Q.Two different liquids of density $\rho_1$ and $\rho_2$ exert the same pressure at a certain point. What will be the ratio of the heights of the respective liquid columns?

Pressure is defined as the normal force a fluid at rest exerts per unit area of any surface it touches, $p = F/A$, with SI unit N/m² (pascal) and dimensional formula $[L^{-1}M^1T^{-2}]$ — the same dimensions as stress, since both are a force per unit area, though pressure, unlike stress, is a scalar with no direction of its own. A small force concentrated on a tiny area can produce an enormous pressure (a knife's edge, a needle's tip), while the same force spread over a large area produces only a small one.

For a liquid of density ρ at rest, the pressure at a depth h below its free surface is $p = h\rho g$, derived by balancing the weight of an imaginary vertical column of the liquid against the force on its own base — a result that turns out to be entirely independent of the shape or cross-sectional area of the container, depending only on the depth and the liquid's density. This is why liquid poured into a set of interconnected vessels of very different shapes always settles to the same level in every vessel (the so-called hydrostatic paradox, resolved once the sideways components of a tapering wall's reaction force on the liquid are properly included) — and it is also why absolute pressure at depth h is the sum of two contributions, $p = p_0 + h\rho g$, where $p_0$ is the atmosphere's own pressure pushing down on the liquid's free surface, and the difference $p - p_0$ is called the gauge pressure. Atmospheric pressure is itself just this same idea applied to the air above us — the weight, per unit area, of the entire column of air overhead, greatest at sea level and falling off with altitude as the height (and density) of the remaining air column above shrinks.