Dilemma of volume displaced by an object I have 2 questions, the second question is based on the answer of the first question
1.)When an object floats, is the volume of the object submerged and displaced water  equal? 
2.)So if volume displaced become equal to volume of object, According to the formula, volume and gravity get cancelled on both sides leaving densities equal. This is not possible because if the density of object is less it floats.
V(object)p(density of the object)g=Buoyant force=V(displaced liquid)p(density of the liquid)g
 A: When an object floats, not all of the object is submerged. There is some portion that remains above the surface.
This means that the total volume of the object is greater than the volume of the object that is submerged. Since, as you said, the volume of the object that is submerged is equal to the volume of displaced water, this means that the total volume of the object is greater than the volume of displaced water.
This is what makes the equality in your question possible - the density of the floating object is less than water, but its total volume is greater than the volume of displaced water. This also means that the closer an object's density is to the density of displaced water, the lower in the water it will float (for example, ice is only slightly less dense than water, so most of an iceberg is submerged - a fact that has led many ships to their demise).
A: 
1.)When an object floats, is the volume of the object submerged and displaced water equal?

Yes, the submerged volume of the object equals the volume of the water displaced.

V(object)p(density of the object)g=Buoyant force=V(displaced
  liquid)p(density of the liquid)g
$$V_{o}ρ_{o}g=V_{l}ρ_{l}g$$

This equation applies if the object floats (is in equilibrium) regardless of how much of the object is submerged. 
The left side of the equation is the total weight of the object. $V_o$ is the total volume of the object, not necessarily the submerged volume of the object. 
The right side of the equation is the buoyant force. $V_l$ is the volume of fluid displaced by the submerged volume of the object. 

2.)So if volume displaced become equal to volume of object, According to the formula, volume and gravity get cancelled on both sides leaving
  densities equal. This is not possible because if the density of object
  is less it floats.

The density of the object does not have to be less than the density of the fluid in order to float. It has to be less than or equal to the density of the fluid. If it is less than the fluid density, the object will float partially submerged. If it is equal to the fluid density, it will float completely submerged, meaning $V_o$ = $V_l$. 
Hope this helps.
A: Archimedes' Principle would be the answer to both of your questions and fix the apparent paradox (or dilemma as the OP calls it).
Archimedes' Principle says that 

The upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.

The other key ingredient you need is that the weight of the object is given by its density multiplied by its volume and $g$. 
With these two simple concepts, we can answer your questions: 
1) Unless the object has the same density as the fluid (ie water), the volume of the fluid displaced is, in general, not equal to the volume of the object. This is because the weight of the fluid displaced is equal to the weight of the object. Note that the equality is not between the volumes but between the weights and this equality ensures that the system is in equilibrium. 
2) You can now see that your assumption that "volume displaced becomes equal to volume of object" is not true so there is no contradiction anywhere. 
