As shown in the figure, a potentiometer wire of resistance 20 Ω and length 300 cm is connected with resistance box (R.B.) and a standard cell of emf 4 V. For a resistance 'R' of resistance box introduced into the circuit, the null point for a cell of 20 mV is found to be 60 cm. The value of 'R' is ___________ Ω .

Solution

Physics sir · 3 steps

The null point tells you the voltage across the whole wire, and that gives the current in the main loop.

1. Potential gradient. At balance, the 20 mV cell matches the drop across 60 cm of wire. So the gradient is k = (20 mV)/(60 cm) = 1/3 mV cm⁻¹.
2. Drop across wire. Over the full 300 cm, the drop is 300 × 1/3 = 100 mV = 0.1 V. The wire is 20 Ω, so the main-loop current is I = 0.1/20 = 5 × 10⁻³ A.

   I = 4/(R + 20) ⇒ R + 20 = 4/(5 × 10⁻³) = 800 Ω

3. So. The same current flows through the 4 V cell, the resistance box and the wire in series. So R = 800 - 20 = 780 Ω.

The common slip: 800. Taking the total loop resistance as R gives 800. The 20 Ω of the wire is part of that total and must be subtracted.

The idea to keep. In a potentiometer, the null point gives the gradient, the gradient times the full length gives the drop across the wire, and that drop over the wire's resistance gives the main-loop current.

Answer from the NTA final answer key. Solution by PrepWiser.