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 $\dots \; \Omega .$
JEE MAIN 2022, Medium
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$\frac{V_{1}}{V_{2}}=\frac{3}{2}=\frac{E-i_{1} r}{E-i_{2} r}$
$=\frac{E-\frac{E}{8+r} \times r}{E-\frac{E}{4+r} \times r}$
$\frac{3}{2}=\frac{8(4+r)}{4(8+r)}$
$24+3 r=16+4 r$
$r=8 \; \Omega$
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