The given circuit shows a uniform straight wire AB of 40 cm length fixed at both ends. In order to get zero reading in the galvanometer G, the free end of J is to be placed from B at: [figure]
- 32 cm
- 8 cm
- 16 cm
- 24 cm
Answer
(4) 24 cm
Solution
For zero deflection the bridge is balanced:
812 = AJJB, with AJ + JB = 40 cm
AJ = 40 820 = 16 cm
JB = 40 - 16 = 24 cm
Answer: (4)