Question

A uniform tube closed at one end, contains a pellet of mercury 10 cm long. When the tube is kept vertically with the closed-end upward, the length of the air column trapped is 20 cm. Find the length of the air column trapped when the tube is inverted so that the closed-end goes down. Atmospheric pressure = 75 cm of mercury.

Answer 1

Thus the length of the air column is h = 15 cm

Explanation:

Case 1: Net pressure on air in volume

V=(P atm−hρg)

= 75 × (ρHg)×g − [10(ρHg)]×g

= ρHg × g × 65

Case 2: Net pressure on air in volume 'V'.

=P(atm) + (ρHg)×g×h

(P1) (V1) = (P2) (V2)

= (ρHg) × g × 65 × A×20

= [(ρHg) × g × 75 +(ρHg) × g × 10] A×h

= 65+20

= 85 h

h = (65×2085)

=15.2 cm  = 15 cm

Thus  the length of the air column is h = 15 cm

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Answer 2

The length of the air column trapped is 15 cm.

Explanation:

Let the cross sectional area of the tube be A.

Initial volume of air  , $V_1 = 20A cm = 0.2A$

Mercury’s length , h = 0.1 m

Let the trapped air pressure be $P_1$  when the tube is turned inverted and vertical.

Already ambient pressure is balanced by mercury vapor and compressed dust. Accordingly,

$\begin{aligned}&\mathrm{P}_{1}+0.1 \mathrm{Pg}=0.75 \rho \mathrm{g}\\&\mathrm{P}_{1}=0.65 \rho \mathrm{g}\end{aligned}$

The pressure acting on the trapped air is given when the tube is inverted with the closed end down  by Atmospheric pressure + Column pressure of mercury

Trapped air pressure = Atmospheric pressure + Column pressure of mercury ( In equilibrium)

$\mathrm{P}_{2}=0.75 \rho \mathrm{g}+0.1 \mathrm{\rhog}=0.85 \mathrm{\rhog}$

If the temperature remains constant we apply the Boyle's law

$\mathrm{P}_{1} \mathrm{V}_{1}=\mathrm{P}_{2} \mathrm{V}_{2}$

Let the trapped air be new height x.

$\begin{aligned}&0.65 \rho g \times 0.2 A=0.85 \rho g x A\\&0.65 \times 0.2=0.85 x\end{aligned}$

0.65×0.2 = 0.85 x  

$\begin{aligned}&\frac{0.65 \times 0.2}{0.85}=x\\&\frac{0.13}{0.85}=x\end{aligned}$

x = 0.15 m

x = 15 cm            

Therefore 15 cm is the length of the air column trapped when the tube is inverted so that the closed end goes down for a uniform tube closed at on end which contains a pellet of mercury 10 cm long.