Meter Bridge-1!

Okay! This question can be solved by a simple elimination procedure:

# Statement $1$ is easily rejected as if you exchange the resistances, then to maintain the condition of Wheatstone bridge, the lengths ${l}_{1}$ and $100-{l}_{1}$ will exchange and you won't get the same neutral point.

# Statement $2$ can be thought of like this. If you move the jockey to the right, you decrease the length ${l}_{1}$ and hence the potential drop across the length ${l}_{1}$ also decreases. This causes the potential of point $B$ to be less than that of the point of contact and hence the current glows towards $B$ through the galvanometer.

# By a similar argument, if you move the jockey to the right, you increase the length ${l}_{1}$ and hence increase the potential drop across it. This causes the potential of point $B$ to be more than the point of contact and hence, the current flows from point $B$ to the bridge wire, through the galvanometer. So, this statement is CORRECT .

# Again, if you increase the value of $R$ , then to maintain the condition of resistances in a Wheatstone bridge, i.e.

$\frac {R} {{R}_{{l}_{1}}} = \frac {S} {{R}_{100-{l}_{1}}}$

So, if you increase $R$ , then to maintain this condition, you need to increase ${R}_{{l}_{1}}$ and to increase it, you need to move the jockey to the right, hence increasing ${l}_{1}$ .