Two wires of the same material and equal length are joined in parallel combination. If one of them has half the thickness of the other

Q: Two wires of the same material and equal length are joined in parallel combination. If one of them has half the thickness of the other and the thinner wire has a resistance of $8\, ohms$, the resistance of the combination is equal to

b $\rho - {\rm{same, }}\,l - {\rm{same, }}\,{A_2} = \frac{1}{4}{A_1}$ (as ${r_2} = \frac{{{r_1}}}{2}$) By using $R = \rho \frac{l}{A} \Rightarrow \frac{{{R_1}}}{{{R_2}}} = \frac{{{A_2}}}{{{A_1}}} \Rightarrow \frac{{{R_1}}}{8} = \frac{1}{4} \Rightarrow {R_1} = 2\,\Omega $ Hence, ${R_{eq}} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \frac{{2 \times 8}}{{(2 + 8)}} = \frac{8}{5}\,\Omega .$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

Two wires of the same material and equal length are joined in parallel combination. If one of them has half the thickness of the other and the thinner wire has a resistance of $8\, ohms$, the resistance of the combination is equal to

A$\frac{5}{8}\,\,ohms$
B$\frac{8}{5}\,\,ohms$
C$\frac{3}{8}\,\,ohms$
D$\frac{8}{3}\,\,ohms$

Medium

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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