Solution 18
1. Initial State:
The tube of length $h$ is open at both ends and half submerged. The pressure inside the tube is atmospheric ($P_0$) because it is open.
The tube of length $h$ is open at both ends and half submerged. The pressure inside the tube is atmospheric ($P_0$) because it is open.
- Length of air column, $l_1 = h/2$
- Pressure of air, $P_1 = P_0 = \rho g h_0$ (given atmospheric pressure is $h_0$ of Hg)
2. Final State:
The top is closed, and the tube is slowly pulled out. Let $x$ be the length of the mercury column remaining in the tube.
The top is closed, and the tube is slowly pulled out. Let $x$ be the length of the mercury column remaining in the tube.
- Length of air column, $l_2 = h – x$
- Pressure of air, $P_2 = P_0 – \rho g x = \rho g (h_0 – x)$
3. Applying Boyle’s Law (Isothermal Process):
Since temperature is constant ($PV = \text{constant}$): $$ P_1 l_1 = P_2 l_2 $$ Substitute the values in terms of Hg column height: $$ h_0 \left(\frac{h}{2}\right) = (h_0 – x)(h – x) $$ $$ \frac{h_0 h}{2} = h_0 h – h_0 x – hx + x^2 $$ Rearranging into a quadratic equation for $x$: $$ x^2 – (h_0 + h)x + \frac{h_0 h}{2} = 0 $$
Since temperature is constant ($PV = \text{constant}$): $$ P_1 l_1 = P_2 l_2 $$ Substitute the values in terms of Hg column height: $$ h_0 \left(\frac{h}{2}\right) = (h_0 – x)(h – x) $$ $$ \frac{h_0 h}{2} = h_0 h – h_0 x – hx + x^2 $$ Rearranging into a quadratic equation for $x$: $$ x^2 – (h_0 + h)x + \frac{h_0 h}{2} = 0 $$
4. Solving the Quadratic:
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$: $$ x = \frac{(h_0 + h) \pm \sqrt{(h_0 + h)^2 – 4(1)(\frac{h_0 h}{2})}}{2} $$ $$ x = \frac{h_0 + h \pm \sqrt{h_0^2 + h^2}}{2} $$ Since the length of mercury $x$ must be less than the tube length $h$, we must select the negative root: $$ x = \frac{h_0 + h – \sqrt{h_0^2 + h^2}}{2} $$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$: $$ x = \frac{(h_0 + h) \pm \sqrt{(h_0 + h)^2 – 4(1)(\frac{h_0 h}{2})}}{2} $$ $$ x = \frac{h_0 + h \pm \sqrt{h_0^2 + h^2}}{2} $$ Since the length of mercury $x$ must be less than the tube length $h$, we must select the negative root: $$ x = \frac{h_0 + h – \sqrt{h_0^2 + h^2}}{2} $$