Question

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

• A. 74.122 cm
• B. 79.152 cm
• C. 42.161 cm
• D. 142.39 cm

Answer 1

Answer: (A)

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