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. Numerical answer type. Work it out, then open the answer.

Answer: 780. k = (20 mV)/(60 cm), so p.d. across 300 cm = 20 × 5 = 100 mV = 0.1 V. I = 0.1/20 = 5 × 10⁻³ A. Total resistance = (4)/(0.005) = 800 Ω, so R = 800 - 20 = 780 Ω.

As shown in the figure, a potentiometer wire of resistance 20 Ω and… | JEE Main 2022 (28 Jul, Shift 1)

Q49JEE Main 2022 · 28 JUL · SHIFT 1Medium

Potentiometer

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

Numerical answer type. Work it out, then open the answer.

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Answer: 780

$k = \frac{20\ \text{mV}}{60\ \text{cm}}$ , so p.d. across 300 cm $= 20 \times 5 = 100\ \text{mV} = 0.1\ \text{V}$ .
$I = \frac{0.1}{20} = 5 \times 10^{-3}\ \text{A}$ .
Total resistance $= \frac{4}{0.005} = 800\ \Omega$ , so $R = 800 - 20 = 780\ \Omega$ .

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