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Q.
A resistance wire has a resistance R. Half of this wire is stretched to double its length and half is twisted to double its thickness, then its new resistance becomes
Current Electricity
Solution:
Let the resistances of both segments be $R _{1}$ and $R _{2}$ then $R_{1}=\frac{\rho \ell}{A^{\prime}}$ where $A^{\prime} \ell=A \frac{\ell}{2}$
$\Rightarrow A^{\prime}=\frac{A}{2}$
so, $R_{1}=\frac{2 \rho \ell}{A}=2 R$
$R_{2}=\frac{\rho \ell / 4}{2 A}=\frac{R}{8}$
Effective resistance $R _{ eq }= R _{1}+ R _{2}=\frac{17 R }{8}$