Question 13: Thermodynamics & KTG

Question

The coefficient of cubical expansion of mercury is $$0.00018 /{ }^{\circ} \mathrm{C}$$ and that of brass $$0.00006 /{ }^{\circ} \mathrm{C}$$. If a barometer having a brass scale were to read $$74.5 \mathrm{~cm}$$ at $$30^{\circ} \mathrm{C}$$, find the true barometric height at $$0^{\circ} \mathrm{C}$$. The scale is supposed to be correct at $$15^{\circ} \mathrm{C}$$.

Accepted Answer

$$\begin{aligned}
	ext { As, } \alpha_{	ext {brass }} & =\frac{\gamma_{	ext {brass }}}{3} \
& =\frac{0.00006}{3}=0.00002 \
& =2 	imes 10^{-5} /{ }^{\circ} \mathrm{C}
\end{aligned}$$
The brass scale is true at $$15^{\circ} \mathrm{C}$$, therefore at $$30^{\circ}$$ its graduations will increase in length and so observed reading will be less than actual reading at $$30^{\circ}$$.
$$	herefore$$ The change in reading
$$\begin{aligned}
\Delta l & =l \alpha_{	ext {brass }}(\Delta T)=74.5 	imes 2 	imes 10^{-5}(30-15) \
& =0 \cdot 02235 \mathrm{~cm}
\end{aligned}$$
$$	herefore$$ Actual reading at $$30^{\circ} \mathrm{C}$$
$$\begin{aligned}
l_{30} & =l_{	ext {observed }}+\Delta l \
& =74.5+0.02235 \
& =74.522 \mathrm{~cm}
\end{aligned}$$
Assuming area of cross-section to be constant, we have
$$V_0 ho_0=V_{30} ho_{30} 	ext { or } a h_0 ho_0=a h_{30} ho_{30}$$
Therefore, True height at $$0^{\circ} \mathrm{C}$$
$$\begin{aligned}
h_0 & =h_{30} \frac{ho_{30}}{ho_0}=\frac{h_{30}}{\left(1+\gamma_{\mathrm{Hg}} \Delta Tight)} \
& =\frac{74.522}{1+0.00018 	imes 30} \
& =\frac{74.522}{1.0054}=74.122 \mathrm{~cm}
\end{aligned}$$

Answer

$$74.122 \mathrm{~cm}$$

Answer

$$79.152 \mathrm{~cm}$$

Answer

$$42.161 \mathrm{~cm}$$

Answer

$$142.39 \mathrm{~cm}$$

Thermodynamics & KTG

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