U-tube

After adding amount of oil the pressure effected by the oil column will be divided equally to the both sides, so the pressure in the side B will be the original pressure plus half of pressure effected by oil column. let X = oil column.

Pressure in both sides are equal and pressure = high * density * g (As g is the gravity constant and i will remove it from the calculations as it is equal in both sides)

Pr at A = Pr at B

$X * 0.85 + ( 7.3 - X ) * 1 = 5 * 1 + 0.5 * X * 0.85$

$X = 4$

The required length = new pressure at B / water density

= $(5 * 1 + 0.5 * X * 0.85)/1 = \boxed {6.7}$

Let the column of oil be $l_2$ , the column of water under the column oil be $l_1$ in the left arm and the column of water above $l_1$ in the right arm be $l_3 = L_B - l_1$ .

Since the volume of water is the same before and after the oil is poured in, assuming the cross-sectional area of the U-tube is uniform throughout, we have $l_1+L_B = 2 l_0 = 10 \ ... (1)$ .

By Newton's third law , the weight of $l_2$ of oil equals to the weight of $l_3$ of water. Therefore, $\rho_{\text{oil}}l_2 g = \rho_{\text{water}}l_3 g$ , where $\rho$ denotes the density of the subscribed liquid and $g$ the acceleration due to gravity. This implies that $l_3 = \dfrac {\rho_{\text{oil}}}{\rho_{\text{water}}}l_2 = 0.85 l_2 \ ... (2)$ .

We also note that $l_1 + l_2 = L_A = 7.3 \ ...(3)$ .

Then we have:

$\begin{aligned} (1): \quad l_1+ {\color{#3D99F6}L_B} & = 10 & \small \color{#3D99F6} \text{Note that }l_1+l_3 = L_B \\ 2l_1 + {\color{#3D99F6}l_3} & = 10 & \small \color{#3D99F6} \text{And }(2): \ l_3 = 0.85 l_2 \\ 2l_1 + {\color{#3D99F6}0.85 l_2} & = 10 \quad ...(4) \end{aligned}$

$\begin{cases} 2\times (3): & 2l_1 + 2l_2 & = 14.6 \\ (4): & 2l_1 + 0.85 l_2 & = 10 \end{cases}$

$\begin{aligned} 2\times (3)-(4): 1.15l_2 & = 4.6 & \implies l_2 = \frac {4.6}{1.15} = 4 \end{aligned}$

$\begin{aligned} (3): \quad l_1 + 4 & = 7.3 & \implies l_1 = 7.3-4 = 3.3 \end{aligned}$

$\begin{aligned} (1): \quad 3.3 + L_B & = 10 & \implies L_B = 10-3.3 = \boxed{6.7} \end{aligned}$

The fluid pressure on both sides would be equal for the fluid to settle. Let's assume the height of the water on the left side is $x$ , then $\begin{aligned}&x\rho_{water}g+(7.3-x)\rho_{oil}g=(10-x)\rho_{water}g \\ &x+(7.3-x)\cdot 0.85 = 10-x \\ &x+(7.3\cdot 0.85)-0.85x=10-x \\ &1.15x=10-6.205 \\ &x=3.3 \implies L_B=10-x=\boxed{6.7} \end{aligned}$

Didn't know how to solve it, but I figured the fluids would equalize so that weight of the column of fluid in A was equal to the column of fluid in B.

Writing out the equation of A-Weight = B-Weight you find that the area and density cancel out leaving you Water-Height-A + 0.85 Oil-Height = Water-Height-B.

You know that the water volume or container volume hasn't changed so Water-Height-A + Water-Height-B = 10cm. You also were given that Water-Height-A + Oil-Height = 7.3cm.

After substitution until only Oil-Height and constants remain, you find it to be 6.7cm.

length of water decreased in left tube = b; intial length of water in left tube =a; length of oil poured =c; L and K are lengths of left and right side of the tube after oil is poured;

L=a-b+c------------(1); K=a+b--------------(2); adding 1 and 2, we get; L+K=2a+c-----------(3); The relation between c and a in densities is c=0.85a; substituting the same in 3 we get; K=2.8c-L=14-7.3; K=6.7 ;

(5 – x) $\rho_{water}$ + (2.3 + x) $\rho_{oil}$ = (5 + x) $\rho_{water}$

5 – x + (2.3 + x)0.85 = 5 + x

1.15x = 2.3 x 0.85, x = 1.9 cm; so L_B = 5 + 1.9 = 6.9 = 69E-1.