Option B is the right answer
We can solve this question by eliminating options
Option (A) False
Let us take a complete graph of $4$ vertices:
In the corresponding line graph, number of edges is
$\sum_{i=0}^{n}$ ${}^{d_{i}}C_{{2}},$ $d_{i}$is degree of each vertex
$=\frac{3\times 2}{2}+\frac{3\times 2}{2}+\frac{3\times 2}{2}+\frac{3\times 2}{2} = 3\times 4=12$
No. of vertices in line graph $=$ No. of edges in original graph $=6.$
So, no. of edges to make complete graph with $6$ vertices $= \frac{6\times 5}{2} = 3\times 5=15$
But we got only $12$ edges for the line graph.
Contradiction.
Option (B) True
Smallest line graph for the original graph with just one edge
which is also connected graph
If a graph is connected with more then one edge, it's line graph will never be disconnected
Option (C) False
This cannot be 2-colorable and hence is not bipartite.
Option (D) False
Because in the line graph degree of a vertex depends on the number of neighbors its corresponding edge has in the original graph.
e.g., $\left [ A, B \right ]$ as a point in the line graph, then the degree of this vertex depends on the degree of $A$ and degree of $B$ in the original graph.
I'm drawing a degree for a point $[AB]$ in the line graph.
So, this is wrong.
Option (E) is false (wrong as proved in the above option (D))