We would like to make a vessel whose volume does not change with … - Vidyadip

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 $\big(\beta_\text{ubrass}=6\times10^{-5}/\text{K and }\beta_\text{uiron}=3.55\times10^{-5}\text{K}\big)$ to create a volume of 100cc. How do you think you can achieve this.

✓

Answer

Volume of vessel, V₀ = 100cm³

= 10^-4 = constant

Volume of iron vessel (V_I) - Volume of brass rod (V_B) = 10^-4m³

⇒ V_I - V_B = 10^-4m³

This condition is possible if

$\beta_\text{I}\text{V}_\text{I}\Delta\text{T}=\beta_\text{B}\text{V}_\text{B}\Delta\text{T}$

$\therefore\ \text{V}_{\text{I}}=\Big(\frac{\beta_\text{B}}{\beta_\text{I}}\Big)\text{V}_\text{B}$

$=\Big(\frac{6\times100^{-5}}{3.55\times10^{-5}}\Big)\text{V}_\text{B}$

$=1.69\text{V}_\text{B}$

From equation (i),

$1.69\text{V}_\text{B}-\text{V}_\text{B}=10^{-4}$

$\Rightarrow\text{V}_\text{B}=\frac{10^{-4}}{0.69}=1.449\times10^{-4}\text{m}^3$

$\Rightarrow\text{V}_\text{B}=\frac{10^{-4}}{0.69}=1449\times10^{-4}\text{m}^3$

$\Rightarrow\text{V}_\text{I}=1.69\text{V}_\text{B}=1.69\times144.9=244.9\text{cm}^3$

Therefore, an iron vessel with a volume of 249.9cm³ fitted with a brass rod of volume 144.9cm³ will serve as a vessel of volume 100cm³, which will not change with temperature.

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