Answer:
Given, \[R=16\Omega ,\] when length = l When the wire is cut into 4 equal parts, then length of each part = 1/4. As \[R\propto l\] \[\left[ \because R=\rho \frac{l}{A} \right]\] The resistance of each part \[R'=\frac{R}{4}=\frac{16}{4}=4\,\Omega \] If a wire is stretched to n times its original length, then its new resistance is given by \[{{R}_{new}}={{n}^{2}}{{R}_{original}}\] Here each part of the wire is stretched to length l i.e. n = 4 \[\therefore \] The resistance of each part \[={{n}^{2}}\times R'={{4}^{2}}\times 4=64\,\Omega \] When 4 such resistances are connected in parallel, the effective resistance is \[\frac{1}{{{R}_{eff}}}=\frac{1}{64}+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}\] \[\frac{1}{{{R}_{eff}}}=\frac{4}{64}\] \[\therefore \] \[{{R}_{eff}}=\frac{64}{4}=16\,\Omega \]
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