Force without acceleration My doubt is very basic and fundamental, by Newton's second law we can say that $F=\frac{dp}{dt}$.
Hence, there can also be possible cases when $F=\frac{dm}{dt}v$, when the body is moving with constant velocity in the presence of a force! Then what is the effect of that force as a whole, what is it doing? We have always thought of force as an agent of acceleration, something that provides acceleration, but here the body is under the influence of a net force and still possess a constant velocity!! This whole idea seems to be absurd and can anyone help me in absorbing this concept.
 A: Yes such a situation is possible, but you are no longer considering point mechanics (where $m$ is by definition constant), but the mechanics of a system consisting of multiple point particles. In other words: to arrive at such an equation with changing mass, you have to analyse a system of point masses, for each of which $F = m\dot v$ (in other words, it all depends on how the mass is gained).
A simple model leading to an equation such as the above is the following. Consider an object, let's say an asteroid, of mass $M$ that moves through space filled with small objects at rest of mass $m$, let's say dust. The small objects are at rest. We assume that if the large object hits a dust particle there will be a completely inelastic collision (idealized to occure instantaneously). In other words we can compute the velocity afterwards by momentum conservation (energy is not conserved, since the non-elastic deformation of the two colliding objects creates heat):
$$ p = Mv = (M+m)v' $$
so the velocity after such an event will be
$$ v' = \frac{M}{M+m} v. $$
Now we can say that $M$ depends on $t$ since the asteroid gains mass $m$ each time it hits a dust particle. Each of these events can be handled as above, the momentum is conserved but the mass of the asteroid changes, in other words, we arrive at the equation
$$ F = \dot p = \partial_t (M(t) v(t)) = \dot M(t) v(t) + M(t) \dot v(t). $$
The force $F$ is assumed to only apply to the asteroid, not the dust.
So if there is a dust trail which the asteroid sweeps up the mass will rise, and it will slow down, unless an external force is applied.
A: This is the idea behind a rocket. Very simplified,  while the rocket looses fuel mass, the exhaust produces thrust
A: The answer of your question itself lies in it. You have written F to be equal to $F=\frac{dm}{dt}v$. It becomes a variable mass system just like a rocket! 
A: A Special Relativistic view :

In the rest system $\:\mathcal{S}_{o}\:$ of a particle, see ($\alpha$), by a mechanism power is transferred to the particle with rate $\:\overset{\boldsymbol{\cdot}}{\mathrm{q}}_{o}\:$. This rate is with respect to the proper time $\:\tau\:$ and this power changes the rest mass $\:m_{o}\:$ of the particle:
\begin{equation}
\overset{\boldsymbol{\cdot}}{\mathrm{q}}_{o}=\dfrac{\mathrm{d}\left(m_{o}c^{2}\right)}{\mathrm{d}\tau}=c^{2}\dfrac{\mathrm{d}m_{o}}{\mathrm{d}\tau}
\tag{B-01}
\end{equation}
In an other inertial system $\:\mathcal{S}\:$ moving with constant 3-velocity  $\:\boldsymbol{-}\mathbf{w}\:$ with respect to  $\:\mathcal{S}_{o}\:$, the particle is moving with constant velocity $\:\mathbf{w}\:$,  see ($\beta$), under the influence of a 'force'
\begin{equation}
\boldsymbol{\mathcal{h}}=\dfrac{\overset{\boldsymbol{\cdot}}{\mathrm{q}}_{o}}{c^{2}}\mathbf{w}=\dfrac{\mathrm{d}m_{o}}{\mathrm{d}\tau}\mathbf{w}=\gamma(w)\dfrac{\mathrm{d}m_{o}}{\mathrm{d} t}\mathbf{w}
\tag{B-02}
\end{equation}
This 'force' $\:\boldsymbol{\mathcal{h}}\:$, although acting on the particle, keeps its velocity $\:\mathbf{w}\:$ constant. So, its 3-acceleration is $\:\mathbf{a}=\mathrm{d}\mathbf{w}/\mathrm{d}t =\boldsymbol{0}\:$ and consequently its 4-acceleration $\:\mathbf{A}=\boldsymbol{0}$. This 'force' is defined as heatlike.
Link :  What does it mean that the electromagnetic tensor is anti-symmetric?.
