A block of ice floats on a liquid of density 1.2in a beaker then level of liquid when ice completely melt

The correct answer is rises.

Let us understand the behavior of the liquid level step-by-step:

Step 1: Determine the volume of the displaced liquid while floating
When a block of ice of mass $m$ floats on a liquid of density ${\rho }_L$, it experiences an upward buoyant force that balances its weight. According to Archimedes' principle, the weight of the floating ice equals the weight of the liquid it displaces:
$m·g=V_{\text{displaced}}·{\rho }_L·g$
where $V_{\text{displaced}}$ is the volume of the liquid displaced by the submerged part of the ice block.

Solving for the displaced volume gives:
$V_{\text{displaced}}=\frac{m}{{\rho }_L}$
This displaced volume determines the initial level of the liquid in the beaker.

Step 2: Find the volume of the water formed after the ice melts
When the ice melts completely, its mass remains conserved as $m$, but it transforms into water. The volume occupied by this melted water ($V_{\text{water}}$) is:
$V_{\text{water}}=\frac{m}{{\rho }_w}$
where ${\rho }_w$ is the density of water (which is $1.0\text{g/cm}{3}^{}$).

Step 3: Compare the two volumes
We are given that the density of the liquid is $1.2\text{g/cm}{3}^{}$, which is greater than the density of water (${\rho }_L>{\rho }_w$). Comparing the volumes:
$V_{\text{water}}=\frac{m}{{\rho }_w}>\frac{m}{{\rho }_L}=V_{\text{displaced}}$
Since $V_{\text{water}}>V_{\text{displaced}}$, the volume of the water produced is greater than the volume of the liquid originally displaced by the ice. Consequently, the liquid level in the beaker will rise.