the above expression.
d2=24.9cm[1+20×10−6/∘C(T2−20∘C)]{d_2} = 24.9{\rm{ cm}}\left[ {1 + 20 \times {{10}^{ - 6}}{\rm{ /}}^\circ {\rm{C}}\left( {{T_2} - 20^\circ {\rm{C}}} \right)} \right]….…(2)
Sphere will pass through the ring when its outer diameter becomes equal to the inner diameter of the ring so we can equate the right-hand sides of equation (1) and equation (2).

25{\rm{ cm}}\left[ {1 + 12 \times {{10}^{ - 5}}{\rm{ /}}^\circ {\rm{C}}\left( {{T_2} - 20^\circ {\rm{C}}} \right)} \right] = 24.9{\rm{ cm}}\left[ {1 + 20 \times {{10}^{ - 6}}{\rm{ /}}^\circ {\rm{C}}\left( {{T_2} - 20^\circ {\rm{C}}} \right)} \right]\\\ \left[ {1 + 12 \times {{10}^{ - 5}}{\rm{ /}}^\circ {\rm{C}}\left( {{T_2} - 20^\circ {\rm{C}}} \right)} \right] = 0.9\left[ {1 + 20 \times {{10}^{ - 6}}{\rm{ /}}^\circ {\rm{C}}\left( {{T_2} - 20^\circ {\rm{C}}} \right)} \right]\\\ 1 + 12 \times {10^{ - 5}}{\rm{ /}}^\circ {\rm{C}}\left( {{T_2} - 20^\circ {\rm{C}}} \right) = 0.9 + 1.992 \times {10^{ - 5}}{\rm{ /}}^\circ {\rm{C}}\left( {{T_2} - 20^\circ {\rm{C}}} \right)\\\ 0.1 + 3960 \times {10^{ - 6}} = 198 \times {10^{ - 6}}{T_2}/^\circ {\rm{C}} \end{array}$$ On further solving the above expression, we get, $$\begin{array}{c} {T_2} = \dfrac{{0.10396}}{{198 \times {{10}^{ - 6}}/^\circ {\rm{C}}}}\\\ = 525.05^\circ {\rm{C}} \end{array}$$ Therefore, $$525.05^\circ {\rm{C}}$$ is the temperature at which the sphere will just pass through the ring. Hence, option (B) is correct. **Note:** We have to take extra care while rearranging the final expression because it is a bit lengthy. Do not confuse the final temperature of the sphere and ring because to pass the sphere through the ring they should be subjected to the same temperature.