A thin tube sealed at both ends is $100\, cm$ long. It lies horizontally, the middle $20\, cm$ containing mercury and two equal ends containing air at standard atmospheric pressure . If the tube is now turned to a vertical position, by what amount will the mercury be displaced ? (Given : cross-section of the tube can be assumed to be uniform) ........ $cm$
JEE MAIN 2013, Diffcult
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Lengths of $M$ and $N$ are $(40-y)$ and $(40+y)$ respecively.
$P_1=$ Preessure of par $M$
$P_2=$ Pressure of part $N$
As temperature is constant
Therefore, we can write $PV =$ constant
For $M : P_0(40 A)=P_1(40-y) A$
$\Rightarrow P_1=\frac{P_0(40)}{(40-y)} \ldots \ldots .(\text { i })$
For $N : P_0(40 A)=P_2(40+y) A$
$\Rightarrow P_2=\frac{P_0 40}{(40+y)}$
Here $A=$ area of cross section of tube
Pressure at lower face $\left(P_1\right)=$ Pressure at upper face $\left(P_2\right)+$ Pressure due to mercury column of height $20 cm$.
$P_1=P_2=20 pg$
Use (i) and (ii) in (iii)
$P_0 \frac{40}{(40-y)}=\frac{P_0 40}{(40+y)}+20 p g$
$\frac{40 P_0}{(40-y)}-\frac{40 P_0}{(40+y)}=20 p g$
$40 \times 76 p g\left[\frac{2 y}{1600-y^2}\right]=20 p g$
$(52)(2 y)=1600-y^2$
$304 y=1600-y^2$
$y^2+304 y-1600=0$
Solving for $y$, we get $y=5.18\; cm$
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