A glass full of water is enclosed by a jar, and the air within the...

Let the height of the water be $h$, the area of the bottom of the glass be $A$, the volume of the water be $V$, and the density of water be $\rho$.

Since the air within the jar is entirely evacuated, the pressure at the free surface of the water is zero. The pressure at the bottom of the glass is:

$P = \rho g h$

The downward force applied by the water on the bottom of the glass is:

$F_b = P \times A = \rho g h A$

Considering the vertical equilibrium of the water, the net upward force exerted by the glass must balance the total weight of the water. Let $F_s$ be the net vertical force exerted by the sides of the glass on the water. The total weight of the water is $W = \rho V g$.

Balancing the forces in the vertical direction gives:

$F_b + F_s = W$

$\rho g h A + F_s = \rho V g$

$F_s = \rho g(V - hA)$

For the force on the bottom ($F_b$) and the force exerted by the sides ($F_s$) to remain unchanged, both the product $hA$ and the term $(V - hA)$ must remain constant.

This condition is satisfied provided the height ($h$), the bottom area ($A$), and the volume ($V$) are all unchanged.