Ncert Exercises & Solutions

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 $(β_{v b r a s s} = 6 \times 10^{- 5} / K a n d β_{v i r o n} = 3.55 \times 10^{- 5} / k)$ to create a volume of 100 cc. How do you think you can achieve this?

Hint: The change in volume of the two materials should remain the same.

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.

Step 1: Find the change in volumes.

Let

V_{i o}, V_{b o}

be the volumes of iron and brass vessel at 0°C.

V_{i}, V_{b}

be the volumes of iron and brass vessel at

θ ° C

γ_{i}, γ_{b}

be the coefficients of volume expansion of iron and brass.
As per the question,

V_{i o} - V_{b o} = 100 c c = V_{i} - V_{b} . . . (i)

V_{i} = V_{i o} (1 + γ_{i} ∆ θ)

V_{b} = V_{b o} (1 + γ_{b} ∆ θ)

V_{i} - V_{b} = (V_{i o} - V_{b o}) + ∆ θ (V_{i o} γ_{i} - V_{b o} γ_{b})

Step 2: Find the individual volumes.

Since,

V_{i} - V_{b}

= constant,

So,

V_{i o} γ_{i} = V_{b o} γ_{b}

\Rightarrow \frac{V_{i o}}{V_{b o}} = \frac{γ_{b}}{γ_{i}} = \frac{\frac{3}{2} β_{b}}{\frac{3}{2} β_{l_{}}} = \frac{6 \times 10^{- 5}}{3.55 \times 10^{- 5}} = \frac{6}{3.55}

\frac{V_{i o}}{V_{b o}} = \frac{6}{3.55} . . . (i i)

Solving Eqs. (i) and (ii), we get,

V_{i o} = 244.9 c c

V_{b o} = 144.9 c c

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