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Question
We would like to make a vessel whose volume does not change with temperature (take a hint from the problem above). We can use brass and iron `(β_(vbrass) = (6 xx 10^(–5))/K and β_(viron) = (3.55 xx 10^(–5))/K)` to create a volume of 100 cc. How do you think you can achieve this.
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Solution
In the previous problem, the difference in the length was constant.
In this problem the difference in volume is constant.
The situation is shown in the diagram.
Let Vio, Vbo be the volume of iron and brass vessel at 0°C
Vi, Vb be the volume of iron and brass vessel at Δθ°C,
γi, γb be the coefficient of volume expansion of iron and brass.
As per the question, Vio – Vbo = 100 cc = Vi – Vb ......(i)
Now, `V_i = V_(io) (1 + γ_iΔθ)`
`V_b = V_(bo) (1 + γ_bΔθ)`
`V_i - V_b = (V_(io) - V_(bo)) + Δθ(V_(io)γ_i - V_(bo)γ_b)`
Since, `V_i - V_b` = constant
So, `V_(io)γ_i - V_(bo)γ_b`
⇒ `V_(io)/V_(bo) = γ_b/γ_i`
= `(3/2 β_b)/(3/2 β_i)`
= `β_b/β_i`
= `(6 xx 10^-5)/(3.55 xx 10^-5)`
= `6/3.55`
`V_(io)/V_(bo) = 6/3.55` ......(ii)
Solving equations (i) and (ii), we get
Vio = 244.9 cc
Vbo = 144.9 cc
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