KCET · Physics · Current Electricity

In a potentiometer experiment, the balancing point with a cell is at a length \( 240 \mathrm{~cm} \). On
shunting the cell with a resistance of \( 2 \Omega \), the balancing length becomes \( 120 \mathrm{~cm} \). The internal
resistance of the cell is

A \( 1 \Omega \)
B \( 04 \Omega \)
C \( 0.5 \Omega \)
D \( 2 \Omega \)

Verified Solution

Answer & Solution

Correct Answer

(D) \( 2 \Omega \)

Step-by-step Solution

Detailed explanation

(D)
Internal resistance of a cell is \( R\left(\frac{l_{1}-l_{2}}{l_{2}}\right)=2\left(\frac{240-120}{120}\right)=2 \Omega \)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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In a potentiometer experiment, the balancing point with a cell is at a length \( 240 \mathrm{~cm} \). Onshunting the cell with a resistance of \( 2 \Omega \), the balancing length becomes \( 120 \mathrm{~cm} \). The internalresistance of the cell is

Step-by-step Solution

In a potentiometer experiment, the balancing point with a cell is at a length \( 240 \mathrm{~cm} \). On
shunting the cell with a resistance of \( 2 \Omega \), the balancing length becomes \( 120 \mathrm{~cm} \). The internal
resistance of the cell is