Question on digital DE-MULTIPLEXER! How does a demultiplexer ignore/discard/block the non-required outputs? 
A demultiplexer channels the input to one of the outputs, but there are several outputs. When one output is selected (depending on the state of the select line(s)) then rest of the outputs are 0/low (am I right?).
Now, if the input is 0/low then the selected and non-selected outputs all becomes 0/low.
How does the demultiplexer perceive the final result in such a case?? 

Please explain in details (with reference to 1:2 demux if possible). Thank You!
It's just AND gates.  So, assuming $in$ is the input, $sel$ is your selector bit, and $out_1$ and $out_2$ are your two possible outputs, you would get:

$out_1 = in \land \lnot sel$

$out_2 = in \land sel$

The idea is that, if $in$ is $0$, out must always be $0$.  If $in$ is $1$, then we have to output it to the proper gate.   If you had a larger demux, you would simply have more selector bits.  So, for a demux with 16 outputs, we would need 4 selector bits: $sel1$, $sel2$, $sel4$, and $sel8$.  Output 10 would thus be a 5-input AND gate:

$out_{10} = in \land (sel_8 \land \lnot sel_4 \land sel_2 \land \lnot sel_1)$

The other outputs would be blocked because they would have different combinations of $sel1$, $sel2$, $sel4$, and $sel8$ ANDed into their outputs.