In a meter bridge the point D is a neutral point (Fig. 2(EP).4).

Options

A. The meter bridge can have no other neutral point for this set of resistances
B. When the jockey contacts a point on meter wire left of D, current flows to B from the wire
C. When the jockey contacts a point on the meter wire to the right of D. current flows from B to the wire through galvanometer
D. When R is increased, the neutral point shifts to left.

Answer

A, B, C

Explanation

When there is a neutral point at D in meter-bridge, then `(R)/(S)=(l_(1))/((100-l_(1))`
For the given values of R and S, there will be only one value of `l_(1)` for which we shall get the neutral point on bridge wire, In this case `V_(A)-V_(B)=V_(A)-V_(D)` or `V_(B)=V_(D)`. Therefore the galvanometer shows no deflection when jockey contacts a point at D. there is no current in galvanometer arm.
When jockey contacts a point `D_(1)` on meter-bridge wire left of D, the resistance of arm `AD_(1)` becomes smaller than previous value. Due to it, `V_(A)-V_(B)gtV_(A)-V_(D_(1))` or `V_(D_(1))gtV_(B)`. Therefore current flows to B from the wire through galvanometer.
When jockey contacts a point `D_(2)` on meter beidge wire right of D, the resistance of arm `AD_(2)` becomes more than first value. Due to it, `V_(A)-V_(D_(2))-V_(A)-V_(B)` or `V_(B)gtV_(D_(2))`
Therefore the current will flow from B to the wire through galvanometer. When R is increased, the neutral point will shift to the right instead of left on bridge wire. Thus, option (a), (b) and (C) are correct.