Question

A narrow tube of length one meter is lying horizontal, with a mercury column of length 10 cm at the center. Both the ends of the tube are closed and, contains air at atmospheric pressure. If now the tube is turned vertical,then find the downward displacement of the mercury column.

• A. 2.96 cm
• B. 3.45 cm
• C. 1.24 cm
• D. 4.45 cm

Answer 1

The correct option is A 2.96 cm


From the given data $\lambda$ = 45 cm, h = 10 cm,
P = 760 mm, Hg = 76 cm Hg
Now, since temperature does not change (room temperature)
$P_1(\lambda +x)\left(area\right)=P\lambda \left(area\right)P_1(\lambda +x)=P\lambda ...\left(1\right)Similarly{P}_2(\lambda -x)=P\lambda ...\left(2\right)andalso{P}_2-P_1=\frac{\text{(mass of mercury column)g}}{\text{(area)}}=\frac{A.h.\left(density\right)g}{A}=h\rho g=hcmofHg...\left(3\right)\text{From (1) and (2)}{P}_2-P_1=P\lambda (\frac{1}{l-x}-\frac{1}{l+x})=\frac{Pl2x}{l^2-x^2}$
Assuming x << $\lambda$, we write ${\lambda }^2-x^2\approx {\lambda }^2$
$\therefore P_2-P_1=\frac{2Px}{l}$
$\therefore x=\frac{(P_2-P-1)l}{2P}=\frac{hl}{2P}=\frac{10\times 45}{2\times 76}\left(cm\right)$
= 2.96 cm