A cylindrical tube, with its base as shown in the figure., is...

Given,

$\beta = \dfrac{{P}_{1} - {P}_{2}}{ρgd} . . . \left(1\right)$

Here we can write now,

${P}_{3} = {P}_{2} + ρ\left(g - \dfrac{a}{\sqrt{2}}\right)d$

${P}_{1} = {P}_{3} - \dfrac{ρa}{\sqrt{2}}d$

${P}_{1} = {P}_{2} + ρgd - \sqrt{2}ρad$

${P}_{1} = {P}_{2} + ρ\left(g - \dfrac{a}{\sqrt{2}}\right)d - \dfrac{ρa}{\sqrt{2}}d$

$\dfrac{{P}_{1} - {P}_{2}}{ρdg} = 1 - \dfrac{\sqrt{2}a}{g}$

Compare it with given equation \left(1\right)$,</p><p>$\beta = 1 - \dfrac{\sqrt{2}a}{g}$</p><p>Now check the options,</p><p>If$a = \dfrac{g}{\sqrt{2}}$,$\beta = 0$.</p><p>If$a = \dfrac{g}{2}$,$\beta = \dfrac{\sqrt{2} - 1}{\sqrt{2}}$.</p><p>Option \left(1\right)$and \left(3\right)$ are correct