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JEE Main 20244 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Morning Shift
Subject
Physics
Chapter
Current Electricity

JEE Main 2024 · 4 April · Morning Shift · Physics · Current Electricity

To measure the internal resistance of a battery, potentiometer is...

To measure the internal resistance of a battery, potentiometer is used. For R=10Ω\mathrm{R}=10 \Omega, the balance point is observed at l=500 cml=500 \mathrm{~cm} and for R=1Ω\mathrm{R}=1 \Omega the balance point is observed at l=400 cml=400 \mathrm{~cm}. The internal resistance of the battery is approximately :

Options

  1. A0.2Ω0.2 \Omega
  2. B0.3Ω0.3 \Omega
  3. C0.4Ω0.4 \Omega
  4. D0.1Ω0.1 \Omega

Correct answer

B. 0.3Ω0.3 \Omega

Step-by-step solution

Let potential gradient be λ\lambda. ∴i×10=λ×500=ε−irs⇒500λ=ε−50λrs\begin{aligned} & \therefore \mathrm{i} \times 10=\lambda \times 500=\varepsilon-\mathrm{ir}_{\mathrm{s}} \\ & \Rightarrow 500 \lambda=\varepsilon-50 \lambda \mathrm{r}_{\mathrm{s}} \end{aligned}
Also, i′×1=λ×400=ε−i′rs⇒400λ=ε−400λr5∴100λ=350λrs⇒rs=1035≈0.3Ω\begin{aligned} & \mathrm{i}^{\prime} \times 1=\lambda \times 400=\varepsilon-\mathrm{i}^{\prime} \mathrm{r}_{\mathrm{s}} \\ & \Rightarrow 400 \lambda=\varepsilon-400 \lambda \mathrm{r}_5 \\ & \therefore 100 \lambda=350 \lambda \mathrm{r}_{\mathrm{s}} \Rightarrow \mathrm{r}_{\mathrm{s}}=\frac{10}{35} \approx 0.3 \Omega \end{aligned}
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