Please help me with this doubt from electromagnetic induction If motional emf and induced emf are the same, how to explain the emf induced in a conducting rod moving with a velocity perpendicular to length of rod and magnetic field(magnetic field is perpendicular to length and velocity) in terms of induced electric field?
There is no specific area present here, so magnetic flux is not varying so how does induced electric field come into play?
 A: Motional emf and induced electric emf are not the same.
A motional EMF is due to magnetic forces and an induced electric EMF is due to electric forces.
And a motional EMF is only sometimes equal to the change in magnetic flux. It requires the wires to be thin and it requires the wires to keep the charges in the wire.
As for a rod moving perpendicular to itself and a constant uniform magnetic field the moving charges feel a magnetic force. The mobile charges are free to move they get the EMF, and the nonmobile charges get strained until the stress from the strain counters the magnetic force.

There is no specific area present here, so magnetic flux is not varying so how does induced electric field come into play?

There isn't an induced electric field. An EMF is not defined as $\int \vec E\cdot \mathrm d \vec \ell$ it is defined as the line integral of the force per unit charge, an example would be $\int \left(\vec E+\vec v\times \vec B\right)\cdot \mathrm d \vec \ell$ and when the loop is stationary then the velocity of the charges is in the same direction as $\mathrm d \vec \ell$ so the magnetic force doesn't contribute. When the loop is moving and the charge moves within the thin wire they can contribute.
For instance in your example the $\vec B$ field could be in the $\hat z$ direction, the velocity could be in the $\hat x$ dircetion and the wire could go in the direction $\mathrm d\vec \ell=\mathrm d \ell \hat y.$

For experts:

can I ask you how Faraday's Law is applied in this situation?  Or put another way where is the loop through which the magnetic flux changes?

If you want to apply the universal flux rule $\left.\mathscr E\right|_{t=t_0}=$ $$\oint_{C(t_0)}\left(\vec E+\vec v\times \vec B\right)\cdot \mathrm d \vec \ell=-\left.\frac{\mathrm d \Phi_B}{\mathrm d t}\right|_{t=t_0}$$ then first you need to be aware it only holds when the wires are thin and the charges stay in the wire. Secondly you can pick any loop you want, but you do have to pick it for every moment of time to compute the flux through it as a function of time.
In particular if you care about the EMF along a bar then your area at any moment could be the set of places the bar has been since $t=0$ so the loop around the area can include the initial position if the bar as one part, the current position of the bar as another part and the locations the ends of the bar have been (between now and then) as the connecting parts. This loop can be used as the instantaneous loop $C(t_0)$ that encloses a surface $S(t_0)$ in the flux integral the line integral $\Phi_B(t)=\iint_{S(t)}\left(\vec B(t)\right)\cdot \hat n \mathrm d S$ for the changing flux. So the area swept out by the bar from $t=0$ to the current time can be the area you use to compute the function of time $\Phi_B(t).$
But the point of that choice is that the magnetic EMF only exists along the bar so the rest didn't matter. You could also just compute the Lorentz Force on the bar per unit charge in the instantaneous direction along the bar. It's the same thing. That's what an EMF is.
And sure, any loop based in time can have two contributions to the force per unit charge. One is the electric field. And around a closed loop that equals the flux of $\frac{\partial B}{\partial t}$ through the instantaneous position of the loop. The other is based on the magnetic force per unit charge. And for a thin wire where the motion is constrained by electric forces to stay in the wire over time, the magnetic for per unit charge.
If you want to think of it as the product rule you could compute $$\Phi_B(t+\Delta t)=\iint_{S(t+\Delta t)}\left(\vec B(t+\Delta t)\right)\cdot \hat n(t+\Delta t)\mathrm d S(t+\Delta t)$$
and $$\Phi_B(t)=\iint_{S(t)}\left(\vec B(t)\right)\cdot \hat n(t)\mathrm d S(t).$$
So when you subtract and divide by $\Delta t$ you can instead look at $$\iint_{S(t)}\left(\frac{\vec B(t+\Delta t)-\vec B(t)}{\Delta t}\right)\cdot \hat n(t)\mathrm d S(t)$$ and $$\frac{1}{\Delta t}\iint_{S(t+\Delta t)}\left(\vec B(t)\right)\cdot\hat n(t+\Delta t) \mathrm d S(t+\Delta t)-\frac{1}{\Delta t}
\iint_{S(t)}\left(\vec B(t)\right)\cdot\hat n(t) \mathrm d S(t).$$
And the limit of their sum equals $\mathrm d \Phi_B/\mathrm d t$ when the wires are thin, there are no magnetic monopoles, and the charges stay in the wire.
A: 
Imagine a loop $ABCC'B'A'$ which is in a magnetic field $\vec B$ as in the left-hand diagram.
Assume that sides $AA'$ and $CC'$ are close together and the magnetic flux through the loop is some value $\Phi_i$.
Suppose that this is an instantaneous picture of the loop at time $t_i$ and that side $CC'$ is moving at constant velocity $\vec v$ in a direction which is at right angle to the magnetic field and the side $CC'$.
The right-hand diagram shows side $CC'$ at a time $t_f$ and the magnetic flux through that loop is now $\Phi_f$.
There is change of magnetic flux through the loop and so Faraday says that there is an emf induced in the loop, $\frac{\Phi_f-\phi_i}{t_f - t_i}$.
If $CC'$ was a conducting rod all that was stated above would still be the same.  
To illustrate a point, in the left-hand diagram I have portions of the loop hanging down.
Those portions could equally well be elastic joining $A$ and $C$ and $A'$ and $C'$ directly which were stretched when side $CC'$ moved.  
Or you could save yourself a lot of trouble and use "The area is the one that is being swept out by the rod" as was stated by @CuriousOne noting that the area must be at right angles to the magnetic field direction $\int_A \vec B \cdot d\vec a$.
A: Great and beautiful question !
Please patiently read through this ! I have tried my best to elucidate it !
The thing is you are talking about two different kinds of induced emfs here ! one is a conservative and the other non-conservative. If you think of the emf in the rod if you start from one end of the rod and move along the entire length the total emf change is non zero. but in the case of the loop present in the magnetic field, with reference to which you are talking about "area", the magnetic field change or area change that you create induces an emf but this emf is non conservative ! meaning if you move along the loop you will come back to the same point, and thus total emf must be zero ! but it isnt zero (or is it ?) ! Because there is a current ! 
Due to this fundamental difference of the two kinds of EMFs being completely different you are faced with this phenomena !
I would suggest you this. think of the EMF formation as a result of the electrons getting pushed by the varying magnetic field around them. thus if u see in both cases you have the electrons getting pushed and thus it becomes obvious that there must be emfs in both cases. If the magnetic flux is changing inside the loop then field lines must be either entering the loop or leaving it. thus this must happen by the field lines intersecting the wire at some point of entering or leaving !
I would also add that in the case of the rod the emf can manifest itself at the two ends ! Well because there are "ends" for it to manifest at ! But there is ideally no current ! This is a voltage that primarily appears as increased potential of one point compared to the other ! But in the loop there are no ends thus there is no measurable EMF ! but there is a current in a loop with has a finite inductance ! This is the other Manifestation of EMF ! This is another Voltage that manifests as an induced current against a finite resistance (impedance) !
