A vessel in the shape of a truncated cone, which is narrower at the...

The pressure $P$ exerted by the liquid on the base of the vessel is given by $P = \rho g h$.

The force $F$ exerted by the liquid on the base is $F = P \times A = \rho g h A$.

Because the vessel is narrower at the top and wider at the base, its actual volume $V$ is less than the volume of a cylinder with the same base area $A$ and height $h$ (i.e., $V < A h$).

The total mass of the liquid is $M = \rho V$.

Since $V < A h$, it follows that $M < \rho A h$, and the true weight of the liquid is $Mg < \rho g h A$.

Therefore, the force exerted on the base is greater than the true weight of the liquid ($F > Mg$). This is a classic example of the hydrostatic paradox.

Answer: $F > Mg$