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The resistance of the meter bridge AB in given figure is 4 Ω. With a cell of emf ε=0.5 V and rheostat resistance Rₕ=2 Ω, the null point is obtained at some point J. When the cell is replaced by another one of emf ε=ε₂ the same null point J is found for Rₕ=6 Ω. The emf ε₂ is [Figure: wire AB with a 6 V driver cell and rheostat Rₕ in series across A and B. The test cell ε is connected from A, through a galvanometer, to a jockey touching the wire at J.]

Asked in JEE Main 11th Jan 1st Shift 2019 · Potentiometer

• (1) 0.5 V
• (2) 0.4 V
• (3) 0.6 V
• (4) 0.3 V

Step-by-step solution

Why the other options are wrong

• (1) Comes from assuming the same null point means the same emf, ignoring the change in wire current.
• (2) No standard slip gives this; it would need the wire current to fall to 0.8 A.
• (3) Comes from quoting the new wire current 0.6 A as the emf.

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