Question about inductor When an inductor is connected with a voltage source we get equal and opposite voltage on inductor against the source voltage. That equal and opposite voltage gradually decreases with time which allows the current(caused by the source voltage) to gradually rise.
My question is what makes the equal and opposite voltage in an inductor to fall gradually ?
 A: You can't just connect a pure inductor to a voltage source. Even in a thought experiment there has to be some resistance. Even if your inductor is a superconductor the voltage source has an internal resistance. 
OK, so to begin with the current is zero, and the inductor has a potential $L{dI \over dt}$ which is equal to the applied voltage $V$. That means ${dI\over dt}=V/L$ and after some small time $t$ a current $Vt/L$ flows. That produces a voltage drop across the resistor, meaning that much less across the inductor. As time goes by the current continues to increase, the voltage across the resistance increases (tending eventually to $V$) and the voltage across the inductor falls (tensing eventually to $0$).
A: 
theoratically its possible to attach the inductor with a voltage
  source

Yes, in the context of ideal circuit theory, it is possible to do so without contradiction.
Let the voltage source have constant voltage across $V_S \gt 0$ and the inductor have inductance $L$.  The inductor is connected to the voltage source at time $t = 0$.  By KVL, the voltage across the inductor is given by
$$v_L(t) = V_S\, u(t)$$
where $u(t)$ is the unit step function.
The circuit current is described by a very simple differential equation:
$$V_S\,u(t) = L \frac{di}{dt}$$
with solution
$$i(t) = \frac{V_S}{L}\, t\, u(t)$$
In other words, the current is zero for $t \le 0$ and increases at a constant rate for $t \ge 0$.
Note that the current is unbounded (does not reach a limiting value) as $t \rightarrow \infty$ which is clearly unphysical.  For a physical circuit, the internal resistance of the voltage source and/or the inductor will limit the current.
A: Assuming the DC voltage source is ideal, the voltage on its terminals is constant. If the real (resistive) coil is connected to those terminals, the voltage across the coil will be the same, constant.
What you probably mean is that as the current increases, the induced EMF on the coil decreases  in magnitude; in the end the current is constant and induced EMF is zero.
It is important to understand the difference between the voltage and EMF. They have the same units, but they are due to different agents: the voltage is due to the voltage source but the induced EMF is due coil wire carrying  time-varying current. When coil is connected to a voltage source, the EMF works against the voltage, so the current is low and only gradually increases to its final value.
