Pressure in fluids

After rearranging $p = h\rho g$ for $h$, the depth comes out in metres — but if the question asks for centimetres, convert (×100) before writing the final answer. Method correct with the answer in the wrong unit still loses the final mark. Re-read the required unit (cm, kPa, mm) at the very end.

Both $p = F/A$ and $p = h\rho g$ need SI units. Length: 45 cm = 0.45 m. Area is the trap — $1\text{ cm}^2 = 10^{-4}\text{ m}^2$, so 25 cm² = 0.0025 m². Dividing area by 100 (as for length) instead of 10 000 makes the pressure 100 times wrong. Convert area with ÷10 000.

$p = h\rho g$ gives only the gauge (excess) pressure due to the fluid column. Absolute pressure is $p_{abs} = p_{atm} + h\rho g$. If the question gives an atmospheric value (e.g. 100 kPa) or says "show that", add it. Stopping at $h\rho g = 25\,000$ Pa instead of 126 000 Pa caps the mark.

To justify equal water jets from holes at the same depth, you must say pressure acts "equally in all directions" at that depth. Noting only that the holes are at "equal depth" is not enough for the mark — depth alone does not explain why a sideways-facing hole matches a downward-facing one.

The definition needs the ratio: pressure is "force per unit area", $p = F/A$, unit Pa. "Force on an area", "force pushing down", or "how much force there is" drop the "per unit area" and score zero. Two elements are required — the ratio (force ÷ area) and the unit pascal.

$p = F/A$ is for a solid pressing on a surface (a book on a table). $p = h\rho g$ is for the pressure at a point inside a fluid from the weight of fluid above it. Same unit (Pa), completely different situations — using $F/A$ to find pressure at depth in a liquid scores zero. Choose by the physical context.