A candle of diameter $d$ is floating on a liquid in a cylindrical container of diameter $D(D>>d)$ as shown in figure. If it is burning at the rate of $2\text{cm/hour}$ then the top of the candle will:

Let the length of the candle be $L$. By buoyancy, ${\rho }_c(A\cdot L)g={\rho }_L(A\cdot h_{sub})g$, where $h_{sub}$ is the submerged length.If the candle density ${\rho }_c$ is half the liquid density ${\rho }_L$, then $h_{sub}=L/2$.If the candle length decreases by $2\text{cm}$, the submerged part decreases by $1\text{cm}$. The top of the candle falls by $2\text{cm}$ due to burning but moves up by $1\text{cm}$ because it's lighter. Net fall $=1\text{cm/hr}$.