question-images/batch-001/bec6281f77c7.webp" data-orsrc="https://uploads-pracbee-ramanuj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/bec6281f77c7.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline" alt="AP EAPCET 2024 - 23th May Morning Shift Physics - Current Electricity Question 25 English Explanation 1">

$$ \frac{R}{R^{\prime}}=\frac{l_1}{100-l_1} \Rightarrow \frac{9}{R^{\prime}}=\frac{l_1}{100-l_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $$

$$ \text { Case-2 } $$

AP EAPCET 2024 - 23th May Morning Shift Physics - Current Electricity Question 25 English Explanation 2

$$ \begin{array}{ll} \therefore & \frac{R^{\prime}}{9}=\frac{l_1+100}{90-l_1} \\ \Rightarrow & \frac{9}{R^{\prime}}=\frac{90-l_1}{l_1+10} \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,...(ii)\end{array} $$

From Eqs. (i) and (ii), we get

$$ \begin{aligned} & & \frac{l_1}{100-l_1} & =\frac{90-l_1}{l_1+10} \\ & \Rightarrow & l_1^2+10 l_1 & =9000-90 l_1-100 l_1+l_1^2 \\ & \Rightarrow & 200 l_1 & =9000 \\ \Rightarrow & & l_1 & =45 \mathrm{~cm} \end{aligned} $$

$\therefore$ From Eq. (i), $\frac{9}{R^{\prime}}=\frac{45}{100-45}$

$$ \Rightarrow \quad R^{\prime}=11 \Omega $$

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