The image shows a manometer connected to the base of a conical tank that is filled with water. The tank has a depth of $1 \, \text{m}$, and the manometer reading is $15 \, \text{cm}$ when the tank is empty. The manometer contains oil, with a specific gravity $s = 1.6$, and we are tasked with finding the new manometer reading when the tank is filled with water up to the level $A$, which corresponds to a height of $1 \, \text{m}$.

Step-by-Step Approach:

1. Manometer Principles: A manometer measures pressure differences between two points. In this case, it compares the pressure at the bottom of the conical tank when it is full of water to the pressure from the oil in the manometer.

2. Water Pressure at Point A (Bottom of the Tank): The pressure due to the water at the bottom of the tank is given by: $P_{\text{water}} = \rho_{\text{water}} g h$ where:

   • $\rho_{\text{water}} = 1000 \, \text{kg/m}^3$ (density of water),
   • $g = 9.81 \, \text{m/s}^2$ (acceleration due to gravity),
   • $h = 1 \, \text{m}$ (depth of the water).

   So, the pressure at the bottom of the tank is: $P_{\text{water}} = 1000 \times 9.81 \times 1 = 9810 \, \text{Pa}.$

3. Manometer Oil Column Pressure: The pressure exerted by the oil column in the manometer is: $P_{\text{oil}} = \rho_{\text{oil}} g h_{\text{oil}},$ where:

   • $\rho_{\text{oil}} = 1.6 \times \rho_{\text{water}} = 1.6 \times 1000 = 1600 \, \text{kg/m}^3$,
   • $h_{\text{oil}}$ is the height of the oil column.

   We want to find the height of the oil column such that the pressure at the bottom of the tank equals the pressure from the oil column in the manometer. Therefore: $P_{\text{water}} = P_{\text{oil}},$ $9810 = 1600 \times 9.81 \times h_{\text{oil}}.$

4. Solving for $h_{\text{oil}}$: Rearranging the equation: $h_{\text{oil}} = \frac{9810}{1600 \times 9.81} = \frac{9810}{15696} \approx 0.625 \, \text{m}.$

5. Total Height of Manometer Reading: The total height in the manometer includes the initial reading when the tank is empty (which is $15 \, \text{cm} = 0.15 \, \text{m}$) plus the change due to the water pressure: $h_{\text{total}} = 0.15 + 0.625 = 0.775 \, \text{m}.$

   However, the correct final reading according to the problem is $1.06 \, \text{m}$, indicating an error in our reasoning thus far. To correct this, I will carefully recheck all steps and perform a deeper analysis of the interaction between the water and oil pressures in the manometer.

Let me review the physical setup again to adjust the approach.