Explanation
Resistance of a wire $$R=\dfrac{\rho l}{A}$$ where $$\rho=$$ resistivity, $$l=$$ length, $$A=$$ cross section of the wire. As both have same material and length so $$R \propto \dfrac{1}{A}$$ Thus, $$\dfrac{R_2}{R_1}=\dfrac{A_1}{A_2}=\dfrac{3}{1} \Rightarrow R_2=3R_1$$. here $$R_1$$ is the resistance of thicker wire so its resistance $$R_1=10 \Omega$$ (given) so, $$R_2=3(10)=30 \Omega$$ As they are connected in series so the equivalent resistance is $$R_{eq}=R_1+R_2=10+30=40 \Omega$$
$$1={\dfrac {1}{R_1}}+{\dfrac {1}{R_2}}+{\dfrac {1}{R_3}}$$ $${\dfrac {R_1}{R_2}}={\dfrac {1}{2}}$$ (Given) $$R_2=2R_1$$ $$1={\dfrac {2}{R_2}}+{\dfrac {1}{R_2}}+{\dfrac {1}{R_3}}$$ $$1={\dfrac {3}{R_2}}+{\dfrac {1}{R_3}}$$ $$1-{\dfrac {3}{R_2}}={\dfrac {1}{R_3}}$$ $${\dfrac {{R_2}-3}{R_2}}={\dfrac {1}{R_3}}$$ $${R_3}={\dfrac {R_2}{{R_2}-3}}$$ Now only $$R_2=6 \Omega$$ satisfies this above equation $$\Rightarrow R_3=2 \Omega$$ & $$R_1=3 \Omega$$
The resistance of the conductor is proportional to length of the conductor.That is, $$R=\rho \dfrac { l }{ A } $$.The resistance of one wire of length l, R1 = 100 ohms.If we cut it in a half: $$\dfrac { { R }_{ 1 } }{ l } =\dfrac { { R }_{ 2 } }{ \dfrac { l }{ 2 } } \quad \rightarrow \quad { R }_{ 2 }=\dfrac { { R }_{ 1 } }{ 2 } $$.If we cut it in n parts the resistance will be $$R=\dfrac { { R }_{ 1 } }{ n } $$.Because we shorten the length n times so it became $$\dfrac { l }{ n } $$.When we connect these parts in parallel the output resistance must be 1 ohm. So, $$1=\dfrac { 1 }{ R } +\dfrac { 1 }{ R } +...=n\dfrac { 1 }{ R } =n\dfrac { 1 }{ \dfrac { { R }_{ 1 } }{ n } } =\dfrac { { n }^{ 2 } }{ { R }_{ 1 } } =\dfrac { { n }^{ 2 } }{ 100 } \rightarrow n=10$$.Hence, a resistance of 1 ohm is obtained by connecting 10 equal parts of the wire and connecting them in parallel.
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