JEE MAIN - Physics (2025 - 28th January Morning Shift - No. 9)
A wire of resistance R is bent into an equilateral triangle and an identical wire is bent into $a$ square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is
$8 / 9$
$9 / 8$
$32 / 27$
$27 / 32$
Explanation
We know, $$R = \rho {l \over A} \Rightarrow R \propto l$$
Side length of triangle $$ = {l \over 3}$$
Side length of square $$ = {l \over 4}$$
So, _28th_January_Morning_Shift_en_9_1.png)
$$ \Rightarrow {\left( {{R_{eq}}} \right)_1} = {{{{2R} \over 3} \times {R \over 3}} \over {{{2R} \over 3} + {R \over 3}}}$$ 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, $${{{{\left( {{R_{eq}}} \right)}_1}} \over {{{\left( {{R_{eq}}} \right)}_2}}} = {{{{2R} \over 9}} \over {{{3R} \over {16}}}} = {{32} \over {27}}$$
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