A thin tube sealed at both ends is 100 cm lo | vedclass

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(B) Let the initial length of air columns at both ends be $l_0 = (100 - 20) / 2 = 40 \ cm$. Let the mercury column be displaced by $y$ when the tube i

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$5.18$

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(B) Let the initial length of air columns at both ends be $l_0 = (100 - 20) / 2 = 40 \ cm$. Let the mercury column be displaced by $y$ when the tube is held vertically.The pressure of the air in the lower part $(P_1)$ and upper part $(P_2)$ changes due to the compression and expansion of air, respectively. Using Boyle's Law $(PV = 	ext{constant})$ at constant temperature:For the lower part: $P_0 (40 A) = P_1 (40 - y) A \Rightarrow P_1 = \frac{40 P_0}{40 - y}$For the upper part: $P_0 (40 A) = P_2 (40 + y) A \Rightarrow P_2 = \frac{40 P_0}{40 + y}$In the vertical position, the pressure at the lower face is equal to the pressure at the upper face plus the pressure due to the $20 \ cm$ mercury column $(h = 20 \ cm)$:$P_1 = P_2 + h ho g$Substituting $P_0 = 76 \ cm$ of $Hg$ (where $P_0 = ho g 	imes 76$):$\frac{40 	imes 76 ho g}{40 - y} = \frac{40 	imes 76 ho g}{40 + y} + 20 ho g$Dividing by $ho g$:$\frac{3040}{40 - y} - \frac{3040}{40 + y} = 20$$3040 \left( \frac{40 + y - (40 - y)}{1600 - y^2} ight) = 20$$3040 \left( \frac{2y}{1600 - y^2} ight) = 20$$304 y = 1600 - y^2$$y^2 + 304 y - 1600 = 0$Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$y = \frac{-304 + \sqrt{304^2 - 4(1)(-1600)}}{2} \approx 5.18 \ cm$.

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