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 $74.5cm$ 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
$\Delta l=l{\alpha }_{\text{brass }}(\Delta T)=74.5\times 2\times 10^{-5}(30-15)$
$=0.02235cm$
$\therefore$ Actual reading at $30^{\circ }C$
$l_{30}=l_{\text{observed }}+\Delta l$
$=74.5+0.02235$
$=74.522cm$
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 $0^{\circ }C$
$h_0=h_{30}\frac{{\rho }_{30}}{{\rho }_0}$
$=\frac{h_{30}}{\left(1+{\gamma }_{Hg}\Delta T\right)}$
$=\frac{74.522}{1+0.00018\times 30}$
$=\frac{74.522}{1.0054}=74.122cm$