A vertical U-tube of uniform inner cross section contains mercury in both sides of its arms. A glycerin (density = 1.3 g/cm³) column of length 10 cm is introduced into one of its arms. Oil of density 0.8 gm/cm³ is poured into the other arm until the upper surfaces of the oil and glycerin are in the same horizontal level. Find the length of the oil column, Density of mercury = 13.6 g/cm³

To find the length of the oil column, we can analyze the pressure at a common reference level in both arms of the U-tube. Let us set up the hydrostatic pressure equilibrium.

Let the cross-sectional area of the U-tube be uniform.
Initially, the U-tube contains only mercury of density ${\rho }_{\text{Hg}}=13.6{\text{g/cm}}^3$.
A glycerin column of length $h_1=10\text{cm}$ and density ${\rho }_1=1.3{\text{g/cm}}^3$ is poured into one arm (let's say the left arm).
Oil of density ${\rho }_2=0.8{\text{g/cm}}^3$ is poured into the other arm (the right arm) until the upper surfaces of the oil and glycerin are at the same horizontal level.
Let $h_2$ be the length of the oil column.

Since the upper surfaces of glycerin (left arm) and oil (right arm) are at the same horizontal level, the difference in the heights of the mercury columns in the two arms must make up for the difference in the heights of the glycerin and oil columns.
Specifically, since the glycerin column has height $h_1=10\text{cm}$ and the oil column has height $h_2$, the height of the mercury column above the interface on the side with the shorter liquid column must compensate for this.
Let us find the difference in levels of mercury. The top of the glycerin column is at the same level as the top of the oil column.
Therefore, the depth of the mercury-glycerin interface below the top surface is $h_1=10\text{cm}$.
The depth of the mercury-oil interface below the top surface is $h_2$.
Thus, the difference in the levels of the mercury interfaces in the two arms is $d=|h_1-h_2|$.

Assuming $h_1>h_2$ (since glycerin is denser than oil, a shorter column of oil would not balance if mercury levels were equal, but let's write the general pressure balance at the lower mercury interface).
Let the lower mercury interface be in the left arm (under the glycerin column of height $h_1=10\text{cm}$).
The pressure at this level in the left arm is:

$P_{\text{left}}=P_0+{\rho }_{1}g{h}_1$

where $P_0$ is the atmospheric pressure.
In the right arm, at the same horizontal level, the column consists of oil of length $h_2$ and mercury of length $(h_1-h_2)$ above this level.
The pressure at this level in the right arm is:

$P_{\text{right}}=P_0+{\rho }_{2}g{h}_2+{\rho }_{\text{Hg}}g(h_1-h_2)$

Equating the pressures $P_{\text{left}}=P_{\text{right}}$:

${\rho }_1{h}_1={\rho }_2{h}_2+{\rho }_{\text{Hg}}(h_1-h_2)$

Let us substitute the given values:
${\rho }_1=1.3{\text{g/cm}}^3$
$h_1=10\text{cm}$
${\rho }_2=0.8{\text{g/cm}}^3$
${\rho }_{\text{Hg}}=13.6{\text{g/cm}}^3$

Substituting these values into the equation:

$1.3\times 10=0.8\times h_2+13.6\times (10-h_2)$

Simplify the equation:

$13=0.8h_2+136-13.6h_2$

Combine the terms containing $h_2$:

$13=136-12.8h_2$

Rearranging the terms:

$12.8h_2=136-13$

$12.8h_2=123$

Solve for $h_2$:

$h_2=\frac{123}{12.8}\approx 9.6\text{cm}$

Thus, the length of the oil column is approximately 9.6 cm, which corresponds to the correct option.