Question

A glass tube in the form of an equilateral triangle of uniform cross section is as shown in figure. It lies in the vertical plane, with base horizontal. The tube is filled with equal volumes of three immiscible liquids whose densities are in arithmetic progression. Determine the length $x$ as shown in figure.

• A. $3\,l$
• B. $2\,l$
• C. $l/3$
• D. $l/2$

Answer 1

Answer: (C)

Pressure at A,
Due to left arm, $p=dg \left(l-x\right)sin \,60^{\circ}+\left(d+2y\right)gx \, sin 60^{\circ}$
Due to right arm, $P = dgx \, sin\,60^{\circ}+\left(d+y\right)g\left(1-x\right)sin 60^{\circ}$
Equating the two values of pressure for equilibrium,
$dg\left(l-x\right) \frac{\sqrt{3}}{2}+\frac{x \sqrt{3}}{2}\left(d+2y\right)g=\frac{x\sqrt{3}}{2}dg+\left(l-x\right)\frac{\sqrt{3}}{2}\left(d+y\right)g$
or $\left(l-x\right)d+x\left(d+2y\right)=xd+\left(l-x\right)\left(d+y\right)$
or $ ld-xd+xd+2xy=xd +ld +ly-xd-xy$
$\Rightarrow 2yx=y\left(l-x\right)$ or $2x=l-x$
$\therefore x=\frac{l}{3}$