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$$ \Rightarrow {\left( {{R_{eq}}} \right)_1} = {{{{2R} \over 3} \times {R \over 3}} \over {{{2R} \over 3} + {R \over 3}}}$$ and $${\left( {{R_{eq}}} \right)_2} = {{{{3R} \over 4} \times {R \over 4}} \over {{{3R} \over 4} + {R \over 4}}}$$

$$ = {{2{R^2}} \over 9} \times {3 \over {3R}} = {{2R} \over 9}$$ and $${\left( {{R_{eq}}} \right)_2} = {{3{R^2}} \over {16}} - R = {{3R} \over {16}}$$

Hence, $${{{{\left( {{R_{eq}}} \right)}_1}} \over {{{\left( {{R_{eq}}} \right)}_2}}} = {{{{2R} \over 9}} \over {{{3R} \over {16}}}} = {{32} \over {27}}$$

19:T532,

हम जानते हैं, $$R = \rho {l \over A} \Rightarrow R \propto l$$

त्रिभुज की भुजा की लंबाई $$ = {l \over 3}$$

वर्ग की भुजा की लंबाई $$ = {l \over 4}$$

अतः, JEE Main 2025 (Online) 28th January Morning Shift Physics - Current Electricity Question 39 Hindi Explanation

$$ \Rightarrow {\left( {{R_{eq}}} \right)_1} = {{{{2R} \over 3} \times {R \over 3}} \over {{{2R} \over 3} + {R \over 3}}}$$ और $${\left( {{R_{eq}}} \right)_2} = {{{{3R} \over 4} \times {R \over 4}} \over {{{3R} \over 4} + {R \over 4}}}$$

$$ = {{2{R^2}} \over 9} \times {3 \over {3R}} =

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