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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2022
  • Medium
  • Numerical

A cell, shunted by a \(8 \; \Omega\) resistance, is balanced across a potentiometer wire of length \(3 \;  m\). The balancing length is \(2 \; m\) when the cell is shunted by \(4 \; \Omega\) resistance. The value of internal resistance of the cell will be \(\dots \; \Omega .\)

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Answer & Solution

Correct Answer

(B) \(8\)

Step-by-step Solution

Detailed explanation

\(\frac{V_{1}}{V_{2}}=\frac{3}{2}=\frac{E-i_{1} r}{E-i_{2} r}\)

\(=\frac{E-\frac{E}{8+r} \times r}{E-\frac{E}{4+r} \times r}\)

\(\frac{3}{2}=\frac{8(4+r)}{4(8+r)}\)

\(24+3 r=16+4 r\)

\(r=8 \; \Omega\)

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